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:  
\(2 < x < 10\)
So possible integer values for x are: \(\{3, 4, 5, 6, 7, 8, 9\}\)   → 7 values

Also, 
\(14 < y < 23\)
So possible integer values for y are: \(\{15, 16, 17, 18, 19, 20, 21, 22\}\)   → 8 values

Let N = x + y. We now look at the possible values of N: 

Maximum value of N = \(x = 9, y = 22 \Rightarrow N = 31\)
Minimum value of N = \(x = 3, y = 15 \Rightarrow N = 18\)

But the question says we are only interested in values where x = 9 (as per the example shown), so: 

If \(x = 9\), then possible values of y: \(\{15, 16, 17, 18, 19, 20, 21, 22\}\)
So corresponding N values: \(\{24, 25, 26, 27, 28, 29, 30, 31\}\)

However, the passage states that x+y (i.e. N) can have only the values: \(\{26, 27, 28, 29, 30, 31\}\)
These are 6 distinct values of N.

∴ The number of different values N can take is 6.