Prime Factorization, GCD, and LCM Alıştırma Soruları — GRE General Test

Soru 1Soru →

Let $n$ be a positive integer whose prime factorization consists only of the prime factors $2$ and $3$ . If $n$ has exactly $12$ positive divisors and $\gcd(n, 36) = 12$ , what is the value of $n$ ?

Cevabı ve açıklamayı göster

Cevap: 96

Cevap

The value of

n

is

96

.

Representing

n = 2^a \times 3^b

, the number of positive divisors is

(a+1)(b+1) = 12

. The greatest common divisor

\gcd(n, 36) = \gcd(2^a \times 3^b, 2^2 \times 3^2) = 2^{\min(a,2)} \times 3^{\min(b,2)} = 12 = 2^2 \times 3^1

. This requires

\min(a,2) = 2 \implies a \ge 2

and

\min(b,2) = 1 \implies b = 1

. Substituting

b = 1

into

(a+1)(1+1) = 12

gives

2(a+1) = 12

, so

a = 5

. Therefore,

n = 2^5 \times 3^1 = 32 \times 3 = 96

.

Adım Adım Çözüm

1

Set up the prime factorization of

n

and the divisor count equation.

n = 2^a \times 3^b

and

(a + 1)(b + 1) = 12

.

Since the prime factors of

n

are only

2

and

3

,

n

must take the form

2^a \times 3^b

, where the number of positive divisors is

(a+1)(b+1)

.

2

Analyze the exponent requirements using the greatest common divisor.

a \ge 2

and

b = 1

.

\gcd(2^a \times 3^b, 2^2 \times 3^2) = 2^{\min(a,2)} \times 3^{\min(b,2)} = 2^2 \times 3^1

. Matching powers gives

\min(a,2) = 2 \implies a \ge 2

, and

\min(b,2) = 1 \implies b = 1

.

3

Solve for exponent

a

and calculate

n

.

a = 5

, giving

n = 2^5 \times 3^1 = 96

.

Substituting

b = 1

into

(a+1)(1+1) = 12

gives

2(a+1) = 12 \implies a = 5

, which satisfies

a \ge 2

.

Anahtar Kavram

Prime exponent rules for GCD and divisor counting

Tahmini Süre:2m 0s

Soru 2Soru →

Let $K = 2^5 \times 3^4 \times 5^3 \times 11^2$ . How many positive integer factors of $K$ are divisible by $300$ but are not divisible by $900$ ?

16

24

36

48

96

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

24

The correct answer 24 is obtained by analyzing the prime factorizations

300 = 2^2 \times 3^1 \times 5^2

and

900 = 2^2 \times 3^2 \times 5^2

. Any factor

f = 2^a \times 3^b \times 5^c \times 11^d

must satisfy

a \in \{2, 3, 4, 5\}

(4 choices),

b = 1

(1 choice, since

b \ge 1

for 300 but

b < 2

for 900),

c \in \{2, 3\}

(2 choices), and

d \in \{0, 1, 2\}

(3 choices). Multiplying these choices gives

4 \times 1 \times 2 \times 3 = 24

.

Adım Adım Çözüm

1

Express any factor of

K

in terms of prime factor exponent constraints.

Any factor

f

of

K

has the form

f = 2^a \times 3^b \times 5^c \times 11^d

, where

0 \le a \le 5

,

0 \le b \le 4

,

0 \le c \le 3

, and

0 \le d \le 2

.

The prime factors of

f

must be subsets of the prime factors of

K

with exponents not exceeding those in

K

.

2

Find the prime factorizations of

300

and

900

.

300 = 2^2 \times 3^1 \times 5^2

and

900 = 2^2 \times 3^2 \times 5^2

.

Divisibility criteria correspond to minimum exponent requirements for each prime factor.

3

Apply the divisibility conditions to determine constraints on each exponent.

For

f

to be divisible by

300

, we need

a \ge 2

,

b \ge 1

, and

c \ge 2

. For

f

to NOT be divisible by

900

, we must have

b < 2

. Thus,

b = 1

.

Combining

b \ge 1

and

b < 2

uniquely restricts

b

to

1

.

4

Count the number of valid choices for each exponent.

a \in \{2, 3, 4, 5\}

(

4

choices),

b \in \{1\}

(

1

choice),

c \in \{2, 3\}

(

2

choices), and

d \in \{0, 1, 2\}

(

3

choices).

The exponent

d

is unrestricted by

300

or

900

, so it can take any valid power present in

K

.

5

Multiply the number of independent choices using the fundamental counting principle.

4 \times 1 \times 2 \times 3 = 24

.

Each exponent choice can be paired independently to form a unique factor.

Anahtar Kavram

Counting Divisors Using Prime Exponent Constraints

Soru 3Soru →

Let $x$ and $y$ be positive integers such that $\gcd(x, y) = 14$ and $\text{lcm}(x, y) = 420$ . Which of the following values could be the sum $x + y$ ? Select all such values.

Geçerli olan tümünü seçin

154

182

210

238

435

Cevabı ve açıklamayı göster

Cevap: 154; 182; 238

Cevap

The possible values for the sum

x + y

are 154, 182, and 238.

By writing

x = 14a

and

y = 14b

with

\gcd(a, b) = 1

, the relation

\text{lcm}(x, y) = 14ab = 420

requires

ab = 30

. The positive coprime factor pairs of 30 are

(1, 30)

,

(2, 15)

,

(3, 10)

, and

(5, 6)

. Multiplying these pairs by 14 gives the possible sums 434, 238, 182, and 154. Therefore, the options equal to 154, 182, and 238 are all correct.

Adım Adım Çözüm

1

Express

x

and

y

in terms of their greatest common divisor.

Let

x = 14a

and

y = 14b

, where

a

and

b

are positive integers such that

\gcd(a, b) = 1

.

Factoring out the greatest common divisor leaves coprime quotient factors

a

and

b

.

2

Relate the least common multiple to

a

and

b

.

\text{lcm}(x, y) = 14ab = 420 \implies ab = \frac{420}{14} = 30$.

The least common multiple of two numbers sharing a GCD of

g

is given by

g \cdot a \cdot b

.

