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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]

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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]

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New post 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!!

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Re: Numbers 3 [#permalink]

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Is there a specific approach to tackle this?
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Re: Numbers 3 [#permalink]

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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]

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New post 12 Feb 2010, 18:45
Nice Explanation!

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Re: Numbers 3 [#permalink]

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New post 14 Jul 2011, 03:13
Bunuel....+1 to you...
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Re: Numbers 3 [#permalink]

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New post 16 Jul 2011, 03:13
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]

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New post 19 Oct 2012, 04:21
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]

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New post 18 Jul 2013, 06:20
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]

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New post 21 Jul 2013, 02:24
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]

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New post 12 Aug 2015, 12:22
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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Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

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New post 01 Dec 2017, 06:06
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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Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11   [#permalink] 01 Dec 2017, 06:06
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