if the sum of five consecutive positive integers is a perfect cube, wh

Question

if the sum of four consecutive odd positive integers is a perfect cube, which of the following can be one of the numbers?
A.253
B.133
C.49
D.343
E.21

chetan2u · Answer

Let the numbers be a-2, a-1, a, a+1, a+2...
 SUM = 5*a

Thus,  5a=x^3 
 Now, what can we say about x? 
 x is 
 x is a multiple of  5^3

So, a is multiple of 25 and some cube..... a=5^2*b^3

Such numbers are 
a) when b=1...a=25....23, 24, 25, 26, 27...Sum= 125=5^3 
b) when b=2... a=25*2^3=200 ...198,199,200,201,202...Sum= 5*200=1000=10^3 
c) when b=3... a=25*3^3=675 ...673,674,675,676,677...Sum= 5*675=3375=15^3

So, none of the choices fit in.

preetamsaha · Answer

sorry , it should be "the sum of four consecutive positive integers" instead of "the sum of five consecutive positive integers". its a big mistake. i will definitely make the question correct.

chetan2u · Answer

Let the numbers be a-3, a-1, a+1, a+3.
 SUM = 4*a

Thus,  4a=x^3 
 Now, what can we say about x? 
 x is 
 x is a multiple of  2^3

So, a =2*cube=2*b^3..... a=\frac{x^3}{4}

Such numbers are 
1) when b=1...a=2....-1,1,3,5...Sum= 8=2^3 ..But all numbers have to be positive, so b>1
2) when b=2... a=2*2^3=16 ...13, 15, 17, 19...Sum= 64=4^3 
3) when b=3... a=2*3^3=54 ....51,53,55,57..but no choices have any of these numbers
4) when b=4... a=2*4^3=128 ....125,127,129,131..but no choices have any of these numbers
5) when b=5... a=2*5^3=250 ....247,249,251,253..253 is given in the choices,.....ANSWER

A

preetamsaha · Answer

chetan2u  I think its a typo. please correct it.

if the sum of five consecutive positive integers is a perfect cube, wh

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