3

Find all coprime pairs

(a, b)

with

a \le b

whose product is 30.

The prime factorization of 30 is

2 \times 3 \times 5

. The valid coprime pairs

(a, b)

are

(1, 30)

,

(2, 15)

,

(3, 10)

, and

(5, 6)

.

Since

\gcd(a, b) = 1

, all factors of 30 split into pairs of coprime integers.

4

Calculate the corresponding values of

x

,

y

, and their sum

x + y

for each pair.

Pair (1, 30):

x = 14, y = 420 \implies x + y = 434

.
Pair (2, 15):

x = 28, y = 210 \implies x + y = 238

.
Pair (3, 10):

x = 42, y = 140 \implies x + y = 182

.
Pair (5, 6):

x = 70, y = 84 \implies x + y = 154

.

Multiplying each pair by the GCD of 14 yields the original integers

x

and

y

.

Anahtar Kavram

Relationship between GCD, LCM, and prime factorization of quotient factors

Soru 4Soru →

A positive integer $n$ has the prime factorization $n = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers. If $n$ is divisible by $36$ and is a divisor of $5{}400$ , which of the following values could be the total number of positive divisors of $n$ ? Select all such values.

Geçerli olan tümünü seçin

18

20

27

32

40

Cevabı ve açıklamayı göster

Cevap: 18; 27; 32

Cevap

18, 27, and 32 are all possible total numbers of positive divisors for

n

.

Prime factorization gives

36 = 2^2 \cdot 3^2

and

5{}400 = 2^3 \cdot 3^3 \cdot 5^2

. For

n = 2^a \cdot 3^b \cdot 5^c

to be a multiple of 36 and a divisor of 5,400 with positive integer exponents,

a \in \{2,3\}

,

b \in \{2,3\}

, and

c \in \{1,2\}

. The total number of divisors is

(a+1)(b+1)(c+1)

. The possible values for this product are 18, 24, 27, 32, 36, and 48. Among the choices, 18, 27, and 32 are valid values.

Adım Adım Çözüm

1

Find the prime factorizations of the boundary numbers 36 and 5,400.

36 = 2^2 \cdot 3^2

and

5{}400 = 2^3 \cdot 3^3 \cdot 5^2

.

Establishing the prime factor bounds determines the range of possible values for exponents

a

,

b

, and

c

.

2

Determine the constraints on exponents

a

,

b

, and

c

.

Since

36 \mid n

,

a \ge 2

,

b \ge 2

, and

c \ge 1

(given

c

is a positive integer). Since

n \mid 5{}400

,

a \le 3

,

b \le 3

, and

c \le 2

. Thus,

a \in \{2, 3\}

,

b \in \{2, 3\}

, and

c \in \{1, 2\}

.

Divisibility rules require prime factor exponents of a multiple to be greater than or equal to those of the divisor, and exponents of a divisor to be less than or equal to those of the multiple.

3

Calculate all possible total divisor counts using the formula

d(n) = (a+1)(b+1)(c+1)

.

Possible factor values are

(a+1) \in \{3, 4\}

,

(b+1) \in \{3, 4\}

, and

(c+1) \in \{2, 3\}

. Evaluating all combinations yields:

3 \cdot 3 \cdot 2 = 18

,

3 \cdot 3 \cdot 3 = 27

,

3 \cdot 4 \cdot 2 = 24

,

3 \cdot 4 \cdot 3 = 36

,

4 \cdot 4 \cdot 2 = 32

, and

4 \cdot 4 \cdot 3 = 48

.

The total number of positive integer divisors is found by adding 1 to each exponent in the prime factorization and multiplying the results.

4

Compare the calculated divisor counts with the options provided.

The values 18, 27, and 32 appear in the calculated set of possible total divisors.

Direct matching identifies all valid options.

Anahtar Kavram

Prime factor exponent bounds and total number of positive divisors formula

Soru 5Soru →

Two positive integers $m$ and $n$ satisfy $\gcd(m, n) = 20$ and $\text{lcm}(m, n) = 4{}200$ . It is given that $m$ is divisible by $7$ and has exactly $24$ positive integer divisors. If $n$ is a multiple of $3$ , what is the value of $n$ ?

60

200

300

420

600

Cevabı ve açıklamayı göster

Cevap: $60$

Cevap

The value of

n

is

60

.

Prime factorization reveals that

\gcd(m,n) = 2^2 \cdot 3^0 \cdot 5^1 \cdot 7^0

and

\text{lcm}(m,n) = 2^3 \cdot 3^1 \cdot 5^2 \cdot 7^1

. The condition that

n

is a multiple of

3

fixes the exponent of

3

in

n

to

1

(so

m

has exponent

0

). The condition that

m

is a multiple of

7

fixes the exponent of

7

in

m

to

1

(so

n

has exponent

0

). For

m = 2^a \cdot 3^0 \cdot 5^c \cdot 7^1

, its divisor count equation

2(a+1)(c+1) = 24

simplifies to

(a+1)(c+1) = 12

. Since

a \in \{2, 3\}

and

c \in \{1, 2\}

, the only valid solution is

a=3

and

c=2

. This leaves

n

with exponents

2

for prime

2

,

1

for prime

3

,

1

for prime

5

, and

0

for prime

7

, giving

n = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^0 = 60

.

Adım Adım Çözüm

1

Express the GCD and LCM in prime factorized form.

\gcd(m, n) = 20 = 2^2 \cdot 3^0 \cdot 5^1 \cdot 7^0

and

\text{lcm}(m, n) = 4{}200 = 2^3 \cdot 3^1 \cdot 5^2 \cdot 7^1

.

The prime factor exponents of

m

and

n

must have minimums equal to the GCD exponents and maximums equal to the LCM exponents.

2

Determine the exponents of prime factors

3

and

7

for

m

and

n

.

Since

n

is divisible by

3

, exponent of

3

in

n

is

1

, so exponent of

3

in

m

is

0

. Since

m

is divisible by

7

, exponent of

7

in

m

is

1

, so exponent of

7

in

n

is

0

.

Each prime exponent in

\text{lcm}(m, n)

must belong to at least one of the numbers.

