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 $N$?

Question

The largest value of $n$, for which $40^n$ divides $60!$, is

Answer 1

Given the inequality $2<x<10$, the possible integer values for x are $\{3, 4, 5, 6, 7, 8, 9\}$, totaling 7 values.

Given the inequality $14<y<23$, the possible integer values for y are $\{15, 16, 17, 18, 19, 20, 21, 22\}$, totaling 8 values.

Let N = x + y. The possible range for N is determined by its minimum and maximum values:

The maximum value of N occurs when $x = 9$ and $y = 22$, resulting in $N = 31$.

The minimum value of N occurs when $x = 3$ and $y = 15$, resulting in $N = 18$.

The problem specifies a condition where $x = 9$. With $x = 9$, the possible values of y remain $\{15, 16, 17, 18, 19, 20, 21, 22\}$.

The corresponding values for N (where $x=9$) are $\{24, 25, 26, 27, 28, 29, 30, 31\}$.

However, the problem further restricts N to the set $\{26, 27, 28, 29, 30, 31\}$. This set contains 6 distinct values.

Therefore, the number of different values N can take is 6.

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 \(N\)?