Number Properties and Arithmetic Alıştırma Soruları — GMAT — Sayfa 6

Soru 101Soru →

If $n$ is a positive integer greater than $20$ such that $\sqrt{2^{28} + 2^{20} + 2^n}$ is an integer, what is the least possible value of $n$ ?

22

24

25

26

34

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Cevap: 25

Cevap

The least possible value of

n

is 25.

The expression inside the radical must be a perfect square. Matching

2^{28} + 2^{20} + 2^n

to

(2^{14} + 2^{10})^2 = 2^{28} + 2 \cdot 2^{14} \cdot 2^{10} + 2^{20} = 2^{28} + 2^{25} + 2^{20}

reveals that

2^n = 2^{25}

, so

n = 25

. Since 25 is greater than 20 and smaller than the other valid solution

n = 34

, it is the least possible value.

Adım Adım Çözüm

1

Set up the perfect square structure for the radical expression.

For

\sqrt{2^{28} + 2^{20} + 2^n}

to be an integer, the expression

2^{28} + 2^{20} + 2^n

must equal

(2^a + 2^b)^2 = 2^{2a} + 2^{a+b+1} + 2^{2b}

for some positive integers

a > b

.

Expanding a binomial power of 2 generates three power-of-2 terms matching the three terms in the expression.

2

Analyze Case 1 where

2^{28}

corresponds to the leading term

2^{2a}

.

2a = 28 \implies a = 14

. The remaining terms

2^{20}

and

2^n

must correspond to

2^{2b}

and

2^{a+b+1}

.

Matching highest powers establishes the value of the parameter

a

.

3

Evaluate sub-cases for matching

2^{20}

.

Subcase 1: If

2b = 20 \implies b = 10

, then the middle term exponent is

a + b + 1 = 14 + 10 + 1 = 25

. Thus

n = 25

.
Subcase 2: If

a + b + 1 = 20 \implies 14 + b + 1 = 20 \implies b = 5

, then

2b = 10

. Thus

n = 10

.

Checking both term assignments determines possible integer values for

n

.

4

Analyze Case 2 where

2^{20}

corresponds to

2^{2b}

and

2^{28}

is the middle term.

If

2b = 20 \implies b = 10

and

a + b + 1 = 28 \implies a + 11 = 28 \implies a = 17

, then

2a = 34

. Thus

n = 34

.

Checking alternative middle term assignments finds all valid values for

n

.

5

Filter by constraint

n > 20

and select the minimum.

The valid values satisfying

n > 20

are

n = 25

and

n = 34

. The least possible value is 25.

The question asks specifically for the least possible value greater than 20.

Anahtar Kavram

Perfect Square Trinomial Expansion with Exponent Rules

Tahmini Süre:2m 0s

Soru 102Soru →

If $n$ is a positive integer with the prime factorization $n = 2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers, and $n$ has exactly $40$ positive integer divisors, what is the minimum possible value of $a + b + c$ ?

7

8

9

11

14

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

The number of positive integer divisors of a number with prime factorization

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

is given by

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

. Setting this product equal to 40 and factoring 40 into three integers greater than 1 yields two distinct sets of factors:

\{2, 2, 10\}

and

\{2, 4, 5\}

. The sum of the factors in

\{2, 4, 5\}

is

2 + 4 + 5 = 11

, which is smaller than

2 + 2 + 10 = 14

. Subtracting 1 from each factor gives the exponents

a, b, c

as

1, 3, 4

, making the minimum sum

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

.

Adım Adım Çözüm

1

Express the number of divisors of

n

in terms of its exponents.

For

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

, the total number of positive divisors is

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

.

The total divisor count formula requires taking the product of each prime factor's exponent increased by 1.

2

Find all combinations of three integers greater than 1 whose product is 40.

Since

a, b, c \ge 1

, we have

a+1, b+1, c+1 \ge 2

. The integer factorizations of 40 into 3 factors

\ge 2

are

(2, 2, 10)

and

(2, 4, 5)

.

We must break down 40 into 3 integer components corresponding to

(a+1)

,

(b+1)

, and

(c+1)

.

3

Calculate the sum

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

for each combination to find the minimum sum.

For

(2, 2, 10)

, the sum is

2 + 2 + 10 = 14

. For

(2, 4, 5)

, the sum is

2 + 4 + 5 = 11

. The minimum sum of factors is 11.

Minimizing

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

directly minimizes

a+b+c

.

4

Subtract 3 to obtain the minimum possible value of

a+b+c

.

a+b+c = 11 - 3 = 8

.

Since

(a+1)+(b+1)+(c+1) = (a+b+c) + 3 = 11

, subtracting 3 yields

a+b+c = 8

.

Anahtar Kavram

Divisor Count Formula from Prime Factorization

Soru 103Soru →

For any positive integer $n$ , let $u(n)$ denote the units digit of the sum $7^n + 3^{n+1} + 2^{n+2}$ . What is the remainder when the sum $u(1) + u(2) + u(3) + \dots + u(100)$ is divided by 7?

6

5

2

0

1

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The remainder when the sum is divided by 7 is 6.

Evaluating the units digit

u(n)

for

n = 1, 2, 3, 4

gives

u(1) = 4

,

u(2) = 2

,

u(3) = 6

, and

u(4) = 0

. This pattern of 4 values repeats indefinitely. The sum of one full cycle of 4 terms is

4 + 2 + 6 + 0 = 12

. Since there are 100 terms in total, there are

100 / 4 = 25

full cycles. The total sum is

25 \times 12 = 300

. Dividing 300 by 7 gives a quotient of 42 with a remainder of 6. Thus, the correct choice is 6.