3

Use the divisor count of

m

to find its remaining exponents for primes

2

and

5

.

Let

m = 2^a \cdot 3^0 \cdot 5^c \cdot 7^1

, where

a \in \{2, 3\}

and

c \in \{1, 2\}

. The number of positive divisors is

(a+1)(0+1)(c+1)(1+1) = 2(a+1)(c+1) = 24

, giving

(a+1)(c+1) = 12

. Testing

a=2 \implies c=3

(invalid range). Testing

a=3 \implies c=2

(valid). Thus

m = 2^3 \cdot 3^0 \cdot 5^2 \cdot 7^1 = 1{}400

.

The total number of positive integer divisors of

p_1^{e_1} p_2^{e_2} \cdots

is given by

(e_1+1)(e_2+1)\cdots

.

4

Determine the prime exponents for

n

and calculate its value.

Since

a=3

,

n

gets

2^2

. Since

c=2

,

n

gets

5^1

. Together with

3^1

and

7^0

,

n = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^0 = 60

.

For each prime

p

, the exponent in

n

must match the bound opposite to

m

to satisfy both

\gcd

and

\text{lcm}

.

Anahtar Kavram

Prime exponent analysis of GCD and LCM alongside the divisor count formula

Soru 6Soru →

Two automated timers are set to chime at regular intervals. The first timer chimes every $12$ minutes, and the second timer chimes every $18$ minutes. If both timers chime simultaneously at 12:00 PM, how many minutes will pass before they next chime at the exact same time?

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

36 minutes

The time interval between simultaneous chimes is given by the least common multiple (LCM) of the two individual intervals,

12

minutes and

18

minutes. Using prime factorization,

12 = 2^2 \times 3

and

18 = 2 \times 3^2

. The LCM takes the highest power of each prime present:

2^2 \times 3^2 = 4 \times 9 = 36

. Thus,

36

minutes will elapse before the timers chime together again.

Adım Adım Çözüm

1

Find the prime factorizations of 12 and 18

12 = 2^2 \times 3

and

18 = 2 \times 3^2

Expressing numbers as products of prime factors allows us to easily compute the least common multiple.

2

Calculate the Least Common Multiple (LCM)

\text{LCM}(12, 18) = 2^{\max(2,1)} \times 3^{\max(1,2)} = 2^2 \times 3^2 = 36

The next simultaneous occurrence of periodic events is given by the LCM of their periods.

Anahtar Kavram

Least Common Multiple (LCM) of periodic events

Soru 7Soru →

Two positive integers $x$ and $y$ have a greatest common divisor (GCD) of $6$ and a least common multiple (LCM) of $108$ . If $x = 18$ , what is the value of $y$ ?

18

36

54

108

648

Cevabı ve açıklamayı göster

Cevap: $36$

Cevap

The value of

y

is

36

.

For any two positive integers

x

and

y

, their product equals the product of their greatest common divisor and least common multiple:

x \cdot y = \text{GCD}(x,y) \cdot \text{LCM}(x,y)

. Substituting

x = 18

,

\text{GCD} = 6

, and

\text{LCM} = 108

gives

18y = 6 \cdot 108 = 648

. Dividing both sides by

18

yields

y = 36

.

Adım Adım Çözüm

1

State the fundamental identity relating GCD, LCM, and two positive integers.

x \times y = \text{GCD}(x,y) \times \text{LCM}(x,y)

For any two positive integers, the product of the numbers is equal to the product of their greatest common divisor and least common multiple.

2

Substitute the known values

x = 18

,

\text{GCD} = 6

, and

\text{LCM} = 108

into the equation.

18 \times y = 6 \times 108

Replace the abstract variables with the specific values provided in the question stem.

3

Multiply the numbers on the right side of the equation.

18 \times y = 648

Simplify the product of the GCD and LCM.

4

Divide both sides by

18

to isolate

y

.

y = \frac{648}{18} = 36

Solve for the unknown integer

y

.

Anahtar Kavram

Relationship between GCD, LCM, and the product of two positive integers

Tahmini Süre:1m 0s

Soru 8Soru →

Which of the following statements about the positive integer $180$ are true? Select all that apply.

Geçerli olan tümünü seçin

The number of distinct prime factors of

180

is

3

.

The greatest common divisor (GCD) of

180

and

75

is

15

.

The number

1

is the smallest prime factor of

180

.

The least common multiple (LCM) of

180

and

30

is

5400

.

The prime factorization of

180^2

can be written as

(2 + 3 + 5)^4

.

Cevabı ve açıklamayı göster

Cevap: The number of distinct prime factors of $180$ is $3$ .; The greatest common divisor (GCD) of $180$ and $75$ is $15$ .

Cevap

The true statements are that

180

has

3

distinct prime factors and that the greatest common divisor of

180

and

75

is

15

.

The prime factorization of

180

is

2^2 \times 3^2 \times 5^1

, showing exactly

3

distinct prime factors (

2, 3, 5

). Additionally, comparing this with

75 = 3^1 \times 5^2

confirms that their greatest common divisor is

3^1 \times 5^1 = 15

.

Adım Adım Çözüm

1

Find the prime factorization of

180

.

180 = 2^2 \times 3^2 \times 5^1

.

Decomposing

180

into prime factors provides the foundation to evaluate statements regarding prime factors, GCD, and LCM.

2

Count the distinct prime factors of

180

.

The distinct prime bases are

2

,

3

, and

5

, giving

3

distinct prime factors.

Exponents represent multiplicity, so counting distinct bases yields the number of distinct prime factors.

3

Calculate the GCD of

180

and

75

.

Since

75 = 3^1 \times 5^2

,

\text{GCD}(180, 75) = 3^{\min(2,1)} \times 5^{\min(1,2)} = 3^1 \times 5^1 = 15

.

The GCD is found by taking the minimum exponent for each shared prime factor.

4

Verify definition of prime numbers and divisibility rules for remaining choices.

1

is not prime (smallest prime factor is

2

), and

\text{LCM}(180, 30) = 180

because

30

is a factor of

180

.

Eliminate false statements based on arithmetic definitions and divisibility properties.

