Easy Number Theory

Ayush's answer is very clever, I repeat it in details:

if units digit of $n^2 + 4\times n$ is 7,

then units digit of $n^2 + 4\times n + 4$ is 1 , (because $7+4 =11$ )

that is the units digit of $(n+2 )^2$ is 1,

hence, the units digit of $n+2$ is 1 ,

then the units digit of $n+2 + 1$ is 2

that is the units digit of $n + 3$ is 2 .

units digit of n^2 + 4n is 7 . So units digit of n^2 + 4n + 4 will be 1 . ( n+ 2 ) ^2 unit digit = 1 So n+2 unit digit - = 1 n+3 unit digit = 2

Brute Force! n = 7 and 9 will give n^2+4n with unit place 7. But n cannot have its unit digit as 7, so n=9. 9+3 = 12. Answer is 2.

n^2 + 4n =7 can be written as

n(n+4) =7

if we will see which digits product gets us unit digit 7 we will get: (9 X 3) and (7 X 1)

9 is not equal to 3+4.....Also 7 is not equal to 1+4

but 9+4=13 and 7+4=11

So, n's unit digit is either 7 or 9

as it is given that n 's unit digit is not equal to 7, the only possible unit digit of n is = 9

Now, n+3: _ _ _ _ _ _ _ _ _ 9 + 3 = _ _ _ _ _ _ _ _ _ 12

So unit digit of n+3 will be 2