Number Properties and Arithmetic Practice Questions — GMAT — Page 4

Question 61Question →

For distinct prime numbers $p$ and $q$ , two positive integers are defined as $a = p^3 q^2$ and $b = p^2 q^4$ . What is the ratio of the least common multiple of $a$ and $b$ to the greatest common divisor of $a$ and $b$ ?

p q^2

p^3 q^4

p^2 q^2

p^5 q^6

p q

Show answer & explanation

Answer: $p q^2$

Answer

p q^2

The greatest common divisor of

a = p^3 q^2

and

b = p^2 q^4

is

p^2 q^2

by taking the minimum exponent for each prime factor. The least common multiple is

p^3 q^4

by taking the maximum exponent for each prime factor. Dividing the LCM by the GCD yields

\frac{p^3 q^4}{p^2 q^2} = p q^2

.

Step-by-Step Solution

1

Find the prime factorization representation of the greatest common divisor (GCD) of

a

and

b

.

\text{GCD}(a, b) = p^{\min(3, 2)} q^{\min(2, 4)} = p^2 q^2

The GCD taking prime factorizations uses the minimum power of each common prime factor.

2

Find the prime factorization representation of the least common multiple (LCM) of

a

and

b

.

\text{LCM}(a, b) = p^{\max(3, 2)} q^{\max(2, 4)} = p^3 q^4

The LCM taking prime factorizations uses the maximum power of each prime factor.

3

Compute the ratio of the LCM to the GCD.

\frac{\text{LCM}(a, b)}{\text{GCD}(a, b)} = \frac{p^3 q^4}{p^2 q^2} = p^{3-2} q^{4-2} = p q^2

Dividing exponential terms with the same base requires subtracting their exponents.

Key Concept

Calculating GCD and LCM using prime factor exponent rules and evaluating their ratio

Question 62Question →

What is the units digit of $8^{21}$ ?

8

4

2

6

0

Show answer & explanation

Answer: 8

Answer

8

The units digits of integer powers of 8 follow a repeating pattern of length 4: 8, 4, 2, 6. Dividing the exponent 21 by 4 yields

21 = 4 \times 5 + 1

, giving a remainder of 1. A remainder of 1 means the units digit is the first element of the cycle, which is 8.

Step-by-Step Solution

1

Determine the units digit pattern for powers of 8.

The sequence of units digits for

8^1, 8^2, 8^3, 8^4, \dots

is

8, 4, 2, 6, 8, 4, 2, 6, \dots

, which repeats every 4 powers.

Units digits of positive integer powers follow a periodic cyclic pattern.

2

Find the remainder when the exponent 21 is divided by the cycle length 4.

21 \div 4 = 5

with a remainder of 1.

The remainder determines the position of the units digit within the 4-step cycle.

3

Identify the units digit corresponding to a remainder of 1.

The first number in the pattern

8, 4, 2, 6

is 8.

A remainder of 1 points to the 1st term of the repeating sequence.

Key Concept

Units Digit Cyclicity

Question 63Question →

For three positive integers $x$ , $y$ , and $z$ , all of whose prime factors belong exclusively to the set $\{2, 3, 5\}$ , their pairwise greatest common divisors are given by $\gcd(x, y) = 60$ , $\gcd(y, z) = 90$ , and $\gcd(x, z) = 150$ . What is the minimum possible value of the sum $x + y + z$ ?

300

465

930

1,350

2,700

Show answer & explanation

Answer: $930$

Answer

The minimum possible value of the sum

x + y + z

is

930

.

Analyzing the prime factorization exponents for each prime factor

p \in \{2, 3, 5\}

establishes minimum lower bounds on exponents. For prime

2

,

\min(x_2, y_2) = 2

forces

x_2 \ge 2

and

y_2 \ge 2

, which in turn forces

z_2 = 1

to satisfy

\min(y_2, z_2) = 1

. Similarly, for prime

3

,

y_3 \ge 2

and

z_3 \ge 2

forces

x_3 = 1

. For prime

5

,

x_5 \ge 2

and

z_5 \ge 2

forces

y_5 = 1

. Evaluating

x = 2^2 \cdot 3^1 \cdot 5^2 = 300

,

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

, and

z = 2^1 \cdot 3^2 \cdot 5^2 = 450

gives

x + y + z = 930

.

Step-by-Step Solution

1

Express the given pairwise GCDs in their prime factorized forms.

\gcd(x,y) = 60 = 2^2 \times 3^1 \times 5^1

\gcd(y,z) = 90 = 2^1 \times 3^2 \times 5^1

\gcd(x,z) = 150 = 2^1 \times 3^1 \times 5^2

The exponent of any prime

p

in

\gcd(a,b)

is

\min(a_p, b_p)

.

2

Determine the minimum required prime exponents for prime

2

.

\min(x_2, y_2) = 2

,

\min(y_2, z_2) = 1

,

\min(x_2, z_2) = 1 \implies x_2 = 2, y_2 = 2, z_2 = 1

Since

\min(x_2, y_2) = 2

, both

x_2 \ge 2

and

y_2 \ge 2

. To satisfy

\min(y_2, z_2) = 1

, we must have

z_2 = 1

. The minimal exponents are

x_2=2, y_2=2, z_2=1

.

3

Determine the minimum required prime exponents for prime

3

.