Anahtar Kavram

Prime Factorization, Greatest Common Divisor (GCD), and Least Common Multiple (LCM)

Soru 9Soru →

What is the sum of all distinct prime factors of $420$ ?

15

17

18

19

21

Cevabı ve açıklamayı göster

Cevap: $17$

Cevap

The sum of all distinct prime factors of

420

is

17

.

The prime factorization of

420

is

2^2 \times 3 \times 5 \times 7

. The distinct prime numbers in this factorization are

2

,

3

,

5

, and

7

. Adding these values together yields

2 + 3 + 5 + 7 = 17

.

Adım Adım Çözüm

1

Find the prime factorization of

420

.

420 = 2^2 \times 3 \times 5 \times 7

Decompose

420

into prime components:

420 = 10 \times 42 = (2 \times 5) \times (2 \times 3 \times 7) = 2^2 \times 3 \times 5 \times 7

.

2

Identify the set of distinct prime factors.

The distinct prime factors are

2, 3, 5,

and

7

.

Exponents are ignored when selecting unique prime factors.

3

Calculate the sum of the distinct prime factors.

2 + 3 + 5 + 7 = 17

Add the unique prime factors together.

Anahtar Kavram

Prime Factorization

Tahmini Süre:45s

Soru 10Soru →

Two positive integers $a$ and $b$ have prime factorizations of the form $a = 2^x \times 3^y \times 5^z$ and $b = 2^u \times 3^v \times 5^w$ , where $x, y, z, u, v, w$ are non-negative integers. The greatest common divisor (GCD) of $a$ and $b$ is $180$ , and the least common multiple (LCM) of $a$ and $b$ is $54,000$ . If $a$ has exactly $24$ positive divisors, what is the value of $b$ ?

Cevabı ve açıklamayı göster

Cevap: 18000

Cevap

18000

Prime factorizing

\text{GCD}(a,b) = 180 = 2^2 \times 3^2 \times 5^1

and

\text{LCM}(a,b) = 54,000 = 2^4 \times 3^3 \times 5^3

reveals that the pair of exponents for bases

2, 3, 5

are

\{2, 4\}

,

\{2, 3\}

, and

\{1, 3\}

respectively. Applying the total divisor formula

(x+1)(y+1)(z+1) = 24

for

a

,

x+1

must divide

24

. Since

x \in \{2, 4\}

,

x

must be

2

(since

5

does not divide

24

). This leaves

(y+1)(z+1) = 8

. With

y \in \{2, 3\}

,

y+1

must be

4

(since

3

does not divide

8

), giving

y = 3

and

z = 1

. Thus,

a = 2^2 \times 3^3 \times 5^1

. The remaining exponent choices for

b

are

u = 4

,

v = 2

, and

w = 3

, giving

b = 2^4 \times 3^2 \times 5^3 = 18,000

.

Adım Adım Çözüm

1

Factorize GCD and LCM into prime bases.

GCD = 2^2 * 3^2 * 5^1; LCM = 2^4 * 3^3 * 5^3.

GCD represents the minimum prime exponents and LCM represents the maximum prime exponents of a and b.

2

Set up candidate exponent pairs for primes 2, 3, and 5.

{x, u} = {2, 4}, {y, v} = {2, 3}, and {z, w} = {1, 3}.

For any prime p, min(exp_p(a), exp_p(b)) = exp_p(GCD) and max(exp_p(a), exp_p(b)) = exp_p(LCM).

3

Solve for x, y, and z using the divisor count formula d(a) = (x+1)(y+1)(z+1) = 24.

x = 2, y = 3, z = 1.

If x = 4, then x+1 = 5, which is not a factor of 24. Thus x = 2. Then (y+1)(z+1) = 8. Since y in {2, 3}, y+1 = 4 implies y = 3 and z = 1.

4

Determine the exponents of b and calculate b.

u = 4, v = 2, w = 3, so b = 2^4 * 3^2 * 5^3 = 18,000.

Complementary values from the sets {x,u}, {y,v}, and {z,w} give the exponents for b.

Anahtar Kavram

Relationship between prime factor exponents in GCD, LCM, and divisor counting functions.

Soru 11Soru →

Let $m$ and $n$ be positive integers whose prime factorizations consist only of the prime factors $2$ , $3$ , and $5$ . The greatest common divisor of $m$ and $n$ is $360$ , and their least common multiple is $108,000$ . If the power of $2$ in the prime factorization of $m$ is strictly greater than the power of $2$ in the prime factorization of $n$ , and $m$ has exactly $36$ positive divisors, what is the total number of positive divisors of $n$ ?

32

48

64

72

96

Cevabı ve açıklamayı göster

Cevap: $64$

Cevap

The total number of positive divisors of

n

is

64

.

By writing the GCD and LCM in prime factorized form,

\text{GCD}(m,n) = 2^3 \times 3^2 \times 5^1

and

\text{LCM}(m,n) = 2^5 \times 3^3 \times 5^3

. Given

a > x

for the exponent of

2

, we obtain

a=5

and

x=3

. Using the divisor count formula for

m

,

6(b+1)(c+1) = 36

, which requires

(b+1)(c+1) = 6

. Testing possible exponent values

b \in \{2,3\}

reveals that only

b=2

yields an integer exponent

c=1

. This uniquely determines

n = 2^3 \times 3^3 \times 5^3

. Calculating

(3+1)(3+1)(3+1)

gives

64

positive divisors.

Adım Adım Çözüm

1

Express the greatest common divisor (GCD) and least common multiple (LCM) in prime factorization form.

\text{GCD}(m, n) = 360 = 2^3 \times 3^2 \times 5^1

and

\text{LCM}(m, n) = 108,000 = 2^5 \times 3^3 \times 5^3

.

Prime factorizations allow direct comparison of the minimum and maximum exponents for each prime factor.

2

Determine the exponent of

2

for both integers.

For prime factor

2

,

\min(a, x) = 3

and

\max(a, x) = 5

. Since the exponent of

2

in

m

is strictly greater than in

n

(

a > x

),

a = 5

and

x = 3

.

The GCD gives the lower exponent bound while the LCM gives the upper exponent bound.

3

Use the number of divisors of

m

to solve for its remaining exponents.

