Tüm alıştırma soruları

Soru 161Soru →

A positive integer $N$ has no prime factors other than $2$ and $3$ . If $N$ is a multiple of $12$ and has exactly $18$ positive divisors, what is the sum of all possible values of $N$ ?

Cevabı ve açıklamayı göster

Cevap: 2028

Cevap

2028

The correct numerical answer is 2028, obtained by determining all exponent combinations of 2 and 3 that yield 18 total factors while guaranteeing divisibility by 12.

Adım Adım Çözüm

1

Set up the prime factorization of

N

with constraints.

N = 2^a \cdot 3^b

with

a \ge 2

and

b \ge 1

.

Since the only prime factors are

2

and

3

, and

12 = 2^2 \cdot 3^1

divides

N

, the exponents must satisfy

a \ge 2

and

b \ge 1

.

2

Apply the total number of divisors formula.

(a+1)(b+1) = 18

.

The number of positive divisors of

p_1^{e_1} p_2^{e_2}

is

(e_1 + 1)(e_2 + 1)

.

3

Identify all valid integer solution pairs for

(a+1, b+1)

.

Valid pairs are

(9,2)

,

(6,3)

, and

(3,6)

, corresponding to

(a,b) = (8,1), (5,2), (2,5)

.

We require

a+1 \ge 3

and

b+1 \ge 2

. Pairs

(18,1)

and

(2,9)

violate these lower bounds.

4

Compute the corresponding values of

N

and find their sum.

768 + 288 + 972 = 2028

.

Evaluating

2^8 \cdot 3^1 = 768

,

2^5 \cdot 3^2 = 288

, and

2^2 \cdot 3^5 = 972

yields a total sum of

2028

.

Anahtar Kavram

Divisor Count Formula & Multiplicativity Constraints

Soru 162Soru →

For how many integer values of $k$ in the range $1 \le k \le 60$ is the expression $3k^2 + 5k + 7$ an even integer?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

0

The expression

3k^2 + 5k + 7

can be evaluated for parity by considering the cases for

k

.
If

k

is even,

3k^2

is even,

5k

is even, and

7

is odd. The sum of two even integers and an odd integer (even + even + odd) is always odd.
If

k

is odd,

3k^2

is odd,

5k

is odd, and

7

is odd. The sum of three odd integers (odd + odd + odd) is always odd.
Because the expression yields an odd integer for every integer

k

, there are no integer values of

k

in the specified range for which the expression is even. Therefore, the correct count is 0.

Adım Adım Çözüm

1

Examine parity by testing even and odd cases for k

If k is even: 3(even)^2 + 5(even) + 7 = even + even + odd = odd. If k is odd: 3(odd)^2 + 5(odd) + 7 = odd + odd + odd = odd.

Covering both cases establishes the parity of the expression for all integer inputs.

2

Count the number of values of k in 1 <= k <= 60 that yield an even result

Since the expression is odd for all integer values of k, zero values yield an even integer.

The question specifically asks for the number of integer values of k that make the expression even.

Anahtar Kavram

Parity Rules for Addition, Multiplication, and Algebraic Expressions

Soru 163Soru →

For any integer $n$ , which of the following expressions must be divisible by $2$ ?

n^2 + n

n^2 + 1

n^3 + 1

(n + 1)^2

2n + 1

Cevabı ve açıklamayı göster

Cevap: $n^2 + n$

Cevap

The expression

n^2 + n

is always divisible by

2

for any integer

n

.

The expression

n^2 + n

factors into

n(n + 1)

. Because

n

and

n + 1

are consecutive integers, one of them must be even. Any integer multiplied by an even integer produces an even number, guaranteeing that

n^2 + n

is divisible by

2

for every integer

n

.

Adım Adım Çözüm

1

Factor the given algebraic expression.

n^2 + n = n(n + 1)

Factoring out

n

reveals the product of two consecutive integers.

2

Analyze the parity of consecutive integers

n

and

n + 1

.

In any pair of consecutive integers

(n, n + 1)

, exactly one number is even.

Even and odd integers alternate sequentially.

3

Determine divisibility by

2

.

Because one of the factors is even (divisible by

2

), their product

n(n + 1)

is always divisible by

2

.

Any integer multiplied by an even number yields an even result.

Anahtar Kavram

The product of any two consecutive integers is always even and therefore divisible by 2.