\min(x_3, y_3) = 1

,

\min(y_3, z_3) = 2

,

\min(x_3, z_3) = 1 \implies x_3 = 1, y_3 = 2, z_3 = 2

Since

\min(y_3, z_3) = 2

, both

y_3 \ge 2

and

z_3 \ge 2

. To satisfy

\min(x_3, y_3) = 1

, we must have

x_3 = 1

. The minimal exponents are

x_3=1, y_3=2, z_3=2

.

4

Determine the minimum required prime exponents for prime

5

.

\min(x_5, y_5) = 1

,

\min(y_5, z_5) = 1

,

\min(x_5, z_5) = 2 \implies x_5 = 2, y_5 = 1, z_5 = 2

Since

\min(x_5, z_5) = 2

, both

x_5 \ge 2

and

z_5 \ge 2

. To satisfy

\min(x_5, y_5) = 1

, we must have

y_5 = 1

. The minimal exponents are

x_5=2, y_5=1, z_5=2

.

5

Calculate the values of

x

,

y

, and

z

and find their sum.

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

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

z = 2^1 \times 3^2 \times 5^2 = 450

x + y + z = 300 + 180 + 450 = 930

Summing the minimal values yields the required minimum total.

Key Concept

Finding integers from pairwise GCD conditions using prime factor exponent minimums:

\text{v}_p(\gcd(a,b)) = \min(\text{v}_p(a), \text{v}_p(b))

.

Estimated Time:2m 30s

Question 64Question →

Three automated security beacons flash at regular intervals of $18$ seconds, $24$ seconds, and $42$ seconds, respectively. If all three beacons flash simultaneously at 8:00:00 AM, how many times will all three beacons flash simultaneously between 8:01:00 AM and 9:00:00 AM, inclusive?

Show answer & explanation

Answer: 7

Answer

7

The least common multiple of 18, 24, and 42 is 504 seconds. The timeframe between 8:01:00 AM and 9:00:00 AM inclusive represents elapsed times from 60 seconds to 3600 seconds after 8:00:00 AM. Dividing 3600 by 504 gives 7 full periods (3528 seconds), all of which are greater than 60 seconds. Therefore, the beacons flash simultaneously 7 times within the specified period.

Step-by-Step Solution

1

Find the Least Common Multiple (LCM) of the three flash intervals.

LCM(18, 24, 42) = 504 seconds

The beacons flash simultaneously at time intervals that are integer multiples of the LCM of their individual cycle times.

2

Convert the target time interval (8:01:00 AM to 9:00:00 AM) into seconds after the initial flash at 8:00:00 AM (t = 0).

Range is from t = 60 seconds to t = 3600 seconds

8:01:00 AM is 60 seconds after 8:00:00 AM, and 1 hour (9:00:00 AM) equals 3600 seconds after 8:00:00 AM.

3

Determine the number of positive multiples of 504 within the interval [60, 3600].

7 multiples (504, 1008, 1512, 2016, 2520, 3024, 3528)

Dividing 3600 by 504 gives 7 with a remainder of 72. Since the first positive multiple (504) is greater than 60, all 7 multiples fall strictly within the specified window.

Key Concept

Application of Least Common Multiple (LCM) to periodic events

Question 65Question →

Two positive integers $m$ and $n$ satisfy $m < n$ . Their greatest common divisor is $18$ and their least common multiple is $1,080$ . If $m$ is not divisible by $27$ and $n$ is not divisible by $5$ , what is the value of $n - m$ ?

Show answer & explanation

Answer: 126

Answer

The value of

n - m

is

126

.

By writing

m = 18a

and

n = 18b

with

\text{GCD}(a,b) = 1

, we find

a \cdot b = 60

. Evaluating all coprime factor pairs

(a, b)

shows that only

(5, 12)

satisfies all constraints (

m = 90

is not a multiple of

27

, and

n = 216

is not a multiple of

5

). The difference is

216 - 90 = 126

.

Step-by-Step Solution

1

Relate GCD, LCM, and coprime factors.

Let

m = 18a

and

n = 18b

, where

\text{GCD}(a, b) = 1

and

a < b

. Then

a \cdot b = \frac{1080}{18} = 60

.

The product of two positive integers equals the product of their GCD and LCM.

2

Determine all coprime factor pairs of

60

.

The coprime pairs

(a, b)

with

a < b

such that

a \cdot b = 60

are

(1, 60)

,

(3, 20)

,

(4, 15)

, and

(5, 12)

.

Since

\text{GCD}(a, b) = 1

, prime factors of

60 = 2^2 \cdot 3 \cdot 5

cannot be shared between

a

and

b

.

3

Apply the given constraints to eliminate invalid pairs.

The pair

(5, 12)

is the only pair satisfying both conditions:

m = 90

is not divisible by

27

, and

n = 216

is not divisible by

5

.

Pairs

(1,60)

and

(4,15)

yield

n

divisible by

5

; pair

(3,20)

yields

m=54

which is divisible by

27

.

4

Compute the final difference

n - m

.

n - m = 216 - 90 = 126

.

Subtract

m

from

n

using the valid values found.

Key Concept

Properties of GCD and LCM, prime factorization, and coprime factor pairs

Estimated Time:2m 0s

Question 66Question →

A clothing store reduced the original price of a jacket by $20\%$ . During a promotional event, the store offered an additional $15\%$ discount off the reduced price. If the final price of the jacket was $\$102$ , what was the original price of the jacket, in dollars?

