Question

if 21y^5 is a multiple of 9 ,where y is a digit ,what is the value of y? Explain..

Answer 1

Its given that 21y^5 is a multiple of 9 
it can be written as 7*3*y^5
since its a multiple of 9 so smallest number to which y is equal to is 3
as 7*3*3^5 = 7*3^6 = 7*(3²) ^3 = 7*9^3
Therefor smallest value of y is 3
It can further be 6,9,12,18......

Answer 2

21y^5 is a multiple of 9  (given)
21 can be written as 18+3
so (18+3)y^5 is a multiple of 9
18y^5 + 3y^5 is a multiple of 9
18y^5 + 3y*y^4  is a multiple of 9
since 18 is a multiple of 9
so 3y should also be a multiple of 9
3y=9n  (where n is any natural no.)
therefore y=3n
y=3,6,9,12,15.......................    ANSWER