Adım Adım Çözüm

1

Determine the units digit cyclicity for each power term.

The units digits of

7^n

follow the 4-term cycle [7, 9, 3, 1]. The units digits of

3^{n+1}

follow [9, 7, 1, 3]. The units digits of

2^{n+2}

follow [8, 4, 2, 6].

Units digits of positive integer powers repeat periodically with a cycle length of 4.

2

Calculate

u(n)

for the first 4 terms to identify the repeating pattern of

u(n)

.

For

n=1

:

7+9+8 = 24 \implies u(1)=4

. For

n=2

:

9+7+4 = 20 \implies u(2)=0

. For

n=3

:

3+1+2 = 6 \implies u(3)=6

. For

n=4

:

1+3+6 = 10 \implies u(4)=0

. The sequence of

u(n)

is

[4, 0, 6, 0]

.

Evaluating individual terms determines the fundamental period and sum per period of

u(n)

.

3

Sum the values of

u(n)

over one complete cycle of 4 terms.

Sum of one cycle =

4 + 0 + 6 + 0 = 10

.

Finding the sum of a single period simplifies finding the total sum over 100 terms.

4

Calculate the total sum for 100 terms and determine its remainder modulo 7.

Since 100 terms contain

100 / 4 = 25

complete cycles, the total sum

S = 25 \times 10 = 250

. Dividing 250 by 7 gives

250 = 7 \times 35 + 5

? Wait:

u(2) = (9+7+4)

units digit is 0. Let's verify:

7^2=49

(9),

3^3=27

(7),

2^4=16

(6).

9+7+6=22 \implies u(2)=2

. Let's re-verify:

u(1)=4, u(2)=2, u(3)=6, u(4)=0

. Sum per cycle =

4+2+6+0 = 12

. Total sum

S = 25 \times 12 = 300

.

300 = 7 \times 42 + 6

. Remainder is 6.

Dividing the total sum of 300 by 7 yields a quotient of 42 and a remainder of 6.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic Sums

Tahmini Süre:1m 30s

Soru 104Soru →

When a positive integer $n$ is divided by $4$ , the remainder is $3$ . What is the units digit of the expression $M = 3^{n+1} + 8^{2n+1} - 7^{n+2}$ ?

2

4

6

8

0

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The units digit of the expression is 6.

Since

n \equiv 3 \pmod{4}

, we substitute into each exponent:

n+1 \equiv 0 \pmod{4}

,

2n+1 \equiv 3 \pmod{4}

, and

n+2 \equiv 1 \pmod{4}

. The units digit cyclicity patterns give

3^4 \to 1

,

8^3 \to 2

, and

7^1 \to 7

. Evaluating the combined units digits yields

1 + 2 - 7 = -4

. Adding 10 to obtain a valid units digit gives

6

.

Adım Adım Çözüm

1

Express the integer

n

in terms of modular arithmetic

n \equiv 3 \pmod{4}

, which means

n = 4k + 3

for some non-negative integer

k

.

Units digits of powers repeat in cycles of 4, so exponent behavior depends on exponents modulo 4.

2

Determine the units digit of

3^{n+1}

Exponent

n+1 = (4k+3)+1 = 4k+4 \equiv 0 \pmod{4}

. The units digit pattern for 3 is

3, 9, 7, 1

. Since exponent mod 4 is 0 (or 4), the units digit is

1

.

The cyclicity pattern for powers of 3 has period length 4.

3

Determine the units digit of

8^{2n+1}

Exponent

2n+1 = 2(4k+3)+1 = 8k+7 \equiv 3 \pmod{4}

. The units digit pattern for 8 is

8, 4, 2, 6

. For exponent mod 4 equal to 3, the units digit is

2

.

The cyclicity pattern for powers of 8 has period length 4.

4

Determine the units digit of

7^{n+2}

Exponent

n+2 = (4k+3)+2 = 4k+5 \equiv 1 \pmod{4}

. The units digit pattern for 7 is

7, 9, 3, 1

. For exponent mod 4 equal to 1, the units digit is

7

.

The cyclicity pattern for powers of 7 has period length 4.

5

Combine the units digits and resolve negative intermediate values

1 + 2 - 7 = -4 \equiv 6 \pmod{10}

. The units digit is

6

.

Units digits must be non-negative integers from 0 to 9, so a negative result requires adding 10.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic

Tahmini Süre:2m 0s

Soru 105Soru →

If $x$ and $y$ are positive integers such that $3^x \cdot 5^y = 675$ , what is the value of $2^{x+y} \cdot 3^{x-y}$ ?

32

36

96

192

288

Cevabı ve açıklamayı göster

Cevap: 96

Cevap

96

Prime factorizing

675

gives

3^3 \cdot 5^2

. Equating the exponents yields

x = 3

and

y = 2

. Substituting these into

2^{x+y} \cdot 3^{x-y}

yields

2^5 \cdot 3^1 = 32 \cdot 3 = 96

.

Adım Adım Çözüm

1

Find the prime factorization of 675.

675 = 27 \times 25 = 3^3 \cdot 5^2

Decomposing 675 into prime factors allows matching the exponents of prime bases 3 and 5.

2

Equate exponents of matching prime bases to determine

x

and

y

.

x = 3

and

y = 2

Since 3 and 5 are prime numbers, the prime factorization of 675 is unique.