Show answer & explanation

Answer: 150

Answer

The original price of the jacket was $150.

To find the original price, express the final price as a decimal multiplier of the original price. A

20\%

discount leaves

80\%

of the original price (

0.80

). An additional

15\%

discount off the reduced price leaves

85\%

of that amount (

0.85

). The overall price multiplier is

0.80 \times 0.85 = 0.68

. Setting

0.68 \times \text{Original Price} = 102

gives an original price of

\frac{102}{0.68} = 150

dollars.

Step-by-Step Solution

1

Determine the combined price multiplier after both successive discounts.

The first discount leaves

80\%

(

0.80

) of the original price, and the second discount leaves

85\%

(

0.85

) of that reduced price. The combined multiplier is

0.80 \times 0.85 = 0.68

.

Successive percentage discounts are multiplicative rather than additive.

2

Set up an equation equating the final discounted price to

102

.

0.68 \times P = 102

, where

P

represents the original price.

The final price is

68\%

of the original base price.

3

Solve for the original price

P

.

P = \frac{102}{0.68} = \frac{10200}{68} = 150

.

Dividing the final amount by the net decimal multiplier yields the original value.

Key Concept

Successive percentage change calculation

Estimated Time:1m 0s

Question 67Question →

A wholesale distributor imported a batch of organic green tea leaves. On Monday, $30\%$ of the initial batch was sold. On Tuesday, $\frac{4}{7}$ of the remaining batch was sold. If the distributor had $126$ kilograms of green tea leaves left at the end of Tuesday, what was the total weight, in kilograms, of the initial batch imported?

294

315

420

700

980

Show answer & explanation

Answer: $420$

Answer

The total weight of the initial batch imported was

420

kilograms.

Selling

30\%

on Monday leaves

70\%

of the initial total. On Tuesday, selling

\frac{4}{7}

of that remaining amount leaves

\frac{3}{7}

of the

70\%

. Calculating

\frac{3}{7} \times 70\% = 30\%

. Since

30\%

of the initial batch is equal to

126

kg, the total initial weight is

126 \div 0.30 = 420

kg.

Step-by-Step Solution

1

Determine the fraction of the initial batch remaining after Monday's sale.

Since

30\%

(

0.30

) was sold,

1 - 0.30 = 0.70

(or

\frac{7}{10}

) of the initial batch remained.

Percentage decreases are calculated relative to the original whole.

2

Determine the fraction of Monday's remaining batch that was left after Tuesday's sale.

Since

\frac{4}{7}

of Monday's remainder was sold,

1 - \frac{4}{7} = \frac{3}{7}

of Monday's remainder was left.

Subsequent fractional sales are relative to the updated intermediate amount.

3

Calculate the final remaining amount as a fraction of the initial total weight

W

.

\text{Final Remaining} = \frac{3}{7} \times \left(\frac{7}{10} W\right) = \frac{3}{10} W = 0.30 W$.

Multiplying successive remaining ratios gives the net remaining fraction of the original batch.

4

Solve for the initial weight

W

using the given remaining weight of

126

kg.

0.30 W = 126 \implies W = \frac{126}{0.30} = 420\text{ kg}.

Dividing the remaining weight by its corresponding decimal fraction yields the total initial weight.

Key Concept

Successive percentage and fractional reductions require applying each change to the updated intermediate base value rather than the original total.

Estimated Time:1m 30s

Question 68Question →

Let $M$ be a positive odd integer that is divisible by $45$ but not by $27$ . If $M$ has exactly $36$ positive divisors, what is the maximum possible number of distinct prime factors of $M$ ?

3

4

5

6

7

Show answer & explanation

Answer: 4

Answer

4

The number of positive divisors of an integer with prime factorization

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)

. Because

M

is divisible by

45 = 3^2 \times 5

but not by

27 = 3^3

, the prime factor

3

must have an exponent of exactly

2

, which contributes a multiplier of

(2+1) = 3

to the total divisor count. Dividing the total

36

divisors by

3

leaves a product of

12

for the remaining terms

(e_i+1)

. To maximize the number of distinct prime factors, we write

12

as a product of as many integers greater than

1

as possible, which is

3 \times 2 \times 2

(3 factors). Adding the prime factor

3

gives a maximum of

1 + 3 = 4

distinct prime factors.

Step-by-Step Solution

1

Determine the prime factorization constraints from the given conditions.

M = 3^2 \cdot 5^a \cdot p_1^{b_1} \cdot p_2^{b_2} \cdots p_r^{b_r}

, where

a \ge 1

,

b_i \ge 1

, and

p_i

are distinct odd primes other than

3

and

5

.

Since

M

is odd,

2

is not a prime factor. Since

M

is divisible by

45 = 3^2 \times 5

but not by

27 = 3^3

, the exponent of

3

must be exactly

2

, and the exponent of

5

is at least

1

.

2

Set up the formula for the number of positive divisors.

f(M) = (2 + 1)(a + 1)(b_1 + 1)(b_2 + 1) \cdots (b_r + 1) = 36

, which simplifies to

(a + 1)(b_1 + 1)(b_2 + 1) \cdots (b_r + 1) = 12

.

The number of divisors of a number

n = q_1^{e_1} q_2^{e_2} \cdots q_k^{e_k}

is

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

.

3

Maximize the number of terms in the product yielding

12

.

The maximum number of factors greater than

1

whose product is

12

is

3

, since

12 = 3 \times 2 \times 2

.

