All practice questions

What is the units digit of $7^{43}$ ?

Show answer & explanation

Answer: 3

Answer

The units digit of

7^{43}

is 3.

The units digits of powers of 7 repeat in a pattern of four terms: 7, 9, 3, 1. Dividing the exponent 43 by 4 yields a remainder of 3. The 3rd term in the repeating pattern is 3, so the units digit of

7^{43}

is 3.

Step-by-Step Solution

1

Determine the cyclicity pattern of the units digit of powers of 7.

The units digits for

7^1, 7^2, 7^3, 7^4, \dots

follow the repeating sequence 7, 9, 3, 1 with a cycle length of 4.

Units digits of positive integer powers follow a periodic pattern.

2

Divide the exponent 43 by the pattern cycle length of 4.

43 \div 4 = 10

remainder 3.

The remainder determines which term in the repeating sequence gives the units digit.

3

Find the units digit corresponding to the 3rd term in the cyclicity sequence.

The 3rd digit in the sequence (7, 9, 3, 1) is 3.

A remainder of 3 corresponds to the 3rd power in the cycle,

7^3

.

Key Concept

Units Digit Cyclicity

Question 182Question →

For all integers $k$ and $m$ , if $k^2 + 3km + 2m^2$ is an odd integer, then $k + m$ must be an odd integer.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

The statement is true because factoring

k^2 + 3km + 2m^2

yields

(k + m)(k + 2m)

. By the fundamental properties of parity, a product of two integers is odd if and only if both factors are odd. Consequently,

k + m

must be odd.

Step-by-Step Solution

1

Factor the algebraic expression

k^2 + 3km + 2m^2

.

k^2 + 3km + 2m^2 = (k + m)(k + 2m)

Factoring into a product of linear terms allows us to apply parity rules for multiplication.

2

Apply parity rules of multiplication to the product

(k + m)(k + 2m)

.

Since

(k + m)(k + 2m)

is odd, both

(k + m)

and

(k + 2m)

must be odd integers.

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

3

Conclude the parity of

k + m

.

k + m

is guaranteed to be an odd integer.

It is one of the two factors required to be odd for the product to be odd.

Key Concept

Parity Rules for Integer Products and Factoring

Question 183Question →

If an integer $n$ is expressed in its prime factorization as $n = 2^3 \times 5^2 \times 11$ , how many distinct prime factors does $n$ have?

2

3

4

6

30

Show answer & explanation

Answer: 3

Answer

The integer has 3 distinct prime factors.

The expression

n = 2^3 \times 5^2 \times 11

shows

n

written as a product of prime powers. The distinct prime factors are the unique prime numbers serving as bases in this expansion: 2, 5, and 11. Therefore, there are exactly 3 distinct prime factors.

Step-by-Step Solution

1

Identify the prime bases in the prime factorization of

n

.

The prime factorization is given as

n = 2^3 \times 5^2 \times 11^1

. The prime bases are

2

,

5

, and

11

.

By definition, the prime factorization expresses a number as a product of prime numbers raised to exponents.

2

Count the number of unique prime bases.

There are 3 unique prime numbers (

2

,

5

, and

11

).

Exponents indicate how many times a prime factor is multiplied, but they do not add new distinct prime factors.

Key Concept

Distinct Prime Factors

Estimated Time:45s

Question 184Question →

Let $N = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers. If $N^2$ has $105$ positive divisors and $\dfrac{N}{2}$ has $16$ positive divisors, what is the number of positive divisors of $\dfrac{N}{15}$ ?

8

9

12

16

18

Show answer & explanation

Answer: $9$

Answer

The number of positive divisors of

\dfrac{N}{15}

is

9

.

The prime factorization of

N^2

is

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

, giving

(2a+1)(2b+1)(2c+1) = 105 = 3 \times 5 \times 7

. This implies that the exponents

a, b, c

are a permutation of

1, 2, 3

. The number of positive divisors of

\dfrac{N}{2} = 2^{a-1} \cdot 3^b \cdot 5^c

is

a(b+1)(c+1) = 16

. Testing

a \in \{1, 2, 3\}

shows that only

a = 2

satisfies

2(b+1)(c+1) = 16

with

\{b, c\} = \{1, 3\}

. Finally,

\dfrac{N}{15} = 2^a \cdot 3^{b-1} \cdot 5^{c-1}

has

(a+1)bc

positive divisors. Substituting

a=2

and

bc = 3

gives

(2+1) \times 3 = 9

, making

9

the correct choice.

Step-by-Step Solution

1

Express the number of positive divisors of

N^2

in terms of

a

,

b

, and

c

.

(2a+1)(2b+1)(2c+1) = 105

For a prime factorization

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

, the total number of positive divisors is

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

. Here

N^2 = 2^{2a} \cdot 3^{2b} \cdot 5^{2c}

.

2

Determine the set of values for the exponents

\{a, b, c\}

.