3

Evaluate the target expression

2^{x+y} \cdot 3^{x-y}

.

2^{3+2} \cdot 3^{3-2} = 2^5 \cdot 3^1 = 32 \cdot 3 = 96

Substitute the determined values of

x

and

y

into the target power expression.

Anahtar Kavram

Prime Factorization and Exponent Properties

Tahmini Süre:1m 30s

Soru 106Soru →

If $n = 3^x \cdot 5^y \cdot 7^z$ , where $x$ , $y$ , and $z$ are positive integers, and $n$ has exactly $24$ positive divisors, what is the minimum possible value of $x + y + z$ ?

5

6

7

9

21

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6

The total number of positive divisors of

n = 3^x \cdot 5^y \cdot 7^z

is

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

. Because

x, y, z

are positive integers, each factor term must be at least 2. Factoring 24 into three integer factors each

\ge 2

yields two possibilities:

(2, 2, 6)

and

(2, 3, 4)

. The sum of the exponents

x+y+z

equals

(x+1)+(y+1)+(z+1) - 3

. For

(2, 2, 6)

, the sum is

2+2+6-3 = 7

. For

(2, 3, 4)

, the sum is

2+3+4-3 = 6

. The minimum possible value is 6.

Adım Adım Çözüm

1

Express the number of divisors using the prime factorization formula.

For

n = 3^x \cdot 5^y \cdot 7^z

, the number of positive divisors is

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

.

The total number of divisors of a prime-factored integer is found by taking the product of each prime factor's exponent plus one.

2

Determine the constraints on the factor terms.

Since

x, y, z

are positive integers (

x, y, z \ge 1

), each term

(x+1), (y+1), (z+1) \ge 2

.

Exponents must be at least 1, so each factor in the product must be at least 2.

3

Find all valid sets of 3 integer factors of 24 that are all

\ge 2

.

The possible triples

(a, b, c)

such that

a \cdot b \cdot c = 24

and

a, b, c \ge 2

are

(2, 2, 6)

and

(2, 3, 4)

.

We factor 24 into three integer components, each at least 2.

4

Calculate

x+y+z

for each factor triple to find the minimum.

For

(2, 2, 6)

:

x+1=2, y+1=2, z+1=6 \implies x+y+z = 1 + 1 + 5 = 7

.
For

(2, 3, 4)

:

x+1=2, y+1=3, z+1=4 \implies x+y+z = 1 + 2 + 3 = 6

.
The minimum value is 6.

Comparing the sums of exponents shows that the factor triple

(2, 3, 4)

minimizes

x+y+z

.

Anahtar Kavram

Divisors from Prime Factorization

Tahmini Süre:2m 0s

Soru 107Soru →

When the integer $N = 4^{43} + 9^{43}$ is divided by $7$ , what is the remainder?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The remainder when

N = 4^{43} + 9^{43}

is divided by

7

is

6

.

Reducing

9 \pmod 7

yields

2

, allowing

N

to be expressed as

2^{86} + 2^{43} \pmod 7

. The powers of

2 \pmod 7

repeat every 3 powers

(2, 4, 1)

. Reducing the exponents

86

and

43

modulo 3 gives remainders of

2

and

1

, corresponding to values of

4

and

2

. Their sum

4 + 2 = 6

is the final remainder.

Adım Adım Çözüm

1

Simplify the base modulo 7 and express in powers of 2

9 \equiv 2 \pmod 7

, so

N = 4^{43} + 9^{43} \equiv 2^{86} + 2^{43} \pmod 7

.

Reducing bases modulo 7 simplifies calculating large exponents.

2

Determine the remainder cyclicity of powers of 2 modulo 7

The cycle length is 3 with pattern

(2, 4, 1)

, because

2^1 \equiv 2

,

2^2 \equiv 4

, and

2^3 \equiv 1 \pmod 7

.

Powers of integers modulo a divisor repeat periodically.

3

Evaluate each term using the exponent modulo the cycle length

86 \equiv 2 \pmod 3 \implies 2^{86} \equiv 2^2 = 4 \pmod 7

, and

43 \equiv 1 \pmod 3 \implies 2^{43} \equiv 2^1 = 2 \pmod 7

.

The position in the cyclicity sequence is dictated by the exponent modulo the period length.

4

Add the individual remainders

4 + 2 = 6 \pmod 7

.

The remainder of a sum equals the sum of the individual remainders.

Anahtar Kavram

Modular arithmetic cyclicity of powers and addition of remainders

Soru 108Soru →

If $4^x + 4^{x+1} + 4^{x+2} = 84 \cdot 2^{10}$ , what is the value of $x$ ?

5

6

7

10

12

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6

Factoring out

4^x

gives

4^x(1 + 4 + 16) = 21 \cdot 4^x

. Dividing both sides of

21 \cdot 4^x = 84 \cdot 2^{10}

by 21 yields

4^x = 4 \cdot 2^{10}

. Converting terms to base 2 produces

2^{2x} = 2^2 \cdot 2^{10} = 2^{12}

. Setting exponents equal gives

2x = 12

, which yields

x = 6

.

Adım Adım Çözüm

1

Factor out the common term

4^x

from the left side of the equation.

4^x(1 + 4^1 + 4^2) = 4^x(1 + 4 + 16) = 21 \cdot 4^x

Expressions adding powers with consecutive exponents can be simplified by factoring out the lowest power.

2

Substitute the factored expression into the equation and solve for

4^x

.