Each factor greater than

1

in

(a+1)(b_1+1)\dots(b_r+1)

corresponds to a distinct prime factor of

M

(other than

3

).

4

Calculate the maximum total number of distinct prime factors of

M

.

Total distinct prime factors =

1 \text{ (for 3)} + 3 \text{ (from the factorization of 12)} = 4

.

The prime factors are

3

,

5

,

p_1

, and

p_2

, giving

4

distinct prime factors in total.

Key Concept

Divisor Count Formula and Prime Factorization Constraints

Estimated Time:2m 0s

Question 69Question →

What is the remainder when the sum $S = 3^1 + 3^2 + 3^3 + \dots + 3^{2025}$ is divided by $100$ ?

Show answer & explanation

Answer: 63

Answer

The remainder when the sum

S

is divided by

100

is

63

.

Using the geometric series formula,

S = \frac{3(3^{2025} - 1)}{2}

. Computing

3^{2025} \pmod{200}

gives

43 \pmod{200}

because

3^{20} \equiv 1 \pmod{200}

and

2025 \equiv 5 \pmod{20}

. Substituting this into the numerator yields

3(43 - 1) = 126 \pmod{200}

. Dividing by

2

produces the remainder

63

modulo

100

.

Step-by-Step Solution

1

Sum the geometric series

S = 3^1 + 3^2 + \dots + 3^{2025}

S = \frac{3(3^{2025} - 1)}{2}

Applying the standard finite geometric series formula

a\frac{r^n - 1}{r - 1}

with

a = 3

and

r = 3

.

2

Set up modular equation for

S \pmod{100}

Evaluate

3(3^{2025} - 1) \pmod{200}

Dividing an integer congruence by

2

requires doubling the modulus to

200

to maintain exact divisibility.

3

Find the cyclicity period of

3^n \pmod{200}

3^{20} \equiv 1 \pmod{200}

3^5 = 243 \equiv 43

,

3^{10} \equiv 43^2 = 1849 \equiv 49

, and

3^{20} \equiv 49^2 = 2401 \equiv 1 \pmod{200}

.

4

Compute

3^{2025} \pmod{200}

3^{2025} \equiv 3^5 \equiv 43 \pmod{200}

Dividing the exponent

2025

by the period length

20

leaves a remainder of

5

.

5

Calculate the numerator

3(3^{2025} - 1) \pmod{200}

and divide by 2

3(43 - 1) = 126 \pmod{200}

, so

S \equiv 63 \pmod{100}

Simplifying

126 / 2 = 63

gives the unique remainder modulo

100

.

Key Concept

Geometric series summation combined with modular arithmetic and cyclicity mod 200

Question 70Question →

At the beginning of the year, a logistics company's fleet consisted of trucks, vans, and cargo planes. Exactly $0.40$ of the total fleet were trucks, and $\frac{3}{8}$ of the total fleet were vans, with the remaining vehicles being cargo planes. Over the course of the year, the number of trucks increased by $25\%$ , the number of vans decreased by $20\%$ , and the number of cargo planes increased by $50\%$ . By what percentage did the total number of vehicles in the company's fleet increase over the year?

11.25\%

13.75\%

15.00\%

18.33\%

55.00\%

Show answer & explanation

Answer: $13.75\%$

Answer

The total number of vehicles in the fleet increased by

13.75\%

.

To find the net percentage increase of the entire fleet, we express each vehicle subgroup as a fraction or decimal of the initial total fleet

T

. Trucks account for

0.40T

, vans account for

\frac{3}{8}T = 0.375T

, and cargo planes account for the remaining

1 - (0.40 + 0.375) = 0.225T

. Applying the respective percentage changes gives an increase of

0.25 \times 0.40T = +0.10T

for trucks, a decrease of

0.20 \times 0.375T = -0.075T

for vans, and an increase of

0.50 \times 0.225T = +0.1125T

for cargo planes. Summing these changes yields a net gain of

0.10T - 0.075T + 0.1125T = 0.1375T

, which corresponds to an overall increase of

13.75\%

.

Step-by-Step Solution

1

Determine the initial proportions of each vehicle type in the fleet.

Trucks represent

0.40

of the fleet. Vans represent

\frac{3}{8} = 0.375

of the fleet. Cargo planes represent

1 - (0.40 + 0.375) = 0.225

of the fleet.

The sum of all component proportions must equal

1.0

(or

100\%

of the initial fleet).

2

Calculate the weighted net change contributed by each vehicle type relative to the initial fleet size

T

.

Truck change:

+25\% \text{ of } 0.40T = 0.25 \times 0.40T = +0.100T

.
Van change:

-20\% \text{ of } 0.375T = -0.20 \times 0.375T = -0.075T

.
Cargo plane change:

+50\% \text{ of } 0.225T = 0.50 \times 0.225T = +0.1125T

.

The net change contributed by a subgroup is its relative proportion multiplied by its specific percentage change.

3

Sum the component changes to find the total overall change.

Total net change

= +0.100T - 0.075T + 0.1125T = +0.1375T

.

Combining the individual net contributions yields the overall change in fleet size.

4

Convert the decimal net change to a percentage.

0.1375 \times 100\% = 13.75\%

.

Multiplying the decimal fraction of total increase by

100\%

gives the overall percentage increase.