Tahmini Süre:45s

Soru 164Soru →

Let $A = 12^6 \cdot 35^4$ and $B = 18^4 \cdot 14^6$ . If $d = \text{gcd}(A, B)$ , how many positive factors of $d^2$ are not factors of $d$ ?

Cevabı ve açıklamayı göster

Cevap: 2072

Cevap

The number of positive factors of

d^2

that are not factors of

d

is 2072.

Decomposing

A

and

B

into prime factors yields

A = 2^{12} \cdot 3^6 \cdot 5^4 \cdot 7^4

and

B = 2^{10} \cdot 3^8 \cdot 7^6

. Taking the minimum power of each common prime gives

d = \text{gcd}(A, B) = 2^{10} \cdot 3^6 \cdot 7^4

, which has

(10+1)(6+1)(4+1) = 385

positive factors. For

d^2 = 2^{20} \cdot 3^{12} \cdot 7^8

, the total number of positive factors is

(20+1)(12+1)(8+1) = 2457

. Subtracting the factors of

d

yields

2457 - 385 = 2072

.

Adım Adım Çözüm

1

Find the prime factorizations of

A

and

B

A = 2^{12} \cdot 3^6 \cdot 5^4 \cdot 7^4

and

B = 2^{10} \cdot 3^8 \cdot 7^6

Converting composite bases into prime factors allows determination of common divisor properties.

2

Determine the greatest common divisor

d = \text{gcd}(A, B)

d = 2^{10} \cdot 3^6 \cdot 7^4

The GCD takes the minimum exponent for each common prime factor between

A

and

B

.

3

Calculate the total number of positive factors of

d

385 positive factors

Adding 1 to each prime exponent of

d

and multiplying gives

(10+1)(6+1)(4+1) = 385

.

4

Find the prime factorization and number of positive factors of

d^2

d^2 = 2^{20} \cdot 3^{12} \cdot 7^8

, which has 2457 positive factors

Squaring

d

doubles all prime exponents. The number of factors is

(20+1)(12+1)(8+1) = 2457

.

5

Subtract the number of factors of

d

from the number of factors of

d^2

2457 - 385 = 2072

Since every factor of

d

is also a factor of

d^2

, the factors of

d^2

that are not factors of

d

equal the total factors of

d^2

minus the factors of

d

.

Anahtar Kavram

Prime Factorization, Greatest Common Divisor (GCD), and the Divisor Count Formula

Soru 165Soru →

If $m$ and $n$ are integers such that $m^2 n + m$ is an odd integer, which of the following expressions MUST be an even integer?

m^2 + n

m + 2n

m(n + 1)

m^2 + n^2 + 1

3m - n

Cevabı ve açıklamayı göster

Cevap: $m^2 + n^2 + 1$

Cevap

The expression

m^2 + n^2 + 1

must be an even integer.

Factoring the given expression gives

m(mn + 1) = \text{odd}

, which implies

m

is odd and

mn + 1

is odd. Consequently,

mn

is even, which forces

n

to be even (since

m

is odd). Substituting

m = \text{odd}

and

n = \text{even}

into the expression

m^2 + n^2 + 1

yields

\text{odd} + \text{even} + 1 = \text{even}

. Thus, this expression must be an even integer.

Adım Adım Çözüm

1

Factor the given expression to analyze its parity.

m^2 n + m = m(mn + 1)

.

Factoring out

m

isolates the product of two factors.

2

Determine the parity of each factor.

Since

m(mn + 1)

is odd, both

m

and

(mn + 1)

must be odd integers.

The product of two integers is odd if and only if both factors are odd.

3

Determine the parity of

n

.

Since

mn + 1

is odd,

mn

must be even. Because

m

is odd,

n

must be even.

For the product

mn

to be even when

m

is odd,

n

must be an even integer (including 0).

4

Evaluate the target expression

m^2 + n^2 + 1

.

m^2

is odd,

n^2

is even, so

m^2 + n^2 + 1 = \text{odd} + \text{even} + \text{odd} = \text{even}

.

Summing two odd integers and one even integer always results in an even integer.

Anahtar Kavram

Parity properties under addition, subtraction, and multiplication

Tahmini Süre:1m 30s

Soru 166Soru →

If $a$ , $b$ , and $c$ are integers such that $a(b + c)$ is an odd integer, which of the following expressions MUST be an even integer?

a b + c

a + b c

a b + b c

a + b + c

a c + b

Cevabı ve açıklamayı göster

Cevap: $a + b + c$

Cevap

The expression

a + b + c

must be an even integer.