21 \cdot 4^x = 84 \cdot 2^{10} \implies 4^x = \frac{84}{21} \cdot 2^{10} = 4 \cdot 2^{10}

Dividing both sides by 21 isolates the exponential term on the left.

3

Convert both sides of the equation to powers of base 2.

(2^2)^x = 2^2 \cdot 2^{10} \implies 2^{2x} = 2^{12}

Expressing both sides with a common base enables equating exponents.

4

Equate exponents and solve for

x

.

2x = 12 \implies x = 6

When bases are equal, the powers must be equal.

Anahtar Kavram

Factoring Exponential Expressions and Base Conversion

Tahmini Süre:1m 30s

Soru 109Soru →

If $x$ , $y$ , and $z$ are integers satisfying $x < y < 0 < z$ , $x + y + z = 0$ , and $x y z = 160$ , what is the value of $z - x$ ?

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

The value of

z - x

is

18

.

By defining positive variables

a = -x

and

b = -y

, the given inequality

x < y < 0

implies

a > b > 0

. Since

x + y + z = 0

,

z = a + b

. The product condition

xyz = 160

translates to

ab(a+b) = 160

. Testing positive integer values reveals that

b = 2

and

a = 8

is the unique solution satisfying

a > b > 0

. This gives

x = -8

,

y = -2

, and

z = 10

, making

z - x = 10 - (-8) = 18

.

Adım Adım Çözüm

1

Set up positive variables for the negative integers

Let

a = -x

and

b = -y

, where

a

and

b

are positive integers with

a > b > 0

. From

x + y + z = 0

, we get

-a - b + z = 0

, so

z = a + b

.

Converting negative integers to positive magnitude variables simplifies sign analysis in products and sums.

2

Substitute variables into the product equation

Substituting

x = -a

,

y = -b

, and

z = a + b

into

x y z = 160

gives

(-a)(-b)(a + b) = 160

, which simplifies to

a b (a + b) = 160

.

Multiplying two negative numbers yields a positive product, simplifying the product expression.

3

Solve for positive integer pairs

(a, b)

with

a > b

Testing integer values of

b

:
- If

b = 1

,

a(a+1) = 160

(no integer solution as

12 \times 13 = 156

).
- If

b = 2

,

2a(a+2) = 160 \implies a(a+2) = 80 \implies a = 8

.
- If

b = 3

,

3a(a+3) = 160 \implies a(a+3) = 53.33

(not an integer).
- If

b = 4

,

4a(a+4) = 160 \implies a(a+4) = 40

(no integer solution).
- If

b \ge 5

,

a > b \implies a \ge 6

, so

a b (a+b) \ge 6 \times 5 \times 11 = 330 > 160

.
Thus, the only valid integer pair is

a = 8

and

b = 2

.

Systematically checking integer factors under inequality constraints guarantees finding all unique solutions.

4

Calculate the target expression

z - x

Since

a = 8

and

b = 2

, we have

x = -8

,

y = -2

, and

z = 8 + 2 = 10

. Therefore,

z - x = 10 - (-8) = 18

.

Evaluating

z - x

using the identified values completes the solution.

Anahtar Kavram

Positive and Negative Number Properties and Inequality Constraints

Soru 110Soru →

If $k$ is a positive integer that is divisible by both 8 and 9, and $k$ has exactly 12 positive integer divisors, what is the value of $k$ ?

36

72

108

144

216

Cevabı ve açıklamayı göster

Cevap: 72

Cevap

72

The value 72 has the prime factorization

2^3 \cdot 3^2

. It is divisible by

8 = 2^3

and

9 = 3^2

, and its number of positive integer divisors is

(3+1)(2+1) = 12

, satisfying all conditions of the problem.

Adım Adım Çözüm

1

Express the divisibility conditions in terms of prime factors.

Since

k

is divisible by

8 = 2^3

and

9 = 3^2

, the prime factorization of

k

must be of the form

k = 2^a \cdot 3^b \cdot p_1^{c_1} \cdots

, where

a \geq 3

and

b \geq 2

.

Any multiple of 8 must contain at least three factors of 2, and any multiple of 9 must contain at least two factors of 3.

2

Write the formula for the total number of positive integer divisors.

The number of positive integer divisors is given by

(a+1)(b+1)(c_1+1)\cdots = 12

.

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

3

Determine the values of the exponents

a

and

b

.

Since

a \geq 3

, we have

a+1 \geq 4

. Since

b \geq 2

, we have

b+1 \geq 3

. Therefore,

(a+1)(b+1) \geq 4 \times 3 = 12

.

Because the product

(a+1)(b+1)

is already at least 12, there can be no additional prime factors, so

a+1 = 4 \implies a = 3

and

b+1 = 3 \implies b = 2

.

4

Calculate the value of

k

.

k = 2^3 \cdot 3^2 = 8 \cdot 9 = 72

.

Evaluating

2^3 \cdot 3^2

gives the unique positive integer satisfying all given conditions.

Anahtar Kavram

Determining integer values using prime factorization constraints and total divisor count rules.

Tahmini Süre:2m 0s

Soru 111Soru →

Set $S$ consists of $n$ consecutive integers. The sum of all elements in set $S$ is equal to $0$ . If the product of the smallest element in set $S$ and the total number of elements $n$ is $-300$ , how many positive integers are in set $S$ ?

