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by anshumishra » Sun Jan 16, 2011 9:45 pm
nehatandon wrote:What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?
A. 125
B. 225
C. 325
D. 425
E. 525
1152 = 2*2*2*2*2*2*2*3*3 = 2^7*3^2

(3!)^3 = (2*3)^3 = 2^3*3^3 ---> not divisible by 1152
(4!)^3 = (2*3*4)^3 = 2^3*3^3*4^3 = 2^9*3^3 = 2^7*3^2(2^2*3^1) = 1152*(2^2*3^1) -> so divisible by 1152 = 1152*n
Hence, all the further terms will be divisible by 1152

So, the given sum = (1!)^3+ (2!)^3 + (3!)^3 +1152*m = 1+8+216+1152*m = 225+1152*m

Hence, the remainder is 225.B
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Anshu

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by ankur.agrawal » Sun Jan 16, 2011 11:43 pm
anshumishra wrote:
nehatandon wrote:What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?
A. 125
B. 225
C. 325
D. 425
E. 525
1152 = 2*2*2*2*2*2*2*3*3 = 2^7*3^2

(3!)^3 = (2*3)^3 = 2^3*3^3 ---> not divisible by 1152
(4!)^3 = (2*3*4)^3 = 2^3*3^3*4^3 = 2^9*3^3 = 2^7*3^2(2^2*3^1) = 1152*(2^2*3^1) -> so divisible by 1152 = 1152*n
Hence, all the further terms will be divisible by 1152

So, the given sum = (1!)^3+ (2!)^3 + (3!)^3 +1152*m = 1+8+216+1152*m = 225+1152*m

Hence, the remainder is 225.B
Absolutely Brilliant.
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by AIM GMAT » Mon Jan 17, 2011 2:27 am
Wow !!! Nice explanation anshumishra . Thanks a lot for a nice method to solve the dangerous looking question.
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by nehatandon » Mon Jan 17, 2011 5:56 am
anshumishra wrote:
nehatandon wrote:What is the remainder when (1!)^3+ (2!)^3 + (3!)^3 +.....(1152!)^3 is divided by 1152?
A. 125
B. 225
C. 325
D. 425
E. 525
1152 = 2*2*2*2*2*2*2*3*3 = 2^7*3^2

(3!)^3 = (2*3)^3 = 2^3*3^3 ---> not divisible by 1152
(4!)^3 = (2*3*4)^3 = 2^3*3^3*4^3 = 2^9*3^3 = 2^7*3^2(2^2*3^1) = 1152*(2^2*3^1) -> so divisible by 1152 = 1152*n
Hence, all the further terms will be divisible by 1152

So, the given sum = (1!)^3+ (2!)^3 + (3!)^3 +1152*m = 1+8+216+1152*m = 225+1152*m

Hence, the remainder is 225.B
wow, you made that look so easy. Excellent ! Thanks.
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by anshumishra » Mon Jan 17, 2011 6:19 am
Thanks AIM GMAT, ankur, Neha !
Thanks
Anshu

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