For the product

a(b + c)

to be odd, both

a

and

(b + c)

must be odd integers. Regrouping the sum

a + b + c

as

a + (b + c)

shows it is the sum of two odd integers, which is guaranteed to be an even integer.

Adım Adım Çözüm

1

Analyze the condition

a(b + c)

is odd

For a product of two integers to be odd, both factors must be odd. Therefore,

a

is odd and

(b + c)

is odd.

The product of an even integer and any integer is always even.

2

Evaluate the parity of

a + b + c

a + b + c = a + (b + c) = \text{odd} + \text{odd} = \text{even}

.

The sum of any two odd integers is always an even integer.

Anahtar Kavram

Parity rules for integer addition and multiplication

Soru 167Soru →

For a positive integer $N = 2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers, the integer $\frac{N}{2}$ has $40$ positive factors, the integer $\frac{N}{3}$ has $36$ positive factors, and the integer $\frac{N}{5}$ has $24$ positive factors. What is the total number of positive factors of $N^2$ ?

96

210

231

462

576

Cevabı ve açıklamayı göster

Cevap: 231

Cevap

231 positive factors

The correct answer is 231. By representing the factor counts of

\frac{N}{2}

,

\frac{N}{3}

, and

\frac{N}{5}

algebraically, we establish a system of equations for the exponents

a, b, c

. Solving this system yields

a=5

,

b=3

, and

c=1

. Squaring

N

doubles each exponent, giving

N^2 = 2^{10} \times 3^6 \times 5^2

. Applying the factor count formula gives

(10+1)(6+1)(2+1) = 231

.

Adım Adım Çözüm

1

Set up equations for the number of positive factors of N/2, N/3, and N/5 using the prime factorization formula.

The total number of factors of an integer

2^x 3^y 5^z

is

(x+1)(y+1)(z+1)

. Thus:

a(b+1)(c+1) = 40

,

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

, and

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

.

Dividing

N

by a prime factor reduces that prime factor's exponent by 1.

2

Express each equation in terms of

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

, the total number of factors of

N

.

(b+1)(c+1) = T - 40

,

(a+1)(c+1) = T - 36

, and

(a+1)(b+1) = T - 24

.

Expanding

(x-1)yz = xyz - yz = T - yz

allows expressing pairwise products in terms of

T

.

3

Multiply the three pairwise product equations to solve for

T

.

[(a+1)(b+1)(c+1)]^2 = (T-40)(T-36)(T-24) \implies T^2 = (T-40)(T-36)(T-24)

. Testing

T = 48

:

48^2 = 2304

and

(8)(12)(24) = 2304

. Thus

T = 48

.

Multiplying

(b+1)(c+1) \times (a+1)(c+1) \times (a+1)(b+1)

yields

T^2

.

4

Determine the individual exponents

a

,

b

, and

c

.

a+1 = \sqrt{\frac{(T-36)(T-24)}{T-40}} = \sqrt{\frac{12 \times 24}{8}} = 6 \implies a = 5

. Similarly,

b+1 = 4 \implies b = 3

, and

c+1 = 2 \implies c = 1

.

Dividing the product of two pairwise terms by the third gives the square of a single term.

5

Calculate the total number of positive factors of

N^2

.

N^2 = 2^{2a} \times 3^{2b} \times 5^{2c} = 2^{10} \times 3^6 \times 5^2

. Number of factors

= (10+1)(6+1)(2+1) = 11 \times 7 \times 3 = 231

.

Squaring

N

doubles each prime factor's exponent.

Anahtar Kavram

Divisibility, Prime Factorization, and Total Positive Factor Counting Formula

Soru 168Soru →

If $x$ , $y$ , and $z$ are integers such that $x^2 + y^2 + z^2$ is an odd integer, then the sum $x + y + z$ must be an odd integer.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is True because the square of any integer preserves the parity of that integer, making the parity of

x^2 + y^2 + z^2

identical to the parity of

x + y + z

.

The statement is true because for all integers

k

, the parity of

k^2

is identical to the parity of

k

. Consequently, the parity of the sum of squares

x^2 + y^2 + z^2

is always identical to the parity of the sum

x + y + z

. If the sum of squares is odd, the sum of the variables must also be odd.

Adım Adım Çözüm

1

Analyze the parity relationship between any integer

k

and its square

k^2

.

If

k

is even,

k^2

is even; if

k

is odd,

k^2

is odd. (Note that zero is an even integer, and

0^2 = 0

is even).