11

13

12

24

25

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12

For a set of consecutive integers to sum to zero, the elements must be symmetric about zero. Therefore, the set has an odd number of terms

n = 2k + 1

ranging from

-k

to

k

. Setting up the product of the smallest term

(-k)

and the term count

(2k + 1)

gives

(-k)(2k + 1) = -300

, which simplifies to

2k^2 + k - 300 = 0

. Factoring this quadratic yields

(2k + 25)(k - 12) = 0

, so

k = 12

. The positive integers in the set are

1

through

12

, totaling 12 elements.

Adım Adım Çözüm

1

Determine the symmetry and structure of set

S

using its sum.

The median of set

S

is

0

, and the set contains an odd number of terms

n = 2k + 1

centered at

0

, expressed as

\{-k, -(k-1), \dots, 0, \dots, k-1, k\}

.

For a set of consecutive integers to sum to

0

, the terms must be symmetric around

0

. Since

0

is an element of the set,

n

must be odd.

2

Set up the quadratic equation using the given product.

The smallest element is

-k

and the total number of elements is

n = 2k + 1

, yielding

(-k)(2k + 1) = -300

, which simplifies to

2k^2 + k - 300 = 0

.

The question specifies that the product of the smallest element and the number of elements is

-300

.

3

Solve the quadratic equation for

k

.

Factoring

(2k + 25)(k - 12) = 0

gives

k = 12

as the only positive integer solution.

Since

k

represents a count of elements strictly above zero,

k

must be a positive integer.

4

Count the number of positive integers in set

S

.

The positive integers are

1, 2, \dots, 12

, which gives a total of

12

positive integers.

Zero is neither positive nor negative, so only the integers from

1

to

k

(

12

) are counted.

Anahtar Kavram

Symmetry and median property of consecutive integers centered at zero.

Soru 112Soru →

If $x$ is a positive integer such that $\sqrt{\frac{8^{10} + 4^{10}}{8^4 + 4^{11}}} = 2^x$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

Converting all terms to base 2 yields

\sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}

. Factoring out

2^{20}

in the numerator and

2^{12}

in the denominator leaves

\sqrt{\frac{2^{20}(2^{10}+1)}{2^{12}(2^{10}+1)}} = \sqrt{2^8} = 2^4

. Thus

2^x = 2^4

, so

x = 4

.

Adım Adım Çözüm

1

Express all terms with prime base 2 using exponent rules.

8^{10} = 2^{30}

,

4^{10} = 2^{20}

,

8^4 = 2^{12}

, and

4^{11} = 2^{22}

.

Converting non-prime bases to a common base enables exponent simplification and factoring.

2

Factor the numerator and denominator by pulling out the lowest power of 2 in each.

Numerator:

2^{20}(2^{10} + 1)

; Denominator:

2^{12}(1 + 2^{10})

.

Factoring isolates common terms in sums of powers so they can be canceled.

3

Cancel the identical factor

(2^{10} + 1)

and compute the radical expression.

\sqrt{\frac{2^{20}}{2^{12}}} = \sqrt{2^8} = 2^4 = 16.

Applying quotient rule for exponents

\frac{2^{20}}{2^{12}} = 2^{8}

and radical rule

\sqrt{2^8} = (2^8)^{1/2} = 2^4

.

4

Set

2^4

equal to

2^x

to solve for

x

.

x = 4.

Since the bases are identical and positive, the exponents must be equal.

Anahtar Kavram

Exponents, Roots, and Powers of Integers

Soru 113Soru →

A positive integer $n$ is divisible by 12 and has exactly 15 positive integer divisors. What is the least possible value of $n$ ?

Cevabı ve açıklamayı göster

Cevap: 144

Cevap

144

To find the least positive integer

n

that is a multiple of 12 and has 15 divisors, we examine the prime factorization of 12 (

2^2 \cdot 3^1

) and the divisor count formula

(e_1+1)(e_2+1)\dots = 15

. The number 15 factors as

5 \times 3

, implying

n

has the form

p^4 \cdot q^2

. Minimizing

n

requires placing the larger exponent on the smaller prime factor, so

p=2

and

q=3

, giving

n = 2^4 \cdot 3^2 = 144

.

Adım Adım Çözüm

1

Determine prime factor requirements for divisibility by 12.

Since

12 = 2^2 \cdot 3^1

,

n

must have prime factors 2 and 3 with exponents

a \ge 2

and

b \ge 1

.

Divisibility requires that all prime factors of the divisor appear in the dividend with equal or higher exponents.

2

Apply the divisor count formula to determine the exponent structure.

The total number of divisors is 15, which factors as

(4 + 1)(2 + 1) = 15

. Thus,

n = p^4 \cdot q^2

.

The number of positive divisors is given by

(e_1 + 1)(e_2 + 1) \dots = 15

.

3

Assign prime factors to minimize

n

.

Assigning the larger exponent 4 to the smaller prime 2 gives

2^4 \cdot 3^2 = 144

.

To minimize a product of prime powers, larger exponents should be paired with smaller prime bases.

Anahtar Kavram

Divisor count formula and prime factorization properties

Soru 114Soru →

If $a$ , $b$ , and $c$ are non-zero real numbers such that $a^3 b^2 c < 0$ , $a b^3 > 0$ , and $\frac{a}{c} < 0$ , which of the following expressions MUST be positive?

