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655-705 (Hard)|   Exponents|   Remainders|      
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virtualanimosity
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 Updated on: 26 Dec 2025, 03:34
Originally posted by virtualanimosity on 04 Nov 2009, 01:07.
Last edited by Bunuel on 26 Dec 2025, 03:34, edited 1 time in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

67% (02:28) correct 33% (02:13) wrong based on 665 sessions

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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 is divided by 1152?

A. 125
B. 225
C. 325
D. 333
E. 335
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B
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Bunuel
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 31 Jan 2010, 07:17
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jeeteshsingh
Is there a specific approach to tackle this?
Show more

No specific approach.

We have the sum of many numbers: \((1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3\) and want to determine the remainder when this sum is divided by 1152.

First we should do the prime factorization of 1152: \(1152=2^7*3^2\).

Consider the third and fourth terms:
\((3!)^3=2^3*3^3\) not divisible by 1152;
\((4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152\) divisible by 1152, and all the other terms after will be divisible by 1152.

We'll get \(\{(1!)^3+ (2!)^3 + (3!)^3\} +\{(4!)^3+...+(1152!)^3\}=225+1152k\) and this sum divided by 1152 will result remainder of 225.
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 04 Nov 2009, 04:19
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The ans has to be 225. all the terms in the sequence after (3!)^3 are divisible by 1152 and hence remainder is 0. Upto (3!)^3, sum of all numbers, i.e. 1+8+216 = 225 which is the remainder!!
General Discussion
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testprep2010
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 12 Feb 2010, 18:45
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Nice Explanation!
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 14 Jul 2011, 03:13
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Bunuel....+1 to you...
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 16 Jul 2011, 03:13
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good question. Thanks Bunuel for sharing the approach for these problem!
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 19 Oct 2012, 04:21
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Did .. the same thing as Bunuel..

But took 8 minutes..

In actual exam.. I would have guessed and moved on

What is the source of this problem?
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 18 Jul 2013, 06:20
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Bunuel
jeeteshsingh
Is there a specific approach to tackle this?

No specific approach.

We have the sum of many numbers: \((1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3\) and want to determine the remainder when this sum is divided by 1152.

First we should do the prime factorization of 1152: \(1152=2^7*3^2\).

Consider the third and fourth terms:
\((3!)^3=2^3*3^3\) not divisible by 1152;
\((4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152\) divisible by 1152, and all the other terms after will be divisible by 1152.

We'll get \(\{(1!)^3+ (2!)^3 + (3!)^3\} +\{(4!)^3+...+(1152!)^3\}=225+1152k\) and this sum divided by 1152 will result remainder of 225.
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Hi Bunnel,
To get the remainder, we dont have to reduce the fraction right?
That is we cant do - 225/1152 = 25/ 128 and get remainder 25?
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 21 Jul 2013, 02:24
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cumulonimbus
Bunuel
jeeteshsingh
Is there a specific approach to tackle this?

No specific approach.

We have the sum of many numbers: \((1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3\) and want to determine the remainder when this sum is divided by 1152.

First we should do the prime factorization of 1152: \(1152=2^7*3^2\).

Consider the third and fourth terms:
\((3!)^3=2^3*3^3\) not divisible by 1152;
\((4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152\) divisible by 1152, and all the other terms after will be divisible by 1152.

We'll get \(\{(1!)^3+ (2!)^3 + (3!)^3\} +\{(4!)^3+...+(1152!)^3\}=225+1152k\) and this sum divided by 1152 will result remainder of 225.

Hi Bunnel,
To get the remainder, we dont have to reduce the fraction right?
That is we cant do - 225/1152 = 25/ 128 and get remainder 25?
Show more

Yes, 225 divided by 1152 yields the remainder of 225. The same way as 2 divided by 4 yields the remainder of 2, not 1 (1:2).
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 12 Aug 2015, 12:22
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starting from 4!^3=24^3 all numbers are divisible by 1152, i.e. remainder equal to 0

Only 1!^3, 2!^3 and 3!^3 are not divisible by 1152 and have remainder equal to 1,8 and 216, respectively.

If sum numbers we can sum their remanders to find total remainder, which is 216+8+1+0=225

B
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 01 Dec 2017, 06:06
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factorize 1152 into 2^7*3^2

see carefully all are cubic factorial

we need to look beyonf (4!)^3 as from here onwards remainder is zero

so consider cubic factorial of 1 , 2 and 3 and sum will give 225. and this number on division by 1152 gives 225.

answer B
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 06 Feb 2021, 03:52
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Hi, where can I find a book or pdf source on reminders with factorials and exponents? (eg. 7^27/5 or 13!+11/4).
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 06 Feb 2021, 03:56
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saahulu
Hi, where can I find a book or pdf source on reminders with factorials and exponents? (eg. 7^27/5 or 13!+11/4).
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6. Remainders



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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 06 Feb 2021, 03:59
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Thank you Bunel!!!!
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What is the remainder when (1!)^3 + (2!)^3 + (3!)^3 + ... + (1152!)^3 24 Feb 2026, 00:11
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