Multiplying an even integer by itself yields an even number, and multiplying an odd integer by itself yields an odd number.

2

Relate the parity of

x^2 + y^2 + z^2

to

x + y + z

.

The expression

x^2 + y^2 + z^2

has the exact same parity as

x + y + z

.

Replacing each squared term with its base term does not change whether the sum is even or odd.

3

Apply the given condition to deduce the final parity.

Since

x^2 + y^2 + z^2

is odd,

x + y + z

must be odd.

Two expressions with identical parity must both be odd if one is given as odd.

Anahtar Kavram

Parity Invariance of Integer Powers

Tahmini Süre:1m 0s

Soru 169Soru →

What is the sum of all the distinct prime factors of $30$ ?

10

11

8

15

72

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

10

The prime factorization of 30 is 2 × 3 × 5. The distinct prime factors are 2, 3, and 5, and their sum is 2 + 3 + 5 = 10.

Adım Adım Çözüm

1

Find the prime factorization of 30

30 = 2 \times 3 \times 5

Decomposing 30 into prime factors identifies all prime numbers that divide 30.

2

Identify the distinct prime factors

The distinct prime factors are 2, 3, and 5.

Prime factors are the prime numbers present in the prime factorization.

3

Calculate the sum of these distinct prime factors

2 + 3 + 5 = 10

The question asks for the sum of all distinct prime factors.

Anahtar Kavram

Prime Factorization and Prime Factors

Tahmini Süre:45s

Soru 170Soru →

For how many integers $x$ in the range $-50 \le x \le 50$ is the value of the expression $x^5 - 5x^3 + 4x + 3$ an even integer?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

There are 0 integers in the specified range for which the expression evaluates to an even integer.

Factoring

x^5 - 5x^3 + 4x

yields

(x - 2)(x - 1)x(x + 1)(x + 2)

, which is the product of 5 consecutive integers. The product of consecutive integers is always even for any integer

x

. Adding 3 (an odd integer) to an even integer results in an odd integer for all values of

x

. Consequently, zero integers in the given range result in an even value.

Adım Adım Çözüm

1

Factor the variable portion of the polynomial

x^5 - 5x^3 + 4x + 3

.

The expression rewrites as

(x - 2)(x - 1)x(x + 1)(x + 2) + 3

.

Factoring helps identify structural properties such as consecutive terms.

2

Analyze the parity of the product of 5 consecutive integers

(x - 2)(x - 1)x(x + 1)(x + 2)

.

The product is always an even integer for all integer values of

x

.

Any sequence of 5 consecutive integers includes multiple even numbers. Even when

x = 0

, the product equals 0, which is an even integer.

3

Determine the parity of the full expression by adding the constant term 3.

Even integer + 3 (odd integer) = odd integer.

The sum of an even integer and an odd integer is always an odd integer.

4

Count the number of integer values of

x

in

-50 \le x \le 50

that yield an even integer.

0 integers.

Since the expression evaluates to an odd integer for every integer

x

, no integer value can produce an even integer.

Anahtar Kavram

Parity of consecutive integer products and addition rules for even and odd integers.

Tahmini Süre:2m 0s

Soru 171Soru →

What is the greatest common divisor (GCD) of $2^3 \times 3^2 \times 5$ and $2^2 \times 3^3$ ?

36

72

180

1080

18

Cevabı ve açıklamayı göster

Cevap: $36$

Cevap

The greatest common divisor (GCD) of

2^3 \times 3^2 \times 5

and

2^2 \times 3^3

is

36

.

The greatest common divisor (GCD) is calculated by taking the lowest power of each prime factor common to both expressions. For

2^3 \times 3^2 \times 5

and

2^2 \times 3^3

, the common prime factors are

2

and

3

. Taking the smallest exponents gives

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

.

Adım Adım Çözüm

1

Identify the common prime factors in both numbers.

The prime factors present in both numbers are

2

and

3

. The factor

5

is only present in the first number.

GCD requires prime factors that are shared by all target numbers.

2

Select the minimum exponent for each common prime factor.

For

2

:

\min(3, 2) = 2

. For

3

:

\min(2, 3) = 2

. For

5

:

\min(1, 0) = 0

.

The GCD contains only the prime powers that divide both numbers completely.

3

Multiply the resulting prime power factors to evaluate the GCD.

\text{GCD} = 2^2 \times 3^2 = 4 \times 9 = 36

.

Evaluating the prime factor powers gives the final integer value.

