Number Properties and Arithmetic Alıştırma Soruları — GMAT

Soru 1Soru →

For all integers $a$ , $b$ , and $c$ , if the expression $a^3 b - b^3 c + c^3 a$ is an odd integer, then the product $(a - b)(b - c)(c - a)$ must be divisible by 4.

Doğru

Yanlış

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Cevap: False

Cevap

The statement is False.

The statement claims that the product must be divisible by 4 for all integer inputs where the given expression is odd. However, setting two variables to odd integers and one variable to an even integer (such as

a=3, b=1, c=0

) makes the initial expression odd (

27

) while producing a product of

-6

, which is not divisible by 4.

Adım Adım Çözüm

1

Simplify the expression modulo 2 to establish necessary parity conditions.

Since

x^3 \equiv x \pmod 2

for any integer

x

, the parity of

x^3 y

is identical to the parity of

xy

. Thus,

a^3b - b^3c + c^3a \equiv ab - bc + ca \pmod 2

.

Reducing powers modulo 2 determines which parity combinations of

a, b,

and

c

yield an odd result.

2

Analyze all possible parity combinations for the three variables.

Case 1: All three are odd

\implies odd - odd + odd = odd

.
Case 2: Two are odd and one is even (e.g.,

a, b

odd,

c

even)

\implies odd - even + even = odd

.
Case 3: One is odd and two are even

\implies even - even + even = even

.
Case 4: All three are even

\implies even - even + even = even

.

Identifying that both Case 1 and Case 2 produce an odd value is essential for finding edge cases.

3

Test Case 2 using specific integer values to check whether

(a - b)(b - c)(c - a)

must be divisible by 4.

Set

a = 3

(odd),

b = 1

(odd), and

c = 0

(even). The given expression equals

3^3(1) - 1^3(0) + 0^3(3) = 27

, which is odd. Evaluating the product gives

(3 - 1)(1 - 0)(0 - 3) = (2)(1)(-3) = -6

.

Testing specific valid integers disproves a universal 'must be' claim.

4

Evaluate the divisibility of the resulting product by 4.

-6

is not divisible by 4 because

-6 / 4 = -1.5

, which is not an integer.

Finding a valid case where the product is not divisible by 4 proves the statement is false.

Anahtar Kavram

Parity rules for algebraic expressions and edge-case evaluation using zero as an even integer

Soru 2Soru →

A rectangular tabletop measuring $126\text{ centimeters}$ by $180\text{ centimeters}$ is to be completely covered by non-overlapping, identical square tiles of the maximum possible side length, such that no tiles need to be cut. How many such square tiles are required to cover the entire tabletop?

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Cevap: 70

Cevap

70 tiles are required to cover the tabletop.

To cover a

126\text{ cm} \times 180\text{ cm}

surface with identical square tiles of maximum side length without cutting, the side length of each square tile must be the greatest common divisor of

126

and

180

, which is

18\text{ cm}

. Dividing the surface dimensions by

18\text{ cm}

gives

7

tiles along one side and

10

tiles along the other, yielding a total of

70

tiles.

Adım Adım Çözüm

1

Determine the prime factorizations of the tabletop dimensions 126 and 180.

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

and

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

Prime factorization allows systematic extraction of the greatest common divisor.

2

Calculate the Greatest Common Divisor (GCD) of 126 and 180.

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

The maximum side length of a square tile that fits evenly without cutting is the GCD of the two dimensions.

3

Find the total number of tiles required.

\text{Total tiles} = \left(\frac{126}{18}\right) \times \left(\frac{180}{18}\right) = 7 \times 10 = 70

The total area divided by the area of one tile gives the number of tiles needed.

Anahtar Kavram

Greatest Common Divisor (GCD) application in geometric spatial partitioning

Soru 3Soru →

Set $S$ consists of all consecutive integers from $-m$ to $n$ , inclusive, where $m$ and $n$ are positive integers with $n > m$ . If set $S$ contains exactly $25$ integers and the sum of all integers in set $S$ is $75$ , what is the value of $m$ ?

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Cevap: 9

Cevap

The value of

m

is

9

.

To find

m

, use the two fundamental properties of consecutive integer sets: term count and set sum. The number of terms from

-m

to

n

inclusive is

n - (-m) + 1 = n + m + 1 = 25

, which yields

n + m = 24

. The sum of an evenly spaced set is the product of the number of terms and the arithmetic mean of the smallest and largest terms:

25 \times \frac{-m + n}{2} = 75

, which simplifies to

n - m = 6

. Subtracting

n - m = 6

from

n + m = 24

gives

2m = 18

, so

m = 9

.

Adım Adım Çözüm

1

Set up the equation for the number of terms in the set.

n + m = 24

The number of integers from

-m

to

n

inclusive is

n - (-m) + 1 = n + m + 1 = 25

.