\frac{b - c}{a}

\frac{a b}{c}

c(a + b)

\frac{a + c}{b}

b(c - a)

Cevabı ve açıklamayı göster

Cevap: $\frac{b - c}{a}$

Cevap

\frac{b - c}{a}

Deducing the signs of

a

,

b

, and

c

reveals two possible cases: either

a > 0, b > 0, c < 0

or

a < 0, b < 0, c > 0

. In Case 1,

b - c > 0

and

a > 0

, so

\frac{b - c}{a} > 0

. In Case 2,

b - c < 0

and

a < 0

, so

\frac{b - c}{a} > 0

. Therefore,

\frac{b - c}{a}

is strictly positive in all cases.

Adım Adım Çözüm

1

Analyze the sign conditions from the given inequalities.

From

a b^3 > 0

,

a

and

b

must have the same sign. From

\frac{a}{c} < 0

,

a

and

c

must have opposite signs. From

a^3 b^2 c < 0

, since

b^2 > 0

for non-zero

b

, we have

a^3 c < 0

, which confirms

a

and

c

have opposite signs.

Odd powers preserve the sign of a variable, whereas even powers are strictly positive for non-zero real numbers.

2

Determine the two possible sign scenarios for

(a, b, c)

.

Case 1:

a > 0, b > 0, c < 0

.
Case 2:

a < 0, b < 0, c > 0

.

Since

a

and

b

share the same sign and

c

has the opposite sign, these are the only two valid assignments.

3

Evaluate the sign of the numerator and denominator of

\frac{b - c}{a}

in Case 1.

In Case 1 (

a > 0, b > 0, c < 0

):

b - c = \text{positive} - \text{negative} = \text{positive}

. Denominator

a > 0

. Ratio

\frac{\text{positive}}{\text{positive}} > 0

.

Subtracting a negative number from a positive number yields a positive result.

4

Evaluate the sign of the numerator and denominator of

\frac{b - c}{a}

in Case 2.

In Case 2 (

a < 0, b < 0, c > 0

):

b - c = \text{negative} - \text{positive} = \text{negative}

. Denominator

a < 0

. Ratio

\frac{\text{negative}}{\text{negative}} > 0

.

Dividing two negative values produces a positive quotient.

Anahtar Kavram

Positive and Negative Number Properties in Inequalities

Tahmini Süre:2m 0s

Soru 115Soru →

A positive integer $n$ has exactly four distinct prime factors, the three smallest of which are $2$ , $3$ , and $5$ . If $n$ is divisible by $360$ and has exactly $48$ positive divisors, what is the minimum possible value of $n$ ?

Cevabı ve açıklamayı göster

Cevap: 2520

Cevap

The minimum possible value of

n

is 2520.

To minimize

n

, we analyze its prime factorization

n = 2^a \times 3^b \times 5^c \times p^d

, where

p

is the fourth distinct prime factor. Divisibility by

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

requires

a \ge 3

,

b \ge 2

, and

c \ge 1

. The number of positive divisors is given by

(a+1)(b+1)(c+1)(d+1) = 48

. To minimize

n

, we pick the smallest prime greater than 5, which is

p = 7

, and set

d = 1

. This simplifies the divisor equation to

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

. Since

a+1 \ge 4

,

b+1 \ge 3

, and

c+1 \ge 2

, the minimal product of these terms is

4 \times 3 \times 2 = 24

. This uniquely determines

a = 3

,

b = 2

, and

c = 1

. Substituting these values gives

n = 2^3 \times 3^2 \times 5^1 \times 7^1 = 2520

.

Adım Adım Çözüm

1

Determine the prime factorization of the divisor requirement.

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

.

Divisibility by 360 imposes lower bounds on the exponents of the prime factors 2, 3, and 5 in

n

.

2

Formulate the general prime factorization for

n

and state exponent constraints.

n = 2^a \times 3^b \times 5^c \times p^d

with

a \ge 3

,

b \ge 2

,

c \ge 1

,

d \ge 1

, and prime

p > 5

.

n

has four distinct prime factors, three of which are 2, 3, and 5.

3

Apply the divisor counting formula to set up an algebraic equation.

(a+1)(b+1)(c+1)(d+1) = 48

.

The number of positive divisors of

n = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}

is given by

(e_1+1)(e_2+1)\dots(e_k+1)

.

4

Minimize

n

by choosing optimal values for

p

and

d

.

p = 7

and

d = 1

, leading to

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

.

To make

n

as small as possible, the fourth prime

p

should be the smallest available prime (

7

) and its exponent

d

should be minimized (

1

).

5

Solve for exponents

a

,

b

, and

c

under the given inequality constraints.

a = 3

,

b = 2

,

c = 1

.

Since

a+1 \ge 4

,

b+1 \ge 3

, and

c+1 \ge 2

, the minimum possible product

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

is

4 \times 3 \times 2 = 24

. Hence,

a=3, b=2, c=1

is the unique solution.

6

Calculate the value of

n

.

n = 2^3 \times 3^2 \times 5^1 \times 7^1 = 2520

.

Multiplying out the prime factors yields the smallest integer matching all conditions.

Anahtar Kavram

Prime Factorization and Divisor Counting Constraints

Tahmini Süre:2m 0s

Soru 116Soru →

Set $A$ consists of $k$ consecutive odd integers, and Set $B$ consists of $k$ consecutive even integers, where $k > 1$ . The smallest integer in Set $B$ is $3$ greater than the median of Set $A$ . If all integers in Set $A$ are positive, the sum of all integers in Set $A$ is $145$ , and the median of Set $B$ is $36$ , what is the smallest integer in Set $A$ ?

21

23

25

27

29

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The smallest integer in Set A is 25.