Anahtar Kavram

To find the Greatest Common Divisor (GCD) of numbers written in prime factorized form, take the product of each common prime factor raised to its smallest exponent across all the factorizations.

Tahmini Süre:1m 0s

Soru 172Soru →

What is the least common multiple (LCM) of $12$ and $18$ ?

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

The least common multiple of

12

and

18

is

36

.

To find the least common multiple of

12

and

18

, determine the prime factorization of each number (

12 = 2^2 \times 3^1

and

18 = 2^1 \times 3^2

). The LCM is calculated by taking the maximum power of each prime factor present:

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

.

Adım Adım Çözüm

1

Express both numbers in terms of their prime factorizations.

12 = 2^2 \times 3^1

and

18 = 2^1 \times 3^2

Prime factorization separates each number into basic prime components.

2

Select the highest power of each prime factor that appears in either factorization.

The highest power of

2

is

2^2

, and the highest power of

3

is

3^2

.

The least common multiple must contain enough factors to be divisible by both original numbers.

3

Calculate the product of these highest prime powers.

\text{LCM}(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36

Multiplying the chosen powers yields the minimum integer divisible by both

12

and

18

.

Anahtar Kavram

Finding the Least Common Multiple (LCM) via Prime Factorization

Tahmini Süre:45s

Soru 173Soru →

For how many ordered pairs of positive integers $(a, b)$ , where $1 \le a \le 15$ and $1 \le b \le 15$ , is the value of the expression $(a^2 + a + 1)(b^3 + b + 7)$ an even integer?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

The expression evaluates to an odd integer for all positive integer pairs

(a, b)

, so there are 0 ordered pairs for which the value is an even integer.

The factor

(a^2 + a + 1)

simplifies to

a(a + 1) + 1

. Because

a(a + 1)

is the product of two consecutive integers, it is always even, making

a(a + 1) + 1

odd for every integer

a

. The factor

(b^3 + b + 7)

is always odd because

b^3

and

b

have matching parities (their sum is always even), so adding 7 yields an odd number. Since the product of two odd integers is always odd, the expression is never even, yielding exactly 0 ordered pairs.

Adım Adım Çözüm

1

Examine the parity of the factor

a^2 + a + 1

a^2 + a + 1

is odd for all integer values of

a

The expression

a^2 + a = a(a + 1)

represents the product of two consecutive integers, which is always even. Adding 1 to an even integer results in an odd integer.

2

Examine the parity of the factor

b^3 + b + 7

b^3 + b + 7

is odd for all integer values of

b

Since

b^3

and

b

always share the same parity, their sum

b^3 + b

is always even. Adding 7 to an even integer results in an odd integer.

3

Determine the overall parity of the product

The product

(a^2 + a + 1)(b^3 + b + 7)

is odd for all inputs

The product of two odd integers is strictly an odd integer.

4

Count the number of pairs satisfying the even condition

0 pairs

Because the product is never even, zero pairs satisfy the requirement.

Anahtar Kavram

Parity rules of consecutive integer products and polynomial expressions

Tahmini Süre:1m 30s

Soru 174Soru →

If $x$ , $y$ , and $z$ are integers such that $x + y + z = 0$ and $x^3 y - z$ is an odd integer, which of the following expressions MUST be an even integer?

x y z

x + y

y + z

x - y

x + z

Cevabı ve açıklamayı göster

Cevap: $x y z$

Cevap

The expression representing the product of the three integers,

x y z

, must always be an even integer.

The correct answer is the product of the three variables because logical parity analysis reveals that it is impossible for all three integers to be even, but in all valid cases at least one of the three variables must be an even integer (or zero). A product containing at least one even factor is always even.

Adım Adım Çözüm

1

Express

z

in terms of

x

and

y

using the given sum equality.

z = -(x + y)

.

Since

x + y + z = 0

, substituting

z = -(x + y)

allows evaluation of the given parity condition in terms of

x

and

y

only.

2

Substitute

z = -(x + y)

into the given expression

x^3 y - z

and analyze its parity.

x^3 y - z = x^3 y + x + y

. Since

x^3

and

x

have the same parity for any integer

x

,

x^3 y + x + y

has the same parity as

x y + x + y

.

The parity of an integer raised to a positive integer power is identical to the parity of the base integer.

3

Evaluate the four possible parity combinations for

(x, y)

.

If both

x

and

y

are even, then

x y + x + y = \text{even} + \text{even} + \text{even} = \text{even}

, which contradicts the condition that

x^3 y - z

is odd. Thus,

x

and

y

cannot both be even.