2

Set up the equation for the sum of the integers in the set.

n - m = 6

The sum of an arithmetic progression is given by

\text{number of terms} \times \text{mean} = 25 \times \frac{-m + n}{2} = 75

, leading to

\frac{n - m}{2} = 3

.

3

Solve for

m

using the two linear equations.

m = 9

Subtracting

n - m = 6

from

n + m = 24

yields

2m = 18

, giving

m = 9

.

Anahtar Kavram

Properties of consecutive integer sets: inclusive term counting (

n - \text{start} + 1

) and set sum calculation (

\text{count} \times \text{mean}

).

Tahmini Süre:1m 30s

Soru 4Soru →

If $x$ and $y$ are integers such that $-5 \le x \le -1$ and $2 \le y \le 6$ , what is the minimum possible value of $\frac{x - y}{x + y}$ ?

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Cevap: -11

Cevap

The minimum possible value of the expression is -11.

The correct answer is -11 because the numerator

x - y

is negative for all allowed values of

x

and

y

. To minimize a negative fraction, the denominator

x + y

must be positive and minimized (equal to 1), while the numerator must be as negative as possible. Choosing

x = -5

and

y = 6

gives a numerator of

-11

and a denominator of

1

, producing the minimum value of -11.

Adım Adım Çözüm

1

Evaluate the sign of the numerator

Since

x

is negative (

x \le -1

) and

y

is positive (

y \ge 2

),

x - y

is always negative (

x - y \le -3

).

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

2

Determine the conditions for minimizing a negative fraction

To obtain the minimum (most negative) value, the denominator

x + y

must be a positive integer and as small as possible.

A negative numerator divided by a positive denominator yields a negative quotient. Dividing by a smaller positive number yields a quotient with larger absolute value, making it smaller (more negative).

3

Identify the smallest positive denominator and solve for variables

The smallest positive integer for

x + y

is

1

. Setting

y = 1 - x

gives

x - y = 2x - 1

.

Since

x

and

y

are integers,

x + y

must be an integer.

4

Maximize the magnitude of the negative numerator

Using the lowest boundary

x = -5

gives

y = 6

. The value of the expression is

\frac{-5 - 6}{-5 + 6} = -11

.

Evaluating at

x = -5

and

y = 6

gives numerator

-11

and denominator

1

, resulting in

-11

.

Anahtar Kavram

Minimizing algebraic fractions involving signed numbers

Soru 5Soru →

Let $m = 2^a \cdot 5^3 \cdot 11^b$ and $n = 2^4 \cdot 5^c \cdot 11^2$ , where $a$ , $b$ , and $c$ are positive integers. If the greatest common divisor of $m$ and $n$ is $2^3 \cdot 5^2 \cdot 11^2$ , and the least common multiple of $m$ and $n$ has exactly $180$ positive integer divisors, what is the value of $a + b + c$ ?

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Cevap: 13

Cevap

The value of

a + b + c

is

13

.

To find

a, b, c

, analyze prime factor exponents. The GCD uses minimum exponents:

\min(a,4)=3 \implies a=3

,

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

, and

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

. The LCM uses maximum exponents:

2^4, 5^3, 11^b

. The number of divisors of the LCM is

(4+1)(3+1)(b+1) = 20(b+1) = 180

, yielding

b=8

. Adding

a+b+c = 3+8+2 = 13

.

Adım Adım Çözüm

1

Use the prime exponent rule for greatest common divisor to find

a

and

c

, and bound

b

.

a = 3

,

c = 2

, and

b \ge 2

.

The greatest common divisor takes the minimum exponent for each prime factor:

\min(a, 4) = 3 \implies a = 3

,

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

, and

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

.

2

Determine the prime factorization of

\text{lcm}(m, n)

.

\text{lcm}(m, n) = 2^4 \cdot 5^3 \cdot 11^b

.

The least common multiple takes the maximum exponent for each prime factor:

\max(3, 4) = 4

,

\max(3, 2) = 3

, and

\max(b, 2) = b

since

b \ge 2

.

3

Apply the total number of divisors formula to solve for

b

.

b = 8

.

The number of positive integer divisors of

2^4 \cdot 5^3 \cdot 11^b

is

(4+1)(3+1)(b+1) = 20(b+1) = 180

, which yields

b+1 = 9

so

b = 8

.

4

Calculate the sum

a + b + c

.

a + b + c = 13

.

Summing the values

a=3

,

b=8

, and

c=2

gives

3 + 8 + 2 = 13

.

Anahtar Kavram

GCD and LCM prime factor exponent rules and total divisor formula

Soru 6Soru →

What is the remainder when the expression $7^{45} \cdot 3^{22} + 8^{15}$ is divided by $10$ ?

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Cevap: 5

Cevap

The remainder is 5.

Dividing any number by 10 leaves a remainder equal to the units digit of that number. By analyzing the units digit pattern (cyclicity of period 4) for powers of 7, 3, and 8:

7^{45} \equiv 7^1 = 7 \pmod{10}

,

3^{22} \equiv 3^2 = 9 \pmod{10}

, and

8^{15} \equiv 8^3 = 2 \pmod{10}

. The expression simplifies to

(7 \times 9) + 2 = 63 + 2 = 65

, which has a units digit of 5. Therefore, the remainder when divided by 10 is 5.