Let

m = 2^5 \times 3^b \times 5^c

. The number of positive divisors of

m

is

(5+1)(b+1)(c+1) = 6(b+1)(c+1) = 36

, so

(b+1)(c+1) = 6

. Since

b \in \{2, 3\}

, if

b = 3

then

b+1=4

which does not divide

6

. Thus,

b = 2

and

c = 1

.

The divisor count formula requires integer exponents, uniquely identifying

b = 2

and

c = 1

for

m

.

4

Determine the exponents for

n

and calculate its divisor count.

Since

b = 2

and

\max(b, y) = 3

,

y = 3

. Since

c = 1

and

\max(c, z) = 3

,

z = 3

. Thus

n = 2^3 \times 3^3 \times 5^3

, and its divisor count is

(3+1)(3+1)(3+1) = 64

.

Combining the max/min constraints from LCM/GCD yields all exponents of

n

.

Anahtar Kavram

GCD and LCM Prime Exponent Relations & Divisor Counting Formula

Tahmini Süre:3m 0s

Soru 12Soru →

Let $a$ and $b$ be positive integers such that $\gcd(a, b) = 60$ and $\text{lcm}(a, b) = 75,600$ . Which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

The product

ab

is equal to

4,536,000

.

Neither

a

nor

b

can be divisible by

49

.

There are exactly

8

distinct unordered pairs of positive integers

\{a, b\}

that satisfy the given conditions.

Both

a

and

b

must be multiples of

9

.

The greatest common divisor of

(a + b)

and

60

is equal to

30

.

Cevabı ve açıklamayı göster

Cevap: The product $ab$ is equal to $4,536,000$ .; Neither $a$ nor $b$ can be divisible by $49$ .; There are exactly $8$ distinct unordered pairs of positive integers $\{a, b\}$ that satisfy the given conditions.

Cevap

The statements asserting that

ab = 4,536,000

, that neither number is divisible by

49

, and that there are exactly

8

distinct unordered pairs

\{a, b\}

are all correct.

The product of the GCD and LCM of two numbers always yields their product, confirming

ab = 4,536,000

. The prime factorization of the LCM shows that

7

appears only to the first power, making divisibility by

49

impossible for either number. Finally, assigning the minimum and maximum prime exponents across the four distinct prime factors

(2, 3, 5, 7)

yields

2^4 = 16

ordered pairs, which corresponds to exactly

8

unordered pairs.

Adım Adım Çözüm

1

Apply the product identity for GCD and LCM.

ab = \gcd(a, b) \times \text{lcm}(a, b) = 60 \times 75,600 = 4,536,000

.

The product of two positive integers is always equal to the product of their greatest common divisor and least common multiple.

2

Find the prime factorizations of

\gcd(a, b)

and

\text{lcm}(a, b)

to determine exponent bounds.

\gcd(a, b) = 2^2 \times 3^1 \times 5^1 \times 7^0

and

\text{lcm}(a, b) = 2^4 \times 3^3 \times 5^2 \times 7^1

.

The exponent of each prime factor in

\gcd(a, b)

gives the minimum exponent in

a

and

b

, while the exponent in

\text{lcm}(a, b)

gives the maximum exponent.

3

Evaluate prime factor constraints for prime

7

and prime

3

.

Maximum power of

7

in

a

or

b

is

7^1 = 7

, so neither can be divisible by

49

. For prime

3

, one integer has

3^1

(not divisible by

9

) and the other has

3^3 = 27

.

An integer is divisible by

49 = 7^2

only if its prime factorization contains

7^2

or higher. Similarly, a number is a multiple of

9 = 3^2

only if its prime exponent for

3

is at least

2

.

4

Calculate the total number of distinct unordered pairs

\{a, b\}

.

For each of the

4

prime factors (

2, 3, 5, 7

), there are

2

choices for which integer gets the higher exponent. Total ordered pairs

(a, b) = 2^4 = 16

. Unordered pairs

\{a, b\} = 16 / 2 = 8

.

Since

\min < \max

for all prime factors,

a \neq b

for all pairs, allowing exact division by

2

to account for symmetry.

5

Analyze

\gcd(a + b, 60)

.

Since

60 \mid a

and

60 \mid b

, it follows that

60 \mid (a + b)

, making

\gcd(a + b, 60) = 60

.

If a number

k

divides both

a

and

b

, it must divide their linear combination

a + b

.

Anahtar Kavram

Prime Factorization, GCD-LCM Identities, and Counting Valid Integer Pairs

Soru 13Soru →

A positive integer $n$ has a prime factorization of the form $2^x \times 3^y \times 7^z$ , where $x$ , $y$ , and $z$ are positive integers. The greatest common divisor of $n$ and $420$ is $84$ , and the least common multiple of $n$ and $420$ is $8,820$ . What is the total number of positive integer factors of $n$ ?

8

12

18

27

36

Cevabı ve açıklamayı göster

Cevap: $27$

Cevap

The total number of positive integer factors of

n

is

27

.

By prime factorizing

420

,

84

, and

8,820

, we find that

420 = 2^2 \times 3^1 \times 5^1 \times 7^1

,

\gcd(n, 420) = 2^2 \times 3^1 \times 7^1

, and

\text{lcm}(n, 420) = 2^2 \times 3^2 \times 5^1 \times 7^2

. Comparing the minimum and maximum powers for each prime factor shows that

n = 2^2 \times 3^2 \times 7^2

. Applying the divisor counting formula gives

(2+1)(2+1)(2+1) = 27

.

Adım Adım Çözüm

1

Express all given values in their prime factorized forms.

420 = 2^2 \times 3^1 \times 5^1 \times 7^1

,

\gcd(n, 420) = 84 = 2^2 \times 3^1 \times 7^1

, and

\text{lcm}(n, 420) = 8,820 = 2^2 \times 3^2 \times 5^1 \times 7^2

.

Finding the prime factorizations allows comparison of prime exponent bounds for GCD and LCM.

2

Determine the exponents

x

,

y

, and

z

for

n = 2^x \times 3^y \times 7^z

.