Eliminating the case where both

x

and

y

are even implies that at least one of

x

or

y

must be odd.

4

Determine the parity of

x y z

in all remaining valid cases.

In Case 1 (

x

odd,

y

odd),

z = -(x + y) = -(\text{odd} + \text{odd}) = \text{even}

, so

z

is even and

x y z

is even. In Case 2 (

x

even,

y

odd),

x

is even so

x y z

is even. In Case 3 (

x

odd,

y

even),

y

is even so

x y z

is even. Therefore,

x y z

is even in every valid case.

The product of integers is even whenever at least one factor in the product is an even integer (including zero).

Anahtar Kavram

Parity rules of addition, multiplication, exponentiation, and the property that zero is an even integer.

Tahmini Süre:2m 0s

Soru 175Soru →

For two positive integers $a$ and $b$ , the greatest common divisor is $15$ and the least common multiple is $180$ . If $a = 45$ , what is the value of $b$ ?

12

15

60

135

180

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

60

For any two positive integers

a

and

b

, the product of their greatest common divisor and their least common multiple equals the product of the two numbers, expressed as

\text{GCD}(a, b) \times \text{LCM}(a, b) = a \times b

. Plugging in the given values yields

15 \times 180 = 45 \times b

. Dividing

2700

by

45

results in

60

.

Adım Adım Çözüm

1

Recall the fundamental relation between GCD and LCM for two positive integers.

\text{GCD}(a, b) \times \text{LCM}(a, b) = a \times b

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

2

Substitute the given values into the formula.

15 \times 180 = 45 \times b

We are given

\text{GCD}(a,b) = 15

,

\text{LCM}(a,b) = 180

, and

a = 45

.

3

Solve for

b

.

b = \frac{15 \times 180}{45} = \frac{180}{3} = 60

Dividing both sides by

45

simplifies to

180 \div 3 = 60

.

Anahtar Kavram

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

Tahmini Süre:1m 30s

Soru 176Soru →

If $m$ is a positive integer such that $m!$ is divisible by $3^{10}$ but not by $3^{11}$ , and $(m+5)!$ is divisible by $3^{13}$ but not by $3^{14}$ , what is the total number of distinct positive factors of $m$ ?

6

8

10

12

14

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The total number of distinct positive factors of

m

is 8.

Using Legendre's formula, the highest power of 3 dividing

m!

is 10 when

m

is 24, 25, or 26. Evaluating

(m+5)!

,

29!

(from

m=24

) has

E_3(29!) = 9 + 3 + 1 = 13

, whereas

30!

(from

m=25

) has

E_3(30!) = 10 + 3 + 1 = 14

. Therefore,

m

must equal 24. Since

24 = 2^3 \times 3^1

, its number of distinct positive factors is

(3+1)(1+1) = 8

.

Adım Adım Çözüm

1

Apply Legendre's formula to find the highest power of 3 dividing

m!

.

The exponent of 3 in

m!

is given by

E_3(m!) = \lfloor \frac{m}{3} \rfloor + \lfloor \frac{m}{9} \rfloor + \lfloor \frac{m}{27} \rfloor + \dots = 10

.

Legendre's formula determines the exact power of a prime

p

in

n!

.

2

Test integer values for

m

to satisfy

E_3(m!) = 10

.

For

m=24

:

E_3(24!) = \lfloor \frac{24}{3} \rfloor + \lfloor \frac{24}{9} \rfloor = 8 + 2 = 10

. Thus,

m \in \{24, 25, 26\}

.

Integers 24, 25, and 26 each contain 10 factors of 3 in their factorial products.

3

Apply the second condition

E_3((m+5)!) = 13

to identify the unique value of

m

.

If

m=24

,

m+5=29 \implies E_3(29!) = \lfloor \frac{29}{3} \rfloor + \lfloor \frac{29}{9} \rfloor + \lfloor \frac{29}{27} \rfloor = 9 + 3 + 1 = 13

. For

m=25

or

m=26

,

m+5 \ge 30

, giving

E_3(30!) = 10+3+1 = 14

. Hence,

m = 24

.

Only

m=24

satisfies both power constraints simultaneously.

4

Calculate the number of positive factors of

m = 24

.

24 = 2^3 \times 3^1

. Total positive factors

= (3+1)(1+1) = 4 \times 2 = 8

.

The number of factors is found by adding 1 to each prime exponent and multiplying.