Adım Adım Çözüm

1

Relate remainder modulo 10 to units digit cyclicity.

Finding the remainder when an expression is divided by 10 is equivalent to finding its units digit.

Any positive integer

N

can be expressed as

10k + r

, where

r

is the units digit and the remainder when

N

is divided by 10.

2

Determine the units digit of

7^{45}

.

7^{45} \equiv 7 \pmod{10}

.

Powers of 7 repeat their units digits in a cycle of length 4 (7, 9, 3, 1). Dividing the exponent 45 by 4 gives a remainder of 1, so

7^{45}

has the same units digit as

7^1 = 7

.

3

Determine the units digit of

3^{22}

.

3^{22} \equiv 9 \pmod{10}

.

Powers of 3 repeat their units digits in a cycle of length 4 (3, 9, 7, 1). Dividing the exponent 22 by 4 gives a remainder of 2, so

3^{22}

has the same units digit as

3^2 = 9

.

4

Determine the units digit of the product

7^{45} \cdot 3^{22}

.

7^{45} \cdot 3^{22} \equiv 3 \pmod{10}

.

The product of the units digits is

7 \times 9 = 63

, which has a units digit of 3.

5

Determine the units digit of

8^{15}

.

8^{15} \equiv 2 \pmod{10}

.

Powers of 8 repeat their units digits in a cycle of length 4 (8, 4, 2, 6). Dividing the exponent 15 by 4 gives a remainder of 3, so

8^{15}

has the same units digit as

8^3 = 512

, which ends in 2.

6

Combine the results to find the final remainder modulo 10.

5

Adding the units digit of the first term (3) and the second term (2) gives

3 + 2 = 5

.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic

Soru 7Soru →

If $5^{x} - 5^{x-2} = 600\sqrt{5}$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 4.5

Cevap

The value of

x

is 4.5.

Factoring out

5^{x-2}

converts the left side into

5^{x-2}(25 - 1) = 24 \cdot 5^{x-2}

. Dividing

600\sqrt{5}

by 24 gives

25\sqrt{5}

, which equals

5^{2.5}

. Setting

x - 2 = 2.5

gives

x = 4.5

.

Adım Adım Çözüm

1

Factor out the smallest power of 5, which is

5^{x-2}

, from the left side of the equation.

5^{x-2}(5^2 - 1) = 600\sqrt{5}

Factoring isolates the constant multiplier from the variable power term.

2

Calculate the numerical value inside the parentheses.

24 \cdot 5^{x-2} = 600\sqrt{5}

5^2 - 1 = 25 - 1 = 24

.

3

Divide both sides of the equation by 24.

5^{x-2} = 25\sqrt{5}

Isolating

5^{x-2}

gives

\frac{600\sqrt{5}}{24} = 25\sqrt{5}

.

4

Rewrite

25\sqrt{5}

as a single exponential expression with base 5.

5^{x-2} = 5^2 \cdot 5^{0.5} = 5^{2.5}

Using the product rule for exponents,

5^2 \cdot 5^{1/2} = 5^{2 + 0.5} = 5^{2.5}

.

5

Equate the exponents since the bases are identical.

x - 2 = 2.5 \implies x = 4.5

For any positive base

b \neq 1

, if

b^m = b^n

, then

m = n

.

Anahtar Kavram

Exponents, Roots, and Powers of Integers

Soru 8Soru →

Let $n = 2^a \cdot 3^b \cdot 5$ , where $a$ and $b$ are positive integers. If $n$ has a total of 24 positive divisors and has an equal number of even positive divisors and odd positive divisors, what is the value of $b$ ?

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Cevap: 5

Cevap

5

For

n = 2^a \cdot 3^b \cdot 5^1

, the total number of divisors is

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

, giving

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

. The odd divisors are formed strictly from

3^b \cdot 5^1

, giving

(b + 1)(2)

odd divisors. The even divisors require at least one factor of 2, giving

a(b + 1)(2)

even divisors. Setting even and odd divisor counts equal gives

2a(b + 1) = 2(b + 1)

, which reduces to

a = 1

. Substituting

a = 1

into

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

gives

2(b + 1) = 12

, leading directly to

b = 5

.

Adım Adım Çözüm

1

Set up the formula for total positive divisors of

n

.

(a + 1)(b + 1)(1 + 1) = 24 \implies (a + 1)(b + 1) = 12

The total number of divisors of

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

is

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

.

2

Determine the counts of odd and even positive divisors.

Odd divisors =

2(b + 1)

, Even divisors =

2a(b + 1)

Odd divisors cannot contain any factors of 2 (exponent of 2 is 0). Even divisors must contain at least one factor of 2 (exponent of 2 can be

1, 2, \dots, a

).