For an arithmetic sequence of

k

consecutive odd integers, the mean and median are equal to

\frac{\text{Sum}}{k} = \frac{145}{k}

. The smallest integer in Set

B

is

b_1 = \frac{145}{k} + 3

. Since Set

B

consists of

k

consecutive even integers, its median is

b_1 + k - 1

. Equating this to

36

yields

\frac{145}{k} + k + 2 = 36

, leading to

k^2 - 34k + 145 = 0

, whose roots are

k=5

and

k=29

. If

k=29

, the median of Set

A

is

5

, which implies negative integers exist in Set

A

. Since all integers in Set

A

are positive,

k=5

. With

k=5

, the median of Set

A

is

29

, and the smallest integer is

29 - 2(2) = 25

.

Adım Adım Çözüm

1

Express the median of Set A in terms of k.

M_A = \frac{145}{k}

For any evenly spaced set with an odd number of terms or symmetry, the arithmetic mean equals the median. The mean is the total sum divided by the number of terms

k

.

2

Express the median of Set B in terms of k using the given relationship for the smallest element of Set B.

M_B = \left(\frac{145}{k} + 3\right) + (k - 1) = \frac{145}{k} + k + 2

The smallest element in Set

B

is

b_1 = M_A + 3 = \frac{145}{k} + 3

. Since Set

B

contains

k

consecutive even integers (spacing

d=2

), its median is

b_1 + \frac{2(k-1)}{2} = b_1 + k - 1

.

3

Set the median of Set B to 36 and solve the quadratic equation for k.

k = 5

or

k = 29

Setting

\frac{145}{k} + k + 2 = 36

gives

\frac{145}{k} + k = 34

, which rearranges to

k^2 - 34k + 145 = 0

. Factoring gives

(k-5)(k-29) = 0

.

4

Determine the valid value of k and find the smallest integer in Set A.

Smallest integer in Set A is 25.

If

k = 29

,

M_A = \frac{145}{29} = 5

, and the smallest integer in Set

A

would be

5 - 2(14) = -23

, violating the condition that all integers in Set

A

are positive. Thus

k = 5

, making

M_A = 29

. The 5 consecutive odd integers are

25, 27, 29, 31, 33

, so the smallest integer is

25

.

Anahtar Kavram

Properties of consecutive integer sets, median-mean equivalence in arithmetic sequences, and term indexing.

Soru 117Soru →

If $x$ is a positive integer such that $\frac{6^6 + 6^6 + 6^6 + 6^6 + 6^6 + 6^6}{3^6 + 3^6 + 3^6} = 2^x$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

x

is 7.

Repeated addition converts to multiplication: six terms of

6^6

yield

6 \cdot 6^6 = 6^7

, and three terms of

3^6

yield

3 \cdot 3^6 = 3^7

. Dividing gives

\frac{6^7}{3^7} = \left(\frac{6}{3}\right)^7 = 2^7

. Comparing

2^7

to

2^x

gives

x = 7

.

Adım Adım Çözüm

1

Simplify the numerator by expressing repeated addition as multiplication.

6^6 + 6^6 + 6^6 + 6^6 + 6^6 + 6^6 = 6 \times 6^6 = 6^7

Adding six identical terms of

6^6

is equivalent to multiplying

6^6

by 6. Using the power rule

a^1 \cdot a^n = a^{n+1}

, we get

6^7

.

2

Simplify the denominator by expressing repeated addition as multiplication.

3^6 + 3^6 + 3^6 = 3 \times 3^6 = 3^7

Adding three identical terms of

3^6

is equivalent to multiplying

3^6

by 3, yielding

3^7

.

3

Apply the quotient property of exponents for identical powers.

\frac{6^7}{3^7} = \left(\frac{6}{3}\right)^7 = 2^7

According to exponent laws,

\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n

for any non-zero real numbers

a

and

b

.

4

Equate exponents of equal bases to solve for

x

.

2^x = 2^7 \implies x = 7

Since the bases on both sides of the equation are equal to 2, the exponents must be equal.

Anahtar Kavram

Combining repeated addition into exponential products and dividing powers with equal exponents.

Soru 118Soru →

Let $M = 2^a \times 3^b \times 5^c$ and $N = 2^c \times 3^a \times 5^b$ , where $a$ , $b$ , and $c$ are distinct positive integers. If the greatest common divisor of $M$ and $N$ is $90$ and the least common multiple of $M$ and $N$ is $32,400$ , what is the value of $a + b + c$ ?

6

7

8

9

14

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

a + b + c

is 7.

The correct answer is 7. By applying the fundamental property

M \times N = \gcd(M, N) \times \text{lcm}(M, N)

, we get

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

. Matching exponents gives

a + c = 5

,

a + b = 6

, and

b + c = 3

. Summing these three equations yields

2(a + b + c) = 14

, so

a + b + c = 7

.

Adım Adım Çözüm

1

Express the GCD and LCM of

M

and

N

in terms of their prime factorizations.

\gcd(M,N) = 90 = 2^1 \times 3^2 \times 5^1

and

\text{lcm}(M,N) = 32,400 = 2^4 \times 3^4 \times 5^2

.

Prime factorization allows us to relate the exponents of

M

and

N

directly to their GCD and LCM.

2

Use the identity

M \times N = \gcd(M,N) \times \text{lcm}(M,N)

to multiply the two numbers.