For prime

2

:

\max(x, 2) = 2

and

\min(x, 2) = 2 \implies x = 2

.
For prime

3

:

\max(y, 1) = 2

and

\min(y, 1) = 1 \implies y = 2

.
For prime

7

:

\max(z, 1) = 2

and

\min(z, 1) = 1 \implies z = 2

.

GCD takes the minimum exponent of each prime factor, while LCM takes the maximum exponent.

3

Calculate the total number of positive factors of

n = 2^2 \times 3^2 \times 7^2

.

Total factors

= (x + 1)(y + 1)(z + 1) = (2 + 1)(2 + 1)(2 + 1) = 3 \times 3 \times 3 = 27

.

The number of positive divisors of

p_1^{a} p_2^{b} \dots p_k^{k}

is

(a+1)(b+1)\dots(k+1)

.

Anahtar Kavram

Relationship between prime factorization, greatest common divisor (GCD), least common multiple (LCM), and the number of positive divisors.

Soru 14Soru →

A positive integer $n$ has the prime factorization $2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers. Given that $\gcd(n, 7{}200) = 360$ and $\text{lcm}(n, 1{}080) = 3{}240$ , what is the value of $a + b + c$ ?

6

7

8

9

10

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

By prime factorizing each given term,

7{}200 = 2^5 \cdot 3^2 \cdot 5^2

,

360 = 2^3 \cdot 3^2 \cdot 5^1

,

1{}080 = 2^3 \cdot 3^3 \cdot 5^1

, and

3{}240 = 2^3 \cdot 3^4 \cdot 5^1

. Applying the definition of GCD as taking the minimum exponent gives

\min(a, 5) = 3 \implies a = 3

and

\min(c, 2) = 1 \implies c = 1

. Applying the definition of LCM as taking the maximum exponent gives

\max(b, 3) = 4 \implies b = 4

. Thus,

a + b + c = 3 + 4 + 1 = 8

.

Adım Adım Çözüm

1

Find the prime factorizations of the given integers and the GCD/LCM values.

7{}200 = 2^5 \cdot 3^2 \cdot 5^2

,

360 = 2^3 \cdot 3^2 \cdot 5^1

,

1{}080 = 2^3 \cdot 3^3 \cdot 5^1

, and

3{}240 = 2^3 \cdot 3^4 \cdot 5^1

.

Decomposing into prime factors allows direct comparison of exponents using GCD (minimum exponent) and LCM (maximum exponent) rules.

2

Analyze the GCD condition

\gcd(n, 7{}200) = 360

.

\min(a, 5) = 3 \implies a = 3

,

\min(b, 2) = 2 \implies b \ge 2

, and

\min(c, 2) = 1 \implies c = 1

.

The exponent of each prime factor in

\gcd(x, y)

is the minimum of their respective exponents in

x

and

y

.

3

Analyze the LCM condition

\text{lcm}(n, 1{}080) = 3{}240

.

\max(a, 3) = 3 \implies a \le 3

,

\max(b, 3) = 4 \implies b = 4

, and

\max(c, 1) = 1 \implies c \le 1

.

The exponent of each prime factor in

\text{lcm}(x, y)

is the maximum of their respective exponents in

x

and

y

.

4

Combine the exponent constraints to determine

a

,

b

, and

c

, then calculate their sum.

a = 3

,

b = 4

, and

c = 1

, so

a + b + c = 3 + 4 + 1 = 8

.

Combining

a=3

,

b \ge 2

with

b=4

, and

c=1

with

c \le 1

uniquely specifies

(a,b,c) = (3,4,1)

.

Anahtar Kavram

Prime Exponent Analysis of Greatest Common Divisor and Least Common Multiple

Tahmini Süre:2m 0s

Soru 15Soru →

Three automated production lines—Line A, Line B, and Line C—run continuous maintenance cycles every $45$ minutes, $60$ minutes, and $75$ minutes, respectively. All three lines completed a maintenance cycle simultaneously at 8:00 a.m. on Monday. A supervisory inspection is triggered whenever at least two of the three lines complete a maintenance cycle at the exact same time. Between 8:01 a.m. on Monday and 8:00 a.m. on Tuesday, inclusive, how many supervisory inspections will be triggered?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

16

To find the number of times at least two lines finish maintenance simultaneously, we determine the LCM for each pair of intervals (180, 225, and 300 minutes) and calculate how many multiples of these values occur within 1,440 minutes. Within this duration, there are 8 multiples of 180, 6 multiples of 225, and 4 multiples of 300. The single time all three coincide (at 900 minutes) is included in all three pairwise counts. Subtracting 2 instances of this overlap yields 8 + 6 + 4 - 2 = 16 total supervisory inspections.

Adım Adım Çözüm

1

Prime factorize each interval length

45 = 3^2 × 5; 60 = 2^2 × 3 × 5; 75 = 3 × 5^2

Prime factorization allows systematic calculation of pairwise and triple LCMs.

2

Compute pairwise and triple LCMs

lcm(45, 60) = 180; lcm(45, 75) = 225; lcm(60, 75) = 300; lcm(45, 60, 75) = 900

The LCM determines the interval between simultaneous completions.

3

Count occurrences within the 1,440-minute time window

A & B: 8 times; A & C: 6 times; B & C: 4 times; All three: 1 time

Dividing the total time window by each LCM yields the frequency of simultaneous maintenance events.

4

Apply Principle of Inclusion-Exclusion for sets

8 + 6 + 4 - 2(1) = 16 distinct inspection events

Events where all three lines coincide (at t = 900) are counted in all three pair counts and must be adjusted so that the time point is counted exactly once.

Anahtar Kavram

Least Common Multiple (LCM) and Inclusion-Exclusion Principle

Soru 16Soru →

A rectangular floor measuring $126\text{ cm}$ by $180\text{ cm}$ is to be completely covered with identical square tiles of the largest possible side length, without cutting any tiles or leaving gaps. What is the total number of square tiles required to cover the floor?

18

35

70

140

306

Cevabı ve açıklamayı göster

Cevap: 70

Cevap

The total number of square tiles required is 70.