\{a, b, c\} = \{1, 2, 3\}

The prime factorization of

105

into three factors greater than

1

is uniquely

3 \times 5 \times 7

. Thus, the set of values for

\{2a+1, 2b+1, 2c+1\}

is

\{3, 5, 7\}

, which gives

\{2a, 2b, 2c\} = \{2, 4, 6\}

, so

\{a, b, c\} = \{1, 2, 3\}

.

3

Use the divisor count of

\dfrac{N}{2}

to identify the specific value of

a

.

a = 2

and

\{b, c\} = \{1, 3\}

Since

\dfrac{N}{2} = 2^{a-1} \cdot 3^b \cdot 5^c

, its divisor count is

a(b+1)(c+1) = 16

. Testing values from

\{1, 2, 3\}

for

a

: if

a=2

, then

2(b+1)(c+1) = 16 \implies (b+1)(c+1) = 8

. Since

\{b, c\} = \{1, 3\}

,

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

, which confirms

a = 2

and

\{b, c\} = \{1, 3\}

.

4

Calculate the number of positive divisors of

\dfrac{N}{15}

.

(2+1)(1)(3) = 9

\dfrac{N}{15} = \dfrac{N}{3 \cdot 5} = 2^a \cdot 3^{b-1} \cdot 5^{c-1}

. The number of positive divisors is

(a+1)(b-1+1)(c-1+1) = (a+1)bc

. Substituting

a=2

and

bc = 1 \times 3 = 3

yields

(2+1) \times 3 = 9

.

Key Concept

Divisor Count Formula for Prime Factorized Integers

Estimated Time:2m 0s

Question 185Question →

What is the smallest prime factor of $143$ ?

Show answer & explanation

Answer: 11

Answer

The smallest prime factor of 143 is 11.

To find the smallest prime factor of 143, test prime numbers in increasing order starting from 2. Testing 2, 3, 5, and 7 shows none are factors of 143. Testing 11 yields

143 \div 11 = 13

, which is an integer. Thus,

143 = 11 \times 13

. Since 11 and 13 are both prime, 11 is the smallest prime factor of 143.

Step-by-Step Solution

1

Test divisibility of 143 by small prime numbers in ascending order

143 is not divisible by 2, 3, 5, or 7.

Basic divisibility rules eliminate 2 (odd), 3 (digit sum 8), and 5 (ends in 3). Direct division shows 143 = 7 × 20 + 3.

2

Test divisibility by the next prime number, 11

143 divided by 11 equals 13 exactly.

Since 143 = 11 × 13 and 11 is prime, 11 is a prime factor.

3

Identify the smallest prime factor

The prime factors of 143 are 11 and 13, making 11 the smallest.

No prime number smaller than 11 divides 143.

Key Concept

Prime Factorization

Question 186Question →

A consultancy allocated a total budget for a digital transformation project. In the first phase, $\frac{3}{8}$ of the initial budget was spent. In the second phase, $40\%$ of the remaining budget was spent. If the unspent amount after both phases is $\$27,000$ , what was the initial budget of the project, in dollars?

Show answer & explanation

Answer: 72000

Answer

72,000

To determine the initial budget, track the unspent fraction step by step. After the first phase,

\frac{5}{8}

of the budget remains. In the second phase,

40\%

(or

\frac{2}{5}

) of that remaining fraction is spent, which leaves

60\%

(or

\frac{3}{5}

) of the

\frac{5}{8}

unspent. Multiplying

\frac{3}{5} \times \frac{5}{8} = \frac{3}{8}

. Because

\frac{3}{8}

of the initial budget equals

\$27,000

, the full initial budget is

27,000 \times \frac{8}{3} = 72,000

.

Step-by-Step Solution

1

Find the fraction of the budget remaining after the first phase.

The fraction remaining after Phase 1 is

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

.

The first phase spends

\frac{3}{8}

of the initial total budget.

2

Find the fraction of the total initial budget spent in the second phase.

Phase 2 expenditure =

40\% \times \frac{5}{8} = \frac{2}{5} \times \frac{5}{8} = \frac{2}{8}

.

40\%

is equivalent to

\frac{2}{5}

, which applies to the remaining

\frac{5}{8}

of the budget.

3

Determine the remaining fraction of the total budget after both phases.

Remaining fraction =

\frac{5}{8} - \frac{2}{8} = \frac{3}{8}

.

Subtracting the fraction spent in Phase 2 from the fraction left after Phase 1 yields the unspent portion.

4

Calculate the initial total budget in dollars.

Initial Budget =

\$27,000 \times \frac{8}{3} = \$72,000

.

Setting

\frac{3}{8}

of the total budget equal to the unspent amount of

\$27,000

determines the total.

Key Concept

Sequential fractional and percentage reductions from a base quantity

Question 187Question →

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 188Question →

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 189Question →

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 190Question →

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 191Question →

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 192Question →

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 193Question →

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 194Question →

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 195Question →

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 196Question →

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 197Question →

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 198Question →

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 199Question →

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 200Question →

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.

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution