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What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11

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What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink] New post 04 Nov 2009, 00:07
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What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?

1. 125
2. 225
3. 325
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Re: Numbers 3 [#permalink] New post 04 Nov 2009, 03:19
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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!!
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Re: Numbers 3 [#permalink] New post 31 Jan 2010, 05:01
Is there a specific approach to tackle this?
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Re: Numbers 3 [#permalink] New post 31 Jan 2010, 06:17
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jeeteshsingh wrote:
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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Re: Numbers 3 [#permalink] New post 12 Feb 2010, 17:45
Nice Explanation!
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Re: Numbers 3 [#permalink] New post 14 Jul 2011, 02:13
Bunuel....+1 to you...
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Re: Numbers 3 [#permalink] New post 16 Jul 2011, 02:13
good question. Thanks Bunuel for sharing the approach for these problem!
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Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink] New post 19 Oct 2012, 03:21
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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Re: Numbers 3 [#permalink] New post 18 Jul 2013, 05:20
Bunuel wrote:
jeeteshsingh wrote:
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?
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Re: Numbers 3 [#permalink] New post 21 Jul 2013, 01:24
Expert's post
cumulonimbus wrote:
Bunuel wrote:
jeeteshsingh wrote:
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?


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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink] New post 08 Aug 2014, 05:49
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