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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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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!!

Is there a specific approach to tackle this?
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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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?

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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Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

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08 Aug 2014, 06:49

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Re: What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(11 [#permalink]

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