To cover the rectangular floor with the largest possible identical square tiles, the side length of the square tile must be the greatest common divisor of

126

and

180

. Factoring both numbers gives

126 = 2 \times 3^2 \times 7

and

180 = 2^2 \times 3^2 \times 5

. The greatest common divisor is

2 \times 3^2 = 18\text{ cm}

. Dividing the dimensions by

18\text{ cm}

gives

7

tiles along the length and

10

tiles along the width. Multiplying

7 \times 10

yields a total of

70

tiles.

Adım Adım Çözüm

1

Find the prime factorization of both dimensions of the floor.

126 = 2^1 \times 3^2 \times 7^1

and

180 = 2^2 \times 3^2 \times 5^1

Prime factorization allows systematic determination of the greatest common divisor.

2

Calculate the Greatest Common Divisor (GCD) of 126 and 180 to determine the largest possible square tile side length.

\gcd(126, 180) = 2^1 \times 3^2 = 18\text{ cm}

The square tiles must fit evenly along both the length and width without cutting.

3

Determine the number of tiles along each dimension.

Along length:

126 \div 18 = 7

; along width:

180 \div 18 = 10

Dividing each total length by the tile side length gives the tile count per side.

4

Calculate the total number of tiles required.

7 \times 10 = 70\text{ tiles}

The floor forms a 7 by 10 grid of tiles.

Anahtar Kavram

Greatest Common Divisor (GCD) applications in geometric tiling

Soru 17Soru →

What is the smallest positive integer that is a multiple of 18, 24, and 30, and is also a perfect square?

Cevabı ve açıklamayı göster

Cevap: 3600

Cevap

The smallest positive integer that is a multiple of 18, 24, and 30, and is also a perfect square is 3600.

The least common multiple of 18, 24, and 30 is 360, which factors into

2^3 \cdot 3^2 \cdot 5^1

. For an integer to be a perfect square, all exponents in its prime factorization must be even. Multiplying 360 by

2 \cdot 5 = 10

completes the odd exponents to even values (

2^4 \cdot 3^2 \cdot 5^2

), giving 3600, which is the smallest perfect square divisible by 18, 24, and 30.

Adım Adım Çözüm

1

Find the prime factorization of 18, 24, and 30

18 = 2^1 \cdot 3^2

,

24 = 2^3 \cdot 3^1

,

30 = 2^1 \cdot 3^1 \cdot 5^1

Decomposing numbers into prime factors allows calculation of the LCM and analysis of perfect square conditions.

2

Calculate the LCM of 18, 24, and 30

\text{LCM}(18, 24, 30) = 2^3 \cdot 3^2 \cdot 5^1 = 360

Any common multiple must be a multiple of the LCM of these three numbers.

3

Determine the smallest factor required to make the prime exponents even

Multiply 360 by

2^1 \cdot 5^1 = 10

A perfect square requires all prime exponents to be even; 2 has power 3 and 5 has power 1 in 360, so one more factor of 2 and one more factor of 5 are required.

4

Compute the final result

360 \times 10 = 3600

3600 = 2^4 \cdot 3^2 \cdot 5^2 = 60^2

, which is a perfect square.

Anahtar Kavram

Prime Factorization, LCM, and Exponent Properties of Perfect Squares

Soru 18Soru →

Let $n = 2^a \cdot 3^b \cdot p^c$ , where $a$ , $b$ , and $c$ are positive integers and $p$ is a prime number strictly greater than $5$ . The integer $n$ has exactly $36$ positive divisors, and $\gcd(n, 180) = 36$ . Which of the following statements MUST be true? Select all such statements.

Geçerli olan tümünü seçin

The integer

n

is divisible by

36

.

The exponent

c

cannot exceed

3

.

The sum of the exponents

a + b + c

must equal

7

.

The least common multiple of

n

and

180

is equal to

5n

.

The exponent

a

must equal

2

.

Cevabı ve açıklamayı göster

Cevap: The integer $n$ is divisible by $36$ .; The exponent $c$ cannot exceed $3$ .; The least common multiple of $n$ and $180$ is equal to $5n$ .

Cevap

The statements asserting that

n

is divisible by

36

, that the exponent

c

cannot exceed

3

, and that

\text{lcm}(n, 180) = 5n

must be true.

The statement that

n

is divisible by

36

is true because

a \ge 2

and

b \ge 2

. The statement that

c \le 3

is true because

(a+1)(b+1) \ge 9

, bounding

c+1 \le 4

. The statement that

\text{lcm}(n, 180) = 5n

is true because

\max(a, 2) = a

,

\max(b, 2) = b

, and

180

contributes a factor of

5^1

.

Adım Adım Çözüm

1

Analyze the GCD condition to establish bounds on exponents

a

and

b

.

a \ge 2

and

b \ge 2

.

Since

180 = 2^2 \cdot 3^2 \cdot 5^1

and

\gcd(n, 180) = 36 = 2^2 \cdot 3^2

, the minimum of the exponents of

2

is

2

(so

a \ge 2

) and the minimum of the exponents of

3

is

2

(so

b \ge 2

).

2

Use the divisor count formula

(a+1)(b+1)(c+1) = 36

to find constraints on

c

.

c \le 3

.

Since

a \ge 2 \implies a+1 \ge 3

and

b \ge 2 \implies b+1 \ge 3

, the product

(a+1)(b+1) \ge 9

. Therefore,

c+1 = \frac{36}{(a+1)(b+1)} \le \frac{36}{9} = 4

, which gives

c \le 3

.

3

Evaluate the divisibility of

n

by

36

.

n

is divisible by

36

.

Because

a \ge 2

and

b \ge 2

,

2^2 \cdot 3^2 = 36

divides

2^a \cdot 3^b \cdot p^c = n

.

4

Calculate

\text{lcm}(n, 180)

.

\text{lcm}(n, 180) = 5n

.

\text{lcm}(2^a \cdot 3^b \cdot p^c, 2^2 \cdot 3^2 \cdot 5^1) = 2^{\max(a,2)} \cdot 3^{\max(b,2)} \cdot 5^1 \cdot p^c = 2^a \cdot 3^b \cdot 5 \cdot p^c = 5n

since

a \ge 2

,

b \ge 2

, and

p > 5

.

