Question

Let $N$, $x$ and $y$ be positive integers such that $N=x+y$, $2<x<10$ and $14<y<23$. If $N>25$, then how many distinct values are possible for $N$?

Answer 1

Given, $2<x<10$ and $14<y<23$ $⇒17 <(x+y)<32$ i.e. $17<N<32$

But $N>25$ 

Hence $25<N<32$

$N$ can take $6$ distinct values.

Answer 2

Any value from the set {3, 4, 5, 6, 7, 8, 9} can be used for $2 < x < 10 x$. $14 < y < 23$

Any value from the set {15, 16, 17, 18, 19, 20, 21, 22} can be used for y. 

N, or x+y, can have a maximum value of 9+22 = 31. (at y = 22 and x = 9) 
30 is available at x = 9; y = 21
29 is available at x = 9; y = 20
28 is available at x = 9; y = 19
27 is available at x = 9; y = 18
26 is available at x = 9; y = 17
25 is available at x = 9; y = 16 
However, the intended sum is not x+y=25, which is why x+y have different values {31,30,29,28,27,26}.

So, x+y, and hence, N, can have six different values.