Free Practice Question

PS · Problem Solving

705-805

Divisibility/Multiples/Factors

What is the remainder when $(1!)^3 + (2!)^3 + (3!)^3 + … + (432!)^3$ is divided by 144?

Answer Choices

Correct answer marked below

A

0

B

25

C

49

81

Correct

E

100

See the full step-by-step explanation

You can see the correct answer above. Sign in for free to unlock the complete worked solution.

Sign in Create free account

Track your performance and improve

Get detailed analytics, unlock full explanations, and move up difficulty tiers as you practice.

Start for free Sign in

Explanation

Preview

Official Explanation

Official

Beyond

3!^3

all terms would be divisible by 144 as

4!^3 = 24^3 = 12^3*2^3

and onwards the terms are divisible by 144 (since

144 = 12^2

).

Hence, only

(1!)^3 + (2!)^3 + (3!)^3

needs to be considered.…

Read the full explanation

Sign in for free to see the complete step-by-step solution to this question.

Create free account Sign in

Comments

Sign in to join the discussion

Sign in Create free account

No comments yet

Be the first to share your thoughts and help others understand it better.

Sign in to comment

What is the remainder when $(1!)^3 + (2!)^3 + (3!)^3 + … + (432!)^3$ is divided by 144?

Answer Choices

See the full step-by-step explanation

Track your performance and improve

Official Explanation

Read the full explanation

No comments yet

What it tests

Common trap

Unlock full explanations

What is the remainder when (1!)3+(2!)3+(3!)3+…+(432!)3(1!)^3 + (2!)^3 + (3!)^3 + … + (432!)^3(1!)3+(2!)3+(3!)3+…+(432!)3 is divided by 144?

Answer Choices

See the full step-by-step explanation

Track your performance and improve

Official Explanation

Read the full explanation

No comments yet

What it tests

Common trap

Unlock full explanations

What is the remainder when $(1!)^3 + (2!)^3 + (3!)^3 + … + (432!)^3$ is divided by 144?