5

Test counterexamples for statement constancy regarding

a+b+c=7

and

a=2

.

Neither

a+b+c=7

nor

a=2

is required.

The choice

(a, b, c) = (5, 2, 1)

satisfies

(5+1)(2+1)(1+1) = 36

and

\gcd(n, 180) = 36

, yielding

a+b+c=8 \neq 7

and

a=5 \neq 2

.

Anahtar Kavram

Prime Factorization, Greatest Common Divisor (GCD), and Least Common Multiple (LCM) properties

Soru 19Soru →

Two positive integers $x$ and $y$ satisfy $x < y$ , $\gcd(x, y) = 18$ , and $\text{lcm}(x, y) = 1080$ . If $x$ has exactly $6$ positive divisors, what is the value of $y$ ?

216

270

360

540

1080

Cevabı ve açıklamayı göster

Cevap: $1080$

Cevap

1080

Since

\gcd(x, y) = 18

and

\text{lcm}(x, y) = 1080

, we can express

x = 18m

and

y = 18n

with

\gcd(m, n) = 1

and

m < n

. Substituting into

\text{lcm}(x, y) = 18mn = 1080

gives

mn = 60

. The coprime pairs

(m, n)

with

m < n

are

(1, 60)

,

(3, 20)

,

(4, 15)

, and

(5, 12)

. Testing

x = 18m

for each pair: for

m = 1

,

x = 18 = 2^1 \cdot 3^2

, which has

(1+1)(2+1) = 6

positive divisors. This matches the condition, so

n = 60

and

y = 18 \times 60 = 1080

.

Adım Adım Çözüm

1

Express

x

and

y

in terms of their Greatest Common Divisor

Let

x = 18m

and

y = 18n

, where

m

and

n

are positive integers such that

m < n

and

\gcd(m, n) = 1

.

Since

\gcd(x, y) = 18

, both numbers must be multiples of

18

, and their remaining factor parts must be coprime.

2

Use the LCM formula

\text{lcm}(x, y) = 18mn

to find

mn

18mn = 1080 \implies mn = 60

.

The least common multiple of

18m

and

18n

with

\gcd(m,n)=1

is

18mn

.

3

Find all coprime factor pairs

(m, n)

of

60

with

m < n

The prime factorization of

60

is

2^2 \cdot 3^1 \cdot 5^1

. Coprime pairs

(m, n)

are

(1, 60)

,

(3, 20)

,

(4, 15)

, and

(5, 12)

.

Pairs must be coprime so that

\gcd(18m, 18n) = 18

remains true.

4

Determine the number of positive divisors for

x = 18m

for each candidate pair

For

m = 1

,

x = 18 = 2^1 \cdot 3^2

, which has

(1+1)(2+1) = 6

positive divisors.

The number of positive divisors of

p_1^{a_1} p_2^{a_2} \dots

is given by

(a_1 + 1)(a_2 + 1) \dots

. Only

m = 1

gives exactly

6

divisors.

5

Calculate

y

y = 18n = 18 \cdot 60 = 1080

.

Corresponding to

m = 1

,

n = 60

gives

y = 1080

.

Anahtar Kavram

Prime Factorization, GCD/LCM Coprime Relationships, and Divisor Counting Formula

Alternatif Yöntem

Alternatively, use

\gcd(x, y) \cdot \text{lcm}(x, y) = x \cdot y \implies x \cdot y = 18 \cdot 1080 = 19440

. Since

x

has

6

positive divisors and

x

is a multiple of

18

,

x

must equal

18

because

18 = 2^1 \cdot 3^2

has

(1+1)(2+1) = 6

divisors. Then

y = 19440 / 18 = 1080

.

Tahmini Süre:1m 30s

Soru 20Soru →

Two positive integers $x$ and $y$ have a greatest common divisor (GCD) of $12$ and a least common multiple (LCM) of $360$ . Which of the following could be the value of $x$ ? Indicate all such values.

Geçerli olan tümünü seçin

24

48

60

90

120

Cevabı ve açıklamayı göster

Cevap: $24$ ; $60$ ; $120$

Cevap

The possible values for

x

are

24

,

60

, and

120

.

Any valid value of

x

must be a multiple of

\gcd(x, y) = 12

and a divisor of

\text{lcm}(x, y) = 360

. Expressing

12

as

2^2 \cdot 3^1

and

360

as

2^3 \cdot 3^2 \cdot 5^1

,

x

must be of the form

2^a \cdot 3^b \cdot 5^c

with

2 \leq a \leq 3

,

1 \leq b \leq 2

, and

0 \leq c \leq 1

. The values

24

,

60

, and

120

meet all exponent constraints.

Adım Adım Çözüm

1

Find the prime factorizations of the given GCD and LCM.

\gcd(x, y) = 12 = 2^2 \cdot 3^1 \cdot 5^0

and

\text{lcm}(x, y) = 360 = 2^3 \cdot 3^2 \cdot 5^1

.

Prime factorization allows analysis of exponent constraints for each prime factor.

2

Determine the constraints on any valid integer

x

.

Any valid value of

x

must be a multiple of

12

and a factor of

360

. Specifically,

x = 2^a \cdot 3^b \cdot 5^c

where

2 \leq a \leq 3

,

1 \leq b \leq 2

, and

0 \leq c \leq 1

.

The GCD defines the minimum exponent for each prime factor, and the LCM defines the maximum exponent.

3

Test each provided option against the prime exponent bounds.

24 = 2^3 \cdot 3^1

satisfies the bounds (

a=3, b=1, c=0

).

48 = 2^4 \cdot 3^1

violates

a \leq 3

.

60 = 2^2 \cdot 3^1 \cdot 5^1

satisfies the bounds (

a=2, b=1, c=1

).

90 = 2^1 \cdot 3^2 \cdot 5^1

violates

a \geq 2

.

120 = 2^3 \cdot 3^1 \cdot 5^1

satisfies the bounds (

a=3, b=1, c=1

).

Options satisfying all exponent inequalities are valid possible values of

x

.

Anahtar Kavram

Prime exponent properties of GCD and LCM

Prime Factorization, GCD, and LCM

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Cevap

Adım Adım Çözüm