Key Concept

Weighted Percentage Changes

Alternative Method

Assume a concrete total initial fleet size that works easily with the numbers, such as

T = 800

vehicles. Initial trucks

= 0.40 \times 800 = 320

. Initial vans

= \frac{3}{8} \times 800 = 300

. Initial planes

= 800 - 320 - 300 = 180

. After changes: new trucks

= 320 \times 1.25 = 400

, new vans

= 300 \times 0.80 = 240

, new planes

= 180 \times 1.50 = 270

. New total fleet

= 400 + 240 + 270 = 910

. Fleet increase

= 910 - 800 = 110

. Percentage increase

= \frac{110}{800} \times 100\% = 13.75\%

.

Estimated Time:2m 0s

Question 71Question →

At a logistics distribution hub, a shipment of incoming packages was processed over two shifts. During the morning shift, $0.35$ of the total shipment was processed and dispatched. During the evening shift, $\frac{4}{13}$ of the remaining packages were processed. If $450$ packages remained unprocessed at the end of both shifts, what was the total number of packages in the initial shipment?

Show answer & explanation

Answer: 1000

Answer

1000 packages

To find the initial shipment size, first express

0.35

as the fraction

\frac{7}{20}

. Subtracting this from

1

leaves

\frac{13}{20}

of the initial total after the morning shift. Next, processing

\frac{4}{13}

of these remaining packages leaves

1 - \frac{4}{13} = \frac{9}{13}

of that remaining amount. Multiplying

\frac{9}{13} \times \frac{13}{20}

shows that

\frac{9}{20}

of the original total remains unprocessed. Finally, setting

\frac{9}{20}

of the total equal to

450

packages yields an initial total of

1000

packages.

Step-by-Step Solution

1

Convert decimal portion to fraction and find remaining fraction after morning shift

Remaining fraction after morning shift is

\frac{13}{20}

Since

0.35 = \frac{7}{20}

of the total shipment was processed,

1 - \frac{7}{20} = \frac{13}{20}

of the initial shipment remained.

2

Determine the remaining fraction of the shipment after the evening shift

Final remaining fraction of the initial shipment is

\frac{9}{20}

The evening shift processed

\frac{4}{13}

of the remaining packages, leaving

1 - \frac{4}{13} = \frac{9}{13}

of the remaining packages. Thus,

\frac{9}{13} \times \frac{13}{20} = \frac{9}{20}

of the initial shipment remained.

3

Solve for the total initial number of packages

Total initial packages =

1000

Setting

\frac{9}{20} N = 450

gives

N = 450 \times \frac{20}{9} = 1000

.

Key Concept

Combining decimal-to-fraction conversions with successive remaining fraction calculations

Question 72Question →

If $N = 3^{12} - 1$ , what is the sum of all the distinct prime factors of $N$ ?

Show answer & explanation

Answer: 100

Answer

The sum of all the distinct prime factors of

N

is 100.

Factoring

3^{12} - 1

via difference of squares and sum of cubes yields

(3^6 - 1)(3^6 + 1) = (2^3 \times 7 \times 13)(2 \times 5 \times 73) = 2^4 \times 5 \times 7 \times 13 \times 73

. The distinct prime factors are 2, 5, 7, 13, and 73, which sum to 100.

Step-by-Step Solution

1

Decompose

N = 3^{12} - 1

using the difference of squares identity

a^2 - b^2 = (a - b)(a + b)

.

N = (3^6 - 1)(3^6 + 1)

Breaking down large powers of integers into products of smaller terms allows for systematic prime factor determination.

2

Completely factor the term

(3^6 - 1)

.

3^6 - 1 = (3^3 - 1)(3^3 + 1) = 26 \times 28 = (2 \times 13) \times (2^2 \times 7) = 2^3 \times 7 \times 13

Applying the difference of squares identity iteratively converts the term into small arithmetic integers with obvious prime factorizations.

3

Completely factor the term

(3^6 + 1)

using the sum of cubes identity

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

.

3^6 + 1 = (3^2)^3 + 1 = (3^2 + 1)((3^2)^2 - 3^2 + 1) = 10 \times 73 = 2 \times 5 \times 73

The factor 73 is prime because it is not divisible by any prime numbers less than or equal to

\sqrt{73} \approx 8.54

(namely 2, 3, 5, and 7).

4

Combine the component prime factorizations to list all distinct prime factors of

N

.

N = 2^4 \times 5 \times 7 \times 13 \times 73

, so the set of distinct prime factors is \{2, 5, 7, 13, 73\}.

Each prime base is included exactly once regardless of its exponent.

5

Sum the distinct prime factors.

2 + 5 + 7 + 13 + 73 = 100

Adding the unique prime factors yields the requested value.

Key Concept

Prime Factorization using Algebraic Polynomial Identities

Estimated Time:2m 30s

Question 73Question →

A commercial bakery uses flour, sugar, and butter as the main ingredients by weight to produce a specialized pastry mix. Flour accounts for $0.50$ of the total weight of the mix. Sugar accounts for $\frac{3}{8}$ of the remaining weight of the mix. If the rest of the mix consists of $15$ kilograms of butter, what is the total weight, in kilograms, of the pastry mix?

24

48

60

80

120

Show answer & explanation

Answer: $48$

Answer

The total weight of the pastry mix is

48

kilograms.

