Question

Let C be a positive integer such that C + 7 is divisible by 5. The smallest positive integer n (>2) such that C + n 2 is divisible by 5 is

Answer 1

C+7 is divisible by 5

=>

C+7 = 5k

C = 5k-7

Answer 2

In the question please note its C +n^2 ie n square....
\[C + n ^{2}\]

Answer 3

\(C + n^2 = 5k -7 + n^2\)

Answer 4

clearly \(5k\) is divisible by 5
so you need to wry oly about \(-7+n^2\)

Answer 5

next, think wat digits are possible in in units place for a square number : \(n^2\)

Answer 6

nice qs ...

Answer 7

cosider the last digite of n^2
0 1 4 5 6 9 -
7 7 7 7 7 7
ـــــــــــــــــــــــــــــــــ
7 6 3 2 1 3

Answer 8

the last one 2

Answer 9

so i guss n^2-7 never devide 5 right ?

Answer 10

Yes

Answer 11

\(n^2 - 7 \equiv n^2+3 (\mod{5} ) \)

Answer 12

and n^2 never gives 2 or 7 as last digit

Answer 13

aha right :)