(

2^a \times 3^b \times 5^c) \times (2^c \times 3^a \times 5^b) = (2^1 \times 3^2 \times 5^1) \times (2^4 \times 3^4 \times 5^2)

, which simplifies to

2^{a+c} \times 3^{a+b} \times 5^{b+c} = 2^{1+4} \times 3^{2+4} \times 5^{1+2} = 2^5 \times 3^6 \times 5^3

.

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

3

Equate the exponents for each prime base

2

,

3

, and

5

.

a + c = 5

,

a + b = 6

, and

b + c = 3

.

Since prime bases are unique, exponents of corresponding prime factors on both sides of the equation must be equal.

4

Sum the three equations and solve for

a + b + c

.

(a + c) + (a + b) + (b + c) = 5 + 6 + 3 \implies 2(a + b + c) = 14 \implies a + b + c = 7

.

Adding the three system equations yields twice the desired sum.

Anahtar Kavram

Prime Factorization, GCD and LCM Product Relationship

Tahmini Süre:2m 0s

Soru 119Soru →

Set $S$ consists of $n$ consecutive integers. The arithmetic mean of the $5$ smallest integers in Set $S$ is $-12$ , and the arithmetic mean of the $5$ largest integers in Set $S$ is $24$ . What is the value of $n$ ?

36

37

40

41

42

Cevabı ve açıklamayı göster

Cevap: 41

Cevap

41

For an evenly spaced set of 5 consecutive integers, the arithmetic mean equals the middle term. Therefore, the 3rd smallest term of Set

S

is

-12

, making the 1st term

x = -14

. Similarly, the 3rd term from the end of Set

S

is

24

. Representing the 3rd term from the end as

x + n - 3

and substituting

x = -14

yields

-14 + n - 3 = 24

, which simplifies to

n = 41

.

Adım Adım Çözüm

1

Express the 5 smallest integers and find the first term of the set.

Let the set

S

be represented as

\{x, x+1, x+2, \dots, x+n-1\}

. The 5 smallest integers are

x, x+1, x+2, x+3, x+4

. Their arithmetic mean is the middle term,

x+2

. Setting

x+2 = -12

yields

x = -14

.

In any set of consecutive integers with an odd number of elements, the arithmetic mean equals the median (middle term).

2

Express the 5 largest integers and set up an equation for

n

.

The 5 largest integers in Set

S

are

x+n-5, x+n-4, x+n-3, x+n-2, x+n-1

. Their arithmetic mean is the middle term,

x+n-3

. Setting

x+n-3 = 24

and substituting

x = -14

gives

-14 + n - 3 = 24

.

The 5 largest elements also form an evenly spaced set whose mean is the middle of those 5 terms.

3

Solve for

n

.

n - 17 = 24 \implies n = 41

.

Simplifying the linear equation gives the total number of consecutive integers in Set

S

.

Anahtar Kavram

Arithmetic Mean and Median Equivalence in Consecutive Integer Subsets

Soru 120Soru →

Let $N = 2^x \times 3^y \times 7^z$ , where $x$ , $y$ , and $z$ are positive integers. If $N$ is a multiple of $84$ and $N^2$ has exactly $45$ positive integer divisors, what is the value of $x + y + z$ ?

3

4

5

6

7

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

The prime factorization of 84 is

2^2 \times 3^1 \times 7^1

. Because

N

is a multiple of 84, we must have

x \ge 2

,

y \ge 1

, and

z \ge 1

. The number of positive divisors of

N^2 = 2^{2x} \times 3^{2y} \times 7^{2z}

is

(2x+1)(2y+1)(2z+1) = 45

. Given

x \ge 2

,

2x+1 \ge 5

, while

2y+1 \ge 3

and

2z+1 \ge 3

. The only set of three factors of 45 meeting these criteria is

\{5, 3, 3\}

, giving

x=2, y=1, z=1

. The sum is

2 + 1 + 1 = 4

.

Adım Adım Çözüm

1

Find the prime factorization of 84 to set lower bounds on

x, y, z

.

84 = 2^2 \times 3^1 \times 7^1

, so

x \ge 2

,

y \ge 1

, and

z \ge 1

.

Since

N

is a multiple of 84, its prime factorization must contain at least the prime factors of 84 with at least the same exponents.

2

Express the number of divisors of

N^2

.

N^2 = 2^{2x} \times 3^{2y} \times 7^{2z}

, so the number of positive divisors is

(2x + 1)(2y + 1)(2z + 1) = 45

.

Squaring a number doubles all its prime exponents, and the total number of divisors of

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

is

(e_1 + 1)(e_2 + 1)\dots

.

3

Determine the unique valid integer solution for

(2x+1)

,

(2y+1)

, and

(2z+1)

.

Since

x \ge 2

,

2x + 1 \ge 5

. Since

y, z \ge 1

,

2y+1 \ge 3

and

2z+1 \ge 3

. The factor triples of 45 into three odd factors

\ge 3

is uniquely

5 \times 3 \times 3

.

Factoring 45 gives

45 = 5 \times 3 \times 3

as the only breakdown satisfying

2x+1 \ge 5

.

4

Solve for

x, y, z

and compute their sum.

2x + 1 = 5 \implies x = 2

,

2y + 1 = 3 \implies y = 1

, and

2z + 1 = 3 \implies z = 1

. Thus

x + y + z = 2 + 1 + 1 = 4

.

Equating individual factors yields the exact exponent values.

Anahtar Kavram

Prime Factorization and Divisors of Powers of Integers

Number Properties and Arithmetic

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