The option stating

48

is correct because flour leaves

\frac{1}{2}

of the total mix. Sugar takes

\frac{3}{8}

of that half, leaving

\frac{5}{8}

of the half for butter. Thus, butter accounts for

\frac{5}{16}

of the total mix. Solving

\frac{5}{16} \times W = 15

yields

W = 48

kg.

Step-by-Step Solution

1

Determine the fraction of the total weight remaining after accounting for flour.

Since flour represents

0.50 = \frac{1}{2}

of the total weight

W

, the remaining weight is

1 - \frac{1}{2} = \frac{1}{2}W

.

The base for the sugar component is specified as the remaining weight after flour.

2

Calculate the fraction of the total weight represented by butter.

Sugar takes

\frac{3}{8}

of the remaining weight, so butter takes the remaining

1 - \frac{3}{8} = \frac{5}{8}

of the remaining weight. Therefore, butter is

\frac{5}{8} \times \frac{1}{2}W = \frac{5}{16}W

.

Butter forms the rest of the mixture after sugar is accounted for within the remaining weight.

3

Set up an equation with the given butter weight to solve for total weight

W

.

\frac{5}{16}W = 15 \implies W = 15 \times \frac{16}{5} = 3 \times 16 = 48$ kg.

Equating the algebraic fractional expression to the known numerical weight yields the overall total.

Key Concept

Multi-step successive fraction arithmetic with changing base values

Question 74Question →

Let $n = 13^{25} - 7^{25}$ . When $n$ is divided by $10$ , the remainder is $R_1$ , where $0 \le R_1 < 10$ . When $-n$ is divided by $7$ , the remainder is $R_2$ , where $0 \le R_2 < 7$ . What is the value of $R_1 + R_2$ ?

5

7

10

12

2

Show answer & explanation

Answer: 7

Answer

The value of

R_1 + R_2

is 7.

To find

R_1

, we evaluate

n = 13^{25} - 7^{25} \pmod{10}

. Both 3 and 7 have units digit cyclicity of period 4. Because

25 \equiv 1 \pmod 4

,

13^{25} \equiv 3^1 = 3 \pmod{10}

and

7^{25} \equiv 7^1 = 7 \pmod{10}

. Then

n \equiv 3 - 7 = -4 \equiv 6 \pmod{10}

, so

R_1 = 6

. To find

R_2

, we evaluate

-n = 7^{25} - 13^{25} \pmod 7

. Since

7^{25} \equiv 0 \pmod 7

and

13 \equiv -1 \pmod 7

, we have

-n \equiv 0 - (-1)^{25} = 1 \pmod 7

, so

R_2 = 1

. Summing them yields

R_1 + R_2 = 6 + 1 = 7

.

Step-by-Step Solution

1

Determine

R_1

, the remainder when

n = 13^{25} - 7^{25}

is divided by 10.

R_1 = 6

Working modulo 10,

13 \equiv 3 \pmod{10}

, so

13^{25} \equiv 3^{25} \pmod{10}

. Powers of 3 follow a units digit cycle of period 4 (

3, 9, 7, 1

). Since

25 = 4 \times 6 + 1

,

3^{25} \equiv 3^1 = 3 \pmod{10}

. Similarly, powers of 7 follow a units digit cycle of period 4 (

7, 9, 3, 1

), so

7^{25} \equiv 7^1 = 7 \pmod{10}

. Thus,

n \equiv 3 - 7 = -4 \equiv 6 \pmod{10}

, yielding

R_1 = 6

.

2

Determine

R_2

, the remainder when

-n = 7^{25} - 13^{25}

is divided by 7.

R_2 = 1

Working modulo 7,

7^{25} \equiv 0 \pmod 7

. Since

13 \equiv -1 \pmod 7

,

13^{25} \equiv (-1)^{25} = -1 \pmod 7

. Substituting these gives

-n = 7^{25} - 13^{25} \equiv 0 - (-1) = 1 \pmod 7

. Since

0 \le 1 < 7

,

R_2 = 1

.

3

Calculate the sum

R_1 + R_2

.

6 + 1 = 7

Adding the two non-negative remainders gives

R_1 + R_2 = 6 + 1 = 7

.

Key Concept

Modular arithmetic, units digit cyclicity, and converting negative integer remainders to standard non-negative remainders.

Question 75Question →

Let $n$ be a positive integer with exactly two distinct prime factors. If $n$ is a multiple of $12$ , is not a multiple of $8$ , and has exactly $12$ positive factors, what is the number of positive factors of $n^2$ ?

24

33

35

45

144

Show answer & explanation

Answer: 35

Answer

The number of positive factors of

n^2

is 35.

Since

n

is a multiple of

12 = 2^2 \cdot 3^1

and has exactly two distinct prime factors, its prime factorization must be

n = 2^a \cdot 3^b

. The condition that

n

is not a multiple of

8 = 2^3

forces

a = 2

. Using the factor count formula

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

, we get

(2+1)(b+1) = 12

, which gives

b = 3

. Therefore,

n^2 = (2^2 \cdot 3^3)^2 = 2^4 \cdot 3^6

. The total number of positive factors of

n^2

is

(4+1)(6+1) = 35

.

Step-by-Step Solution

1

Determine the prime factors of

n

n = 2^a \cdot 3^b

, where

a \geq 2

and

b \geq 1

Since

n

has exactly two distinct prime factors and

12 = 2^2 \cdot 3^1

divides

n

, the only prime factors of

n

are

2

and

3

.