3

Equate the number of odd and even divisors to find

a

.

2a(b + 1) = 2(b + 1) \implies a = 1

Dividing both sides by

2(b + 1)

(which is positive since

b \ge 1

) leaves

a = 1

.

4

Solve for

b

using the total divisor relation.

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

Substituting

a = 1

into

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

yields

b = 5

.

Anahtar Kavram

Counting total, odd, and even positive divisors using prime factorization exponents

Soru 9Soru →

If $n$ is a positive integer such that $\frac{6^{n+3} \cdot 15^{2n-1}}{10^{n+1} \cdot 9^{n+1}} = 8{}100$ , what is the value of $n$ ?

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Cevap: 4

Cevap

4

Rewriting all composite bases (

6, 15, 10, 9

) into prime bases (

2, 3, 5

) simplifies the equation to

4 \cdot 3^n \cdot 5^{n-2} = 8,100

. Dividing by 4 yields

3^n \cdot 5^{n-2} = 2,025 = 3^4 \cdot 5^2

, which gives

n = 4

.

Adım Adım Çözüm

1

Express each composite base in the fraction using its prime factor decomposition.

Numerator:

(2 \cdot 3)^{n+3} \cdot (3 \cdot 5)^{2n-1} = 2^{n+3} \cdot 3^{n+3} \cdot 3^{2n-1} \cdot 5^{2n-1} = 2^{n+3} \cdot 3^{3n+2} \cdot 5^{2n-1}

.
Denominator:

(2 \cdot 5)^{n+1} \cdot (3^2)^{n+1} = 2^{n+1} \cdot 5^{n+1} \cdot 3^{2n+2}

.

Breaking composite numbers down into prime factors allows terms with identical bases to be combined using exponent rules.

2

Simplify the fraction by subtracting the denominator exponents from the numerator exponents for each prime base.

\frac{2^{n+3} \cdot 3^{3n+2} \cdot 5^{2n-1}}{2^{n+1} \cdot 3^{2n+2} \cdot 5^{n+1}} = 2^{(n+3)-(n+1)} \cdot 3^{(3n+2)-(2n+2)} \cdot 5^{(2n-1)-(n+1)} = 2^2 \cdot 3^n \cdot 5^{n-2} = 4 \cdot 3^n \cdot 5^{n-2}

.

Applying the quotient rule for exponents:

\frac{a^x}{a^y} = a^{x-y}

.

3

Set the simplified expression equal to

8,100

and solve for

n

.

4 \cdot 3^n \cdot 5^{n-2} = 8,100 \implies 3^n \cdot 5^{n-2} = 2,025

. Prime factorization of

2,025 = 81 \cdot 25 = 3^4 \cdot 5^2

. Equating exponents gives

n = 4

(and

n-2 = 2

, which confirms

n=4

).

Because 3 and 5 are distinct prime bases, the unique prime factorization theorem guarantees that

3^n \cdot 5^{n-2} = 3^4 \cdot 5^2

implies

n = 4

.

Anahtar Kavram

Simplifying exponential expressions using prime factorization and quotient laws

Tahmini Süre:2m 0s

Soru 10Soru →

For any positive integer $n$ , let $f(n)$ denote the product of all distinct prime factors of $n$ . For example, $f(12) = 2 \times 3 = 6$ . What is the value of $f(2^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4)$ ?

Cevabı ve açıklamayı göster

Cevap: 330

Cevap

330

To find

f(2^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4)

, first factor out the greatest common factor

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

. This yields

2^4 \cdot 3^2 \cdot 5^2 (2^2 \cdot 3^2 - 5^2) = 2^4 \cdot 3^2 \cdot 5^2 (36 - 25) = 2^4 \cdot 3^2 \cdot 5^2 \cdot 11

. The distinct prime factors present in this expression are 2, 3, 5, and 11. Multiplying these distinct prime factors gives

2 \times 3 \times 5 \times 11 = 330

.

Adım Adım Çözüm

1

Factor out the greatest common term

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

from

2^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4

2^4 \cdot 3^2 \cdot 5^2 \cdot (2^2 \cdot 3^2 - 5^2)

Factoring out common prime powers simplifies the expression and avoids large calculations.

2

Evaluate the arithmetic expression inside the parentheses

2^2 \cdot 3^2 - 5^2 = 4 \cdot 9 - 25 = 36 - 25 = 11

11 is itself a prime number.

3

Write the full prime factorization of the overall number

2^4 \cdot 3^2 \cdot 5^2 \cdot 11^1

All bases (2, 3, 5, 11) are prime numbers, giving the complete prime factorization.

4

Multiply each distinct prime factor together to evaluate

f(n)

2 \times 3 \times 5 \times 11 = 330

The function

f(n)

takes the product of each unique prime factor exactly once.

