GMATSUCKER wrote:Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?

ajith wrote:

GMATSUCKER wrote:Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?

I cannot think about a formula but, can certainly think about a trick

a)25^3 + 38^3 + 87^3 =90^3

Last digit of 25^3 is 5; 38^3 is 2 and 87^3 is 3 => the sum ends in zero so does 90^3; so cannot rule out this option

b)3^3 + 4^3 + 5^3 = 6^3

This we can calculate 125+64 +27 = 216

c)17^3 + 29^3 = 57^3

Last Digit of 17^3 is 3; last digit of 29^3 is 9; sum ends in 2 but 57^3 must end in 3 => this equation cannot be true

d)1^3 + 6^3 + 8^3 = 9^3

We can calculate 512+216+1 = 729

Hence C

Ian Stewart wrote:

GMATSUCKER wrote:
c)17^3 + 29^3 = 57^3

You do not need any inequality or units digit relationships to see that this cannot be true: on the left side of the equation, we're adding two odd numbers. That must give us an even number, and 57^3 is not even.