2

Apply the divisibility constraints on the exponent

a

a = 2

Since

12

divides

n

,

a \geq 2

. Since

n

is not a multiple of

8 = 2^3

,

a < 3

. Thus

a

must equal

2

.

3

Use the factor count formula to find

b

b = 3

, so

n = 2^2 \cdot 3^3

The number of positive factors of

n

is

(a+1)(b+1) = (2+1)(b+1) = 12 \implies 3(b+1) = 12 \implies b = 3

.

4

Calculate the number of positive factors of

n^2

(4+1)(6+1) = 35

Squaring

n

gives

n^2 = (2^2 \cdot 3^3)^2 = 2^4 \cdot 3^6

. By the factor count formula,

n^2

has

(4+1)(6+1) = 35

positive factors.

Key Concept

Factor Count Formula & Prime Factorization Constraints

Estimated Time:1m 45s

Question 76Question →

If $N = 6^6 + 6^7 + 6^8 + 6^9$ , what is the sum of the exponents of all prime factors in the prime factorization of $N$ ?

14

30

15

8

13

Show answer & explanation

Answer: 14

Answer

14

Factoring out

6^6

gives

N = 6^6(1 + 6 + 36 + 216) = 6^6(259)

. Decomposing composite bases into prime factors yields

6^6 = 2^6 \times 3^6

and

259 = 7^1 \times 37^1

. Thus, the prime factorization is

N = 2^6 \times 3^6 \times 7^1 \times 37^1

. Summing the prime exponents gives

6 + 6 + 1 + 1 = 14

.

Step-by-Step Solution

1

Factor out the common term

6^6

from the expression.

N = 6^6(1 + 6 + 6^2 + 6^3) = 6^6(1 + 6 + 36 + 216) = 6^6 \times 259

Factoring simplifies the sum into a single product of terms.

2

Express

6^6

in terms of its prime factors.

6^6 = (2 \times 3)^6 = 2^6 \times 3^6

The base 6 is composite and must be broken down into prime factors 2 and 3.

3

Determine the prime factorization of 259.

Testing small primes shows

259 = 7 \times 37

, where both 7 and 37 are prime numbers.

259 is not prime and must be decomposed into its prime components

7^1 \times 37^1

.

4

Combine all prime factors to write the complete prime factorization of

N

.

N = 2^6 \times 3^6 \times 7^1 \times 37^1

Writing

N

in standard canonical form reveals all prime exponents.

5

Sum the exponents of all prime factors.

6 + 6 + 1 + 1 = 14

The question asks for the total sum of the exponents of the prime factors.

Key Concept

Prime Factorization of Factored Exponential Sums

Estimated Time:2m 30s

Question 77Question →

At the start of a month, an online bookstore's inventory consisted of Fiction, Non-Fiction, and Textbook titles. Exactly $\frac{3}{8}$ of the total inventory consisted of Fiction titles, and $0.40$ of the remaining inventory consisted of Non-Fiction titles, with the balance consisting of Textbook titles. During the month, the number of Fiction titles increased by $20\%$ , the number of Non-Fiction titles decreased by $25\%$ , and the number of Textbook titles remained unchanged. If the total inventory increased by a net amount of $15$ titles at the end of the month, how many total book titles were in the inventory at the start of the month?

400

600

800

1,200

2,400

Show answer & explanation

Answer: 1,200

Answer

1,200 total book titles

The initial inventory consists of

\frac{3}{8}T

Fiction titles and

0.40 \times \frac{5}{8}T = \frac{1}{4}T

Non-Fiction titles. A

20\%

increase in Fiction adds

\frac{3}{40}T

, while a

25\%

decrease in Non-Fiction subtracts

\frac{1}{16}T

. Combining these gives a net change of

\frac{3}{40}T - \frac{1}{16}T = \frac{6}{80}T - \frac{5}{80}T = \frac{1}{80}T

. Setting

\frac{1}{80}T = 15

gives

T = 1,200

.

Step-by-Step Solution

1

Determine the initial fractional breakdown of each book category relative to the total inventory T

Fiction

= \frac{3}{8}T

. Remaining inventory

= 1 - \frac{3}{8} = \frac{5}{8}T

. Non-Fiction

= 0.40 \times \frac{5}{8}T = \frac{2}{5} \times \frac{5}{8}T = \frac{1}{4}T

.

The question specifies that Non-Fiction is 0.40 of the remaining inventory, not the total inventory.

2

Calculate the net change in titles as a fraction of total initial inventory T

Increase in Fiction

= 20\% \times \frac{3}{8}T = \frac{1}{5} \times \frac{3}{8}T = +\frac{3}{40}T

.
Decrease in Non-Fiction

= 25\% \times \frac{1}{4}T = \frac{1}{4} \times \frac{1}{4}T = -\frac{1}{16}T

.
Net fractional change

= \frac{3}{40}T - \frac{1}{16}T = \frac{6}{80}T - \frac{5}{80}T = +\frac{1}{80}T

.

Percentage changes must be applied to each category's specific share of the total inventory.

3

Equate the net fractional change to the numerical net increase and solve for T

\frac{1}{80}T = 15 \implies T = 15 \times 80 = 1,200.

The overall net increase is given as 15 titles.

Key Concept

Multi-step successive fraction and percentage change with changing base values

Estimated Time:2m 0s

Question 78Question →

What is the value of $(2^3 + 2^3)^2$ ?