Anahtar Kavram

Prime Factorization and Product of Distinct Prime Factors

Soru 11Soru →

A set $S$ consists of $k$ consecutive integers. The sum of the first $m$ integers in set $S$ is $34$ , and the sum of the last $m$ integers in set $S$ is $74$ . If the sum of all $k$ integers in set $S$ is $189$ , what is the value of $k$ ?

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Cevap: 14

Cevap

The total number of integers in set S is 14.

By applying the property that the arithmetic mean of an evenly spaced set is the average of the mean of its first

m

elements and the mean of its last

m

elements, we establish that the mean of the set is

\frac{54}{m}

. Since the total sum is

189

,

k \cdot \frac{54}{m} = 189

, which yields

k = 3.5m

. Substituting this ratio into the difference between the two subset sums

m(k-m) = 40

gives

2.5m^2 = 40

, so

m = 4

and

k = 14

.

Adım Adım Çözüm

1

Set up algebraic expressions for the sums of the first m terms and last m terms.

Let the set be

S = \{a, a+1, \dots, a+k-1\}

. The first

m

terms sum to

S_1 = m a + \frac{m(m-1)}{2} = 34

. The last

m

terms sum to

S_2 = m(a+k-m) + \frac{m(m-1)}{2} = 74

.

Consecutive integer sums can be represented by the starting term and the number of terms.

2

Subtract the sum of the first m terms from the sum of the last m terms.

S_2 - S_1 = m(a+k-m) - ma = m(k-m) = 74 - 34 = 40

.

Subtracting eliminates the initial term

a

and quadratic term

\frac{m(m-1)}{2}

, giving a clean relationship between

m

and

k

.

3

Determine the arithmetic mean of set S using subset averages.

The average of the first

m

terms is

\frac{34}{m}

and the average of the last

m

terms is

\frac{74}{m}

. The average of the entire set is the midpoint of these two averages:

\text{Mean} = \frac{1}{2}\left(\frac{34}{m} + \frac{74}{m}\right) = \frac{54}{m}

.

In any evenly spaced set, the overall median/mean is equal to the average of the lower-bound subset mean and upper-bound subset mean.

4

Relate the total sum to the set size k and overall mean.

S_{total} = k \times \text{Mean} \implies 189 = k \left(\frac{54}{m}\right) \implies \frac{k}{m} = \frac{189}{54} = 3.5 \implies k = 3.5m

.

The sum of a set of consecutive integers is always equal to the number of terms times the mean of the set.

5

Solve for m and k using the system of equations.

Substitute

k = 3.5m

into

m(k-m) = 40 \implies m(2.5m) = 40 \implies 2.5m^2 = 40 \implies m^2 = 16 \implies m = 4

. Thus,

k = 3.5 \times 4 = 14

.

Since

m > 0

, taking the positive square root gives

m = 4

, which leads directly to

k = 14

.

Anahtar Kavram

Average and Sum Equivalences in Evenly Spaced Sets

Soru 12Soru →

At an agricultural cooperative, $0.45$ of the total fruit harvested by weight consisted of apples, and $\frac{3}{8}$ of the remaining fruit weight consisted of pears. If the remaining $550$ kilograms of fruit consisted entirely of oranges, what was the total weight, in kilograms, of the fruit harvested?

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Cevap: 1600

Cevap

1600

Subtracting the apple portion leaves 55% (or 11/20) of the total harvest. Since pears make up 3/8 of this remainder, oranges make up the remaining 5/8 of the 11/20 portion, which simplifies to 11/32 of the total harvest. Setting 11/32 of the total harvest equal to 550 kg yields a total weight of 1,600 kg.

Adım Adım Çözüm

1

Find the fraction of the total harvest left after subtracting the apples.

The remaining portion is 1 - 0.45 = 0.55, which equals 11/20 of the total harvest.

Apples account for 0.45 of the harvest.

2

Calculate the portion of the harvest represented by oranges.

Since pears are 3/8 of the remainder, oranges are 5/8 of the remainder. Thus, oranges are (5/8) * (11/20) = 11/32 of the total harvest.

The non-apple harvest consists only of pears and oranges.

3

Solve for the total weight using the given weight of oranges.

Total weight = 550 * (32 / 11) = 1600 kg.

11/32 of the total weight equals 550 kilograms.

Anahtar Kavram

Multi-step arithmetic combining decimals and fractions to solve remaining-quantity word problems

Soru 13Soru →

What is the exponent of $3$ in the prime factorization of the integer $S = 25! + 26! + 27!$ ?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The exponent of

3

in the prime factorization of

S

is

16

.

Factoring

25!

from the sum gives

S = 25!(1 + 26 + 26 \times 27) = 25!(729) = 25! \times 3^6

. Applying Legendre's formula to

25!

yields

\lfloor 25/3 \rfloor + \lfloor 25/9 \rfloor = 8 + 2 = 10

factors of

3

. Adding the

6

factors of

3

from

729 = 3^6

gives a total exponent of

10 + 6 = 16

.

Adım Adım Çözüm

1

Factor out

25!

from the sum

S = 25! + 26! + 27!