Anahtar Kavram

Legendre's Formula and Prime Factorization Divisor Counting

Tahmini Süre:2m 0s

Soru 177Soru →

For two positive integers $m$ and $n$ , their prime factorizations are given by $m = 2^3 \times 3^a \times 5^b$ and $n = 2^c \times 3^3 \times 5^1$ , where $a$ , $b$ , and $c$ are positive integer exponents. If the greatest common divisor of $m$ and $n$ is $\text{GCD}(m, n) = 360$ and their least common multiple is $\text{LCM}(m, n) = 108,000$ , what is the value of $a + b + c$ ?

7

8

9

10

11

Cevabı ve açıklamayı göster

Cevap: $10$

Cevap

The correct value of

a + b + c

is

10

.

First, find the prime factorizations of

360

and

108,000

:
-

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

-

108,000 = 2^5 \times 3^3 \times 5^3

Using the properties of GCD and LCM:
- For base

2

:

\text{GCD}

power is

3

and

\text{LCM}

power is

5

. Since

m

has

2^3

,

n

must have

2^c

where

c = 5

.
- For base

3

:

\text{GCD}

power is

2

and

\text{LCM}

power is

3

. Since

n

has

3^3

,

m

must have

3^a

where

a = 2

.
- For base

5

:

\text{GCD}

power is

1

and

\text{LCM}

power is

3

. Since

n

has

5^1

,

m

must have

5^b

where

b = 3

.

Summing these values gives

a + b + c = 2 + 3 + 5 = 10

.

Adım Adım Çözüm

1

Express GCD and LCM in prime factorized 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

.

Converting given numbers into prime powers allows direct comparison with the exponents of

m

and

n

.

2

Apply prime exponent min/max rules to determine

a

,

b

, and

c

.

For prime

2

:

\max(3, c) = 5 \implies c = 5

. For prime

3

:

\min(a, 3) = 2 \implies a = 2

. For prime

5

:

\max(b, 1) = 3 \implies b = 3

.

\text{GCD}

uses the minimum exponent for each prime factor, while

\text{LCM}

uses the maximum exponent.

3

Sum the derived exponents

a

,

b

, and

c

.

a + b + c = 2 + 3 + 5 = 10

.

The question asks for the sum

a + b + c

.

Anahtar Kavram

Prime Factor Exponent Rule for GCD and LCM

Alternatif Yöntem

Use the product formula

\text{GCD}(m, n) \times \text{LCM}(m, n) = m \times n

. Multiplying gives

360 \times 108,000 = 38,880,000 = 2^8 \times 3^5 \times 5^4

. Multiplying

m \times n = (2^3 \times 3^a \times 5^b) \times (2^c \times 3^3 \times 5^1) = 2^{3+c} \times 3^{a+3} \times 5^{b+1}

. Equating powers:

3+c=8 \implies c=5

,

a+3=5 \implies a=2

,

b+1=4 \implies b=3

. Thus

a+b+c = 2+3+5 = 10

.

Tahmini Süre:2m 0s

Soru 178Soru →

If $k$ is a positive integer that is divisible by both $6$ and $10$ , which of the following expressions MUST yield an integer?

\dfrac{k}{12}

\dfrac{k}{20}

\dfrac{k}{15}

\dfrac{k}{45}

\dfrac{k}{60}

Cevabı ve açıklamayı göster

Cevap: $\dfrac{k}{15}$

Cevap

The expression

\dfrac{k}{15}

MUST yield an integer.

Since

k

is divisible by both

6

and

10

,

k

must be a multiple of

\text{LCM}(6, 10) = 30

. Therefore,

k = 30m

for some positive integer

m

. Dividing

k

by

15

gives

\dfrac{30m}{15} = 2m

, which is guaranteed to be an integer for any positive integer

m

.

Adım Adım Çözüm

1

Determine the least common multiple (LCM) of

6

and

10

.

\text{LCM}(6, 10) = 30

.

Since

k

is divisible by both

6

and

10

,

k

must be a multiple of their LCM.

2

Express

k

in terms of its smallest possible base multiple.

k = 30m

for some positive integer

m

.

This represents all possible integer values for

k

.

3

Determine which choice has a denominator that divides

30

.

Since

15

divides

30

,

\dfrac{k}{15} = \dfrac{30m}{15} = 2m

, which is always an integer.

Any divisor of

30

will divide

k

for all valid values of

k

.