2^4

2^7

2^8

2^{12}

2^{18}

Show answer & explanation

Answer: $2^8$

Answer

The correct answer is

2^8

.

Combining the terms inside the parentheses gives

2^3 + 2^3 = 2 \cdot 2^3 = 2^4

. Squaring

2^4

using the power rule

(a^m)^n = a^{m \cdot n}

produces

2^{4 \cdot 2} = 2^8

.

Step-by-Step Solution

1

Simplify the expression inside the parentheses

2^3 + 2^3 = 2 \cdot (2^3) = 2^1 \cdot 2^3 = 2^4

Adding two identical terms is equivalent to multiplying the term by 2, which allows combining powers of the same base.

2

Apply the power of a power exponent rule to the outer square

(2^4)^2 = 2^{4 \times 2} = 2^8

When raising a power to another power, multiply the exponents:

(a^m)^n = a^{m \cdot n}

.

Key Concept

Combining like exponential terms and applying power rules

Estimated Time:45s

Question 79Question →

For positive integers $a$ , $b$ , and $c$ , let $x = 2^a \times 3^3 \times 5^b$ and $y = 2^3 \times 3^c \times 5^1$ . If $\text{GCD}(x, y) = 360$ and $\text{LCM}(x, y) = 270,000$ , what is the value of $a + b + c$ ?

8

9

10

11

12

Show answer & explanation

Answer: 10

Answer

10

First, express the given GCD and LCM in prime factor form:

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

and

270,000 = 2^4 \times 3^3 \times 5^4

. For any two numbers, the exponent of each prime factor in their GCD is the minimum of their individual exponents, while the exponent in their LCM is the maximum. Comparing prime 2:

\max(a, 3) = 4

, so

a = 4

. Comparing prime 3:

\min(3, c) = 2

, so

c = 2

. Comparing prime 5:

\max(b, 1) = 4

, so

b = 4

. Therefore,

a + b + c = 4 + 4 + 2 = 10

.

Step-by-Step Solution

1

Find the prime factorizations of GCD(x, y) and LCM(x, y)

360 = 2^3 × 3^2 × 5^1 and 270,000 = 2^4 × 3^3 × 5^4

Expressing GCD and LCM in prime factor form allows direct comparison of prime exponents.

2

Apply prime exponent rules for GCD (minimum) and LCM (maximum)

For prime 2: min(a, 3) = 3 and max(a, 3) = 4, so a = 4.
For prime 3: min(3, c) = 2 and max(3, c) = 3, so c = 2.
For prime 5: min(b, 1) = 1 and max(b, 1) = 4, so b = 4.

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

3

Calculate the sum a + b + c

4 + 4 + 2 = 10

Substitute the evaluated exponent values to find the requested total.

Key Concept

GCD and LCM via Prime Factorization Exponents

Estimated Time:1m 30s

Question 80Question →

A positive integer $n$ has a prime factorization of the form $2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are non-negative integers. If $n$ is a multiple of $20$ but not a multiple of $40$ , is not divisible by $9$ , and has exactly $12$ positive integer divisors, what is the sum of all possible values of $n$ ?

Show answer & explanation

Answer: 560

Answer

The sum of all possible values of

n

is 560.

To determine the sum of all possible values of

n

, first analyze the given conditions on the prime exponents of

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

. Divisibility by

20 = 2^2 \cdot 5^1

requires

a \ge 2

and

c \ge 1

. Since

n

is not divisible by

40 = 2^3 \cdot 5^1

, the exponent

a

must equal 2. Since

n

is not divisible by

9 = 3^2

, the exponent

b

must be strictly less than 2, meaning

b

can be 0 or 1. The total number of positive integer divisors is given by

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

. Substituting

a = 2

gives

3(b+1)(c+1) = 12

, or

(b+1)(c+1) = 4

. If

b = 0

, then

c+1 = 4 \implies c = 3

, which gives

n = 2^2 \cdot 3^0 \cdot 5^3 = 500

. If

b = 1

, then

c+1 = 2 \implies c = 1

, which gives

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

. The sum of these two valid values is

500 + 60 = 560

.

Step-by-Step Solution

1

Analyze the divisibility constraints to find values/ranges for exponents

a

,

b

, and

c

.

a = 2

,

c \ge 1

, and

b \in \{0, 1\}

.

Divisibility by 20 (

2^2 \cdot 5^1

) requires

a \ge 2

and

c \ge 1

. Non-divisibility by 40 (

2^3 \cdot 5^1

) restricts

a < 3

, so

a = 2

. Non-divisibility by 9 (

3^2

) restricts

b < 2

.

2

Apply the divisor count formula

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

.

(b+1)(c+1) = 4

.

Plugging

a = 2

into

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

yields

3(b+1)(c+1) = 12

, which simplifies to

(b+1)(c+1) = 4

.

3

Evaluate the cases for

b = 0

and

b = 1

.

The valid values for

n

are

500

and

60

.

When

b = 0

,

c+1 = 4 \implies c = 3

, yielding

n = 2^2 \cdot 3^0 \cdot 5^3 = 500

. When

b = 1

,

c+1 = 2 \implies c = 1

, yielding

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

.

4

Calculate the sum of all valid integers

n

.

560

500 + 60 = 560

.

Key Concept

Divisor Count Formula and Divisibility Constraints

Number Properties and Arithmetic

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Alternative Method

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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Common Mistakes

Answer

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