S = 25! \left(1 + 26 + 26 \times 27\right)

Factoring out the greatest common factorial term

25!

converts the sum into a product of

25!

and an integer factor.

2

Simplify the expression inside the parentheses

1 + 26 + 702 = 729 = 3^6

Using algebraic simplification,

1 + 26(1 + 27) = 1 + 26(28) = 729 = 27^2 = (3^3)^2 = 3^6

.

3

Compute the exponent of

3

in

25!

using Legendre's formula

E_3(25!) = \lfloor \frac{25}{3} \rfloor + \lfloor \frac{25}{9} \rfloor + \lfloor \frac{25}{27} \rfloor = 8 + 2 + 0 = 10

Legendre's formula counts the total prime factors of

3

contributed by all multiples of

3, 9, 27, \dots

up to

25

.

4

Combine the exponent of

3

from

25!

and the factor

729

Total exponent of

3 = 10 + 6 = 16

Since

S = 25! \times 3^6 = (3^{10} \times k) \times 3^6 = 3^{16} \times k

(where

3 \nmid k

), the total exponent of

3

is

10 + 6 = 16

.

Anahtar Kavram

Exponent of a prime in factorial expressions using Legendre's formula and algebraic factoring

Soru 14Soru →

A sequence of positive integers $a_n$ is defined by $a_n = 2^n + 5^n$ for all integers $n \ge 1$ . What is the remainder when $a_{100}$ is divided by $7$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The remainder when

a_{100}

is divided by

7

is

4

.

By reducing

5 \equiv -2 \pmod 7

, we observe that

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even. The powers of

2 \pmod 7

follow a 3-step cycle (

2, 4, 1

). Dividing the exponent

100

by

3

leaves a remainder of

1

, meaning

2^{100} \equiv 2^1 = 2 \pmod 7

. Adding the remainders for both terms yields

2 + 2 = 4

.

Adım Adım Çözüm

1

Use modular arithmetic to simplify the base

5 \pmod 7

.

5 \equiv -2 \pmod 7

, which implies

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even.

Converting

5

to

-2

allows both terms to be expressed using powers of

2

.

2

Determine the remainder cycle of powers of

2

when divided by

7

.

2^1 \equiv 2 \pmod 7

,

2^2 \equiv 4 \pmod 7

, and

2^3 = 8 \equiv 1 \pmod 7

. The pattern repeats every

3

powers.

Finding the period of cyclicity simplifies evaluating large powers.

3

Divide the exponent

100

by the cycle length

3

.

100 = 3 \times 33 + 1

, leaving a remainder of

1

. Thus,

2^{100} \equiv 2^1 = 2 \pmod 7

.

The remainder of the exponent modulo the cycle length determines the equivalent reduced power.

4

Combine the remainders for

2^{100}

and

5^{100}

.

a_{100} = 2^{100} + 5^{100} \equiv 2 + 2 = 4 \pmod 7

.

Adding the individual modular values gives the overall remainder.

Anahtar Kavram

Modular arithmetic cyclicity and negative congruences

Soru 15Soru →

A set $S$ consists of consecutive integers. The arithmetic mean of all the positive integers in set $S$ is $18.5$ , and the arithmetic mean of all the negative integers in set $S$ is $-12$ . How many integers are in set $S$ ?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

The total number of integers in set

S

is

60

.

The positive integers in set

S

are

1, 2, \dots, 36

, which have an arithmetic mean of

\frac{1 + 36}{2} = 18.5

. The negative integers in set

S

are

-23, -22, \dots, -1

, which have an arithmetic mean of

\frac{-23 + (-1)}{2} = -12

. Since set

S

consists of consecutive integers spanning from

-23

to

36

, it includes

23

negative integers,

36

positive integers, and the integer

0

. The total number of elements is

23 + 1 + 36 = 60

.

Adım Adım Çözüm

1

Find the largest positive integer in set

S

.

The largest positive integer is

36

, meaning there are

36

positive integers in

S

.

The positive integers in

S

must form a consecutive sequence starting at

1

up to some maximum integer

m

. The average of consecutive integers from

1

to

m

is

\frac{1 + m}{2}

. Setting

\frac{1 + m}{2} = 18.5

yields

1 + m = 37

, so

m = 36

.

2

Find the smallest negative integer in set

S

.

The smallest negative integer is

-23

, meaning there are

23

negative integers in

S

.

The negative integers in

S

must form a consecutive sequence ending at

-1

down to some minimum integer

-k

. The average of consecutive integers from

-k

to

-1

is

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

. Setting

\frac{-k - 1}{2} = -12

yields

-k - 1 = -24

, so

k = 23

.

3

Calculate the total number of elements in set

S

.

Set

S

contains

60

integers.

Because set

S

contains consecutive integers ranging from negative to positive values, it must also contain

0

. The total count is

23\text{ (negative integers)} + 1\text{ (the integer zero)} + 36\text{ (positive integers)} = 60

.