Anahtar Kavram

Least Common Multiple (LCM) and Divisibility Properties

Tahmini Süre:1m 0s

Soru 179Soru →

What is the number of distinct prime factors of the greatest common divisor (GCD) of $60$ and $90$ ?

2

3

4

5

30

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

The prime factorization of 60 is

2^2 \times 3 \times 5

and the prime factorization of 90 is

2 \times 3^2 \times 5

. The GCD is obtained by taking the lowest exponent for each common prime base:

\text{GCD}(60, 90) = 2^1 \times 3^1 \times 5^1 = 30

. The distinct prime factors of 30 are 2, 3, and 5, yielding a count of 3.

Adım Adım Çözüm

1

Find the prime factorizations of 60 and 90.

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

and

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

Expressing numbers as products of prime factors is the standard method for finding their GCD.

2

Determine the GCD by taking the lowest power of each common prime factor.

\text{GCD}(60, 90) = 2^{\min(2,1)} \times 3^{\min(1,2)} \times 5^{\min(1,1)} = 2^1 \times 3^1 \times 5^1 = 30

The GCD takes the common prime bases raised to their minimum respective exponents.

3

Identify and count the distinct prime factors of 30.

The distinct prime factors of 30 are 2, 3, and 5, giving a total of 3 prime factors.

By prime definition, 1 is excluded from the list of prime factors.

Anahtar Kavram

Greatest Common Divisor (GCD) and Prime Factorization

Tahmini Süre:1m 0s

Soru 180Soru →

For two positive integers $x$ and $y$ , their greatest common divisor is $\gcd(x, y) = 12$ and their least common multiple is $\text{lcm}(x, y) = 10,800$ . If $x > y$ , $x$ is not divisible by $9$ , and $y$ is not divisible by $25$ , what is the value of $x + y$ ?

732

972

1,308

1,356

1,500

Cevabı ve açıklamayı göster

Cevap: 1,308

Cevap

The sum of the two integers

x + y

is

1,308

.

By prime factorizing the given GCD (

2^2 \cdot 3^1

) and LCM (

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

), the minimum and maximum exponents for each prime factor are determined. The constraint that

x

is not divisible by

9

forces the exponent of

3

in

x

to be

1

, so the exponent of

3

in

y

must be

3

. The constraint that

y

is not divisible by

25

forces the exponent of

5

in

y

to be

0

, so the exponent of

5

in

x

must be

2

. Finally, the condition

x > y

requires the exponent of

2

in

x

to be

4

and in

y

to be

2

. Thus,

x = 1,200

and

y = 108

, giving

x + y = 1,308

.

Adım Adım Çözüm

1

Express the GCD and LCM in their prime factorizations.

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

and

\text{lcm}(x, y) = 10,800 = 2^4 \cdot 3^3 \cdot 5^2

.

Decomposing into prime factors allows direct determination of the minimum and maximum powers of each prime factor present in

x

and

y

.

2

Determine the prime exponents for

x

and

y

using the given divisibility constraints.

For prime factor

3

:

\min(b_1, b_2) = 1

and

\max(b_1, b_2) = 3

. Since

x

is not divisible by

9 = 3^2

,

b_1 = 1

, which forces

b_2 = 3

. For prime factor

5

:

\min(c_1, c_2) = 0

and

\max(c_1, c_2) = 2

. Since

y

is not divisible by

25 = 5^2

,

c_2 < 2

, which forces

c_1 = 2

and

c_2 = 0

.

The GCD gives the minimum exponent of each prime factor across both numbers, while the LCM gives the maximum exponent.

3

Apply the condition

x > y

to assign the powers of

2

.

The exponents for

2

must be

\{2, 4\}

. If

x

takes exponent

4

and

y

takes exponent

2

,

x = 2^4 \cdot 3^1 \cdot 5^2 = 1,200

and

y = 2^2 \cdot 3^3 \cdot 5^0 = 108

. Since

1,200 > 108

, this satisfies

x > y

.

If

x

took exponent

2

and

y

took exponent

4

, then

x = 300

and

y = 432

, violating

x > y

.

4

Calculate

x + y

.

x + y = 1,200 + 108 = 1,308

.

Adding the uniquely determined values of

x

and

y

gives the final requested sum.

Anahtar Kavram

Prime factor exponent min/max rules for GCD and LCM:

\gcd(x,y)

takes the minimum exponent of each prime factor, while

\text{lcm}(x,y)

takes the maximum exponent.

Tahmini Süre:2m 30s

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Alternatif Yöntem

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