Anahtar Kavram

Evenly spaced set averages and classification of zero in consecutive integer sets.

Tahmini Süre:2m 0s

Soru 16Soru →

Set $S$ consists of $n$ consecutive even integers. The arithmetic mean of all the integers in set $S$ is $45$ . If the difference between the largest integer and the smallest integer in set $S$ is $34$ , what is the value of the largest integer in set $S$ ?

Cevabı ve açıklamayı göster

Cevap: 62

Cevap

The largest integer in set

S

is

62

.

In any set of evenly spaced numbers, such as consecutive even integers, the arithmetic mean is equal to the average of the smallest term and the largest term. Since the arithmetic mean is 45, the sum of the smallest term and the largest term must be

45 \times 2 = 90

. Combining this with the given fact that the difference between the largest term and the smallest term is 34 creates a system of equations: Smallest + Largest = 90 and Largest - Smallest = 34. Adding these two equations cancels out the smallest term, resulting in 2 * Largest = 124, which gives 62 for the largest integer.

Adım Adım Çözüm

1

Express the arithmetic mean of the set in terms of the smallest element (

F

) and largest element (

L

).

F + L = 90

For any evenly spaced set of numbers, the arithmetic mean is equal to the average of the first and last elements:

\frac{F + L}{2} = 45

.

2

Set up the equation for the difference between the largest and smallest elements.

L - F = 34

The question specifies that the largest integer exceeds the smallest integer by 34.

3

Solve the system of two linear equations for

L

.

L = 62

Adding

(F + L) + (L - F) = 90 + 34

eliminates

F

, leaving

2L = 124

, which yields

L = 62

.

Anahtar Kavram

Equivalence of arithmetic mean to the average of the first and last terms in an evenly spaced set

Soru 17Soru →

A positive integer $N$ has exactly $12$ positive integer divisors. If $N$ is a multiple of $18$ , and exactly $\frac{2}{3}$ of the positive integer divisors of $N$ are even, what is the value of $N$ ?

36

72

108

144

216

Cevabı ve açıklamayı göster

Cevap: 108

Cevap

108

The integer 108 prime factorizes as 2² × 3³. It has (2+1)(3+1) = 12 total positive divisors. Its odd divisors are the 4 divisors of 3³ (namely 1, 3, 9, 27), leaving 12 - 4 = 8 even divisors. Thus, exactly 8/12 = 2/3 of its divisors are even, and 108 is divisible by 18.

Adım Adım Çözüm

1

Determine the number of even and odd divisors of N.

Number of even divisors = (2/3) × 12 = 8; Number of odd divisors = 12 - 8 = 4.

The problem states that 2/3 of the 12 total divisors are even.

2

Express N as 2^a × M, where M is odd.

Total divisors = (a + 1) × (divisors of M) = 12. Since divisors of M are the odd divisors of N, (a + 1) × 4 = 12, so a + 1 = 3 and a = 2.

Odd divisors of N come entirely from the odd part M.

3

Find M such that M is odd, divisible by 9 (since N is a multiple of 18 = 2 × 3²), and has exactly 4 divisors.

M must be of the form 3^k. Since divisors count is k + 1 = 4, k = 3, so M = 3³ = 27.

If M had two prime factors p × q, each exponent would be 1, which cannot be divisible by 3².

4

Calculate N = 2^a × M.

N = 2² × 27 = 4 × 27 = 108.

Combining the even and odd prime factor components.

Anahtar Kavram

Divisibility and Counting Divisors via Prime Factorization

Tahmini Süre:1m 30s

Soru 18Soru →

Let $d(k)$ denote the number of positive divisors of a positive integer $k$ . What is the smallest positive integer $n$ that is a multiple of $72$ and satisfies $d(n) = 35$ ?

Cevabı ve açıklamayı göster

Cevap: 5184

Cevap

The smallest positive integer

n

that satisfies all conditions is 5184.

The prime factorization of

72

is

2^3 \times 3^2

. Any multiple

n

of

72

must be written as

2^a \times 3^b \times dots

with

a \ge 3

and

b \ge 2

. The total number of positive divisors

d(n) = (a+1)(b+1) dots = 35

. The integer

35

can be factored into integers greater than

1

only as

7 \times 5

. Thus,

n

must have exactly two distinct prime factors, which must be

2

and

3

. The required exponents are

a, b \in \{6, 4\}

. To minimize

n

, we assign the larger exponent

6

to the smaller base

2

, yielding

n = 2^6 \times 3^4 = 64 \times 81 = 5184

.

Adım Adım Çözüm

1

Express 72 in prime factor form to determine minimum required prime exponents.

72 = 2^3 \times 3^2

, meaning

n

must have prime factorization

2^a \times 3^b \dots

where

a \ge 3

and

b \ge 2

.

Any multiple of 72 must contain at least three factors of 2 and two factors of 3.

2

Analyze the divisor count condition

d(n) = 35

.

Since

35 = 7 \times 5

, the number of prime factors of

n

can be at most 2, with exponent increments

(a+1)(b+1) = 7 \times 5

.

The number of positive divisors for

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

is

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

, and 35 factors into integers greater than 1 only as 35 or

7 \times 5

.

3

Determine the required exponents for the prime factors of

n

.

The exponents of the prime factors must be

7-1 = 6

and

5-1 = 4

.

Since

n

must contain both 2 and 3 as prime factors,

n

cannot have only 1 prime factor, so it has exactly two prime factors (2 and 3).

4

Optimize the exponent assignment to minimize

n

.

Assign exponent 6 to base 2 and exponent 4 to base 3, giving

n = 2^6 \times 3^4 = 64 \times 81 = 5184

.

Assigning the larger exponent to the smaller prime base minimizes the total product while meeting the conditions

a \ge 3

and

b \ge 2

.

Anahtar Kavram

Divisor counting formula and prime factor exponent allocation under divisibility constraints

Tahmini Süre:2m 0s

Soru 19Soru →

For any integer $n$ , which of the following expressions MUST be an even integer?

(n + 1)^3 - (n - 1)^3

n^3 - n + 1

(n + 1)^2 + n^2

3n^2 + 3n + 1

n^2 + n - 3

Cevabı ve açıklamayı göster

Cevap: $(n + 1)^3 - (n - 1)^3$

Cevap

The expression

(n + 1)^3 - (n - 1)^3

MUST be an even integer for all integer values of

n

.

Simplifying the expression

(n + 1)^3 - (n - 1)^3

yields

6n^2 + 2

, which equals

2(3n^2 + 1)

. Since

3n^2 + 1

is always an integer for any integer

n

, multiplying by 2 guarantees the expression is an even integer for all values of

n

, including zero and negative integers.

Adım Adım Çözüm

1

Expand both cubic terms algebraically.

(n + 1)^3 = n^3 + 3n^2 + 3n + 1

and

(n - 1)^3 = n^3 - 3n^2 + 3n - 1

.

Expanding the terms allows combining like terms to simplify the expression.

2

Subtract the expanded expression

(n - 1)^3

from

(n + 1)^3

.

(n^3 + 3n^2 + 3n + 1) - (n^3 - 3n^2 + 3n - 1) = 6n^2 + 2

.

The cubic terms (

n^3

) and linear terms (

3n

) cancel out.

3

Factor out a common factor of 2 from the simplified expression.

6n^2 + 2 = 2(3n^2 + 1)

.

Any integer that can be expressed as

2k

, where

k

is an integer, is by definition even. Since

n

is an integer,

3n^2 + 1

is an integer, making

2(3n^2 + 1)

an even integer for all

n

.

Anahtar Kavram

Parity rules for algebraic expressions and consecutive integer products

Alternatif Yöntem

Instead of algebraic expansion, test values of

n

with different parities, including

n = 0

(even) and

n = 1

(odd). For

n = 0

,

(0 + 1)^3 - (0 - 1)^3 = 1 - (-1) = 2

(even). For

n = 1

,

(1 + 1)^3 - (1 - 1)^3 = 8 - 0 = 8

(even). Because

n + 1

and

n - 1

always have the same parity, their cubes also share the same parity, so their difference is always even.

Tahmini Süre:1m 30s

Soru 20Soru →

If $a$ and $b$ are integers such that $a^2b + a$ is an odd integer, which of the following expressions must be an even integer?

a + b

a^2 - b

a + b + 1

ab + a

a^2 + b^2

Cevabı ve açıklamayı göster

Cevap: $a + b + 1$

Cevap

The expression

a + b + 1

must be an even integer.

Factoring

a^2b + a

gives

a(ab + 1)

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

a

is odd and

ab + 1

is odd. If

ab + 1

is odd, then

ab

must be even. Since

a

is odd,

b

must be even. Evaluating the expression

a + b + 1

:

a

(odd) +

b

(even) + 1 (odd) equals an even integer.

Adım Adım Çözüm

1

Factor the given algebraic expression to analyze its parity components.

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

.

Factoring allows us to analyze the parity of individual factors whose product is given as odd.

2

Determine the parity of

a

and the factor

(ab + 1)

.

Both

a

and

(ab + 1)

must be odd integers.

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

3

Determine the parity of

b

using the parities established in Step 2.

Since

ab + 1

is odd,

ab

must be even. Because

a

is odd,

b

must be even (including zero).

An odd number multiplied by an even number yields an even number, and adding 1 converts it to an odd number.

4

Evaluate the parity of

a + b + 1

.

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

.

Adding two odd numbers (

a

and 1) together with an even number (

b

) results in an even integer.

Anahtar Kavram

Parity rules for integer addition and multiplication: Odd × Odd = Odd, Odd × Even = Even, Odd + Even = Odd, Odd + Odd = Even.

Number Properties and Arithmetic

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Alternatif Yöntem

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