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 $N$? [This Question was asked as TITA]

Question

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

Answer 1

Given the conditions: $ N = x + y $, where $ 2 < x < 10 $, $ 14 < y < 23 $, and $ N > 25 $.

Step 1: Define the ranges for $x$ and $y$
The possible integer values for $x$ are: $ \{3, 4, 5, 6, 7, 8, 9\} $.
The possible integer values for $y$ are: $ \{15, 16, 17, 18, 19, 20, 21, 22\} $.

Step 2: Determine pairs of $x$ and $y$ such that $N = x + y > 25$
We calculate the sums $N$ for valid combinations of $x$ and $y$, ensuring $N$ is strictly greater than 25.

For $x = 3$, the minimum $y$ is 15, $N_{min} = 3 + 15 = 18$. No valid $y$ yields $N > 25$.
For $x = 4$, minimum $y$ is 15, $N_{min} = 4 + 15 = 19$. To get $N > 25$, $y$ must be at least $25 - 4 + 1 = 22$. Valid $y$: $ \{22\} $. $N = 4 + 22 = 26$.
For $x = 5$, minimum $y$ is 15, $N_{min} = 5 + 15 = 20$. To get $N > 25$, $y$ must be at least $25 - 5 + 1 = 21$. Valid $y$: $ \{21, 22\} $. $N = 5 + 21 = 26$, $N = 5 + 22 = 27$.
For $x = 6$, minimum $y$ is 15, $N_{min} = 6 + 15 = 21$. To get $N > 25$, $y$ must be at least $25 - 6 + 1 = 20$. Valid $y$: $ \{20, 21, 22\} $. $N = 6 + 20 = 26$, $N = 6 + 21 = 27$, $N = 6 + 22 = 28$.
For $x = 7$, minimum $y$ is 15, $N_{min} = 7 + 15 = 22$. To get $N > 25$, $y$ must be at least $25 - 7 + 1 = 19$. Valid $y$: $ \{19, 20, 21, 22\} $. $N = 7 + 19 = 26$, $N = 7 + 20 = 27$, $N = 7 + 21 = 28$, $N = 7 + 22 = 29$.
For $x = 8$, minimum $y$ is 15, $N_{min} = 8 + 15 = 23$. To get $N > 25$, $y$ must be at least $25 - 8 + 1 = 18$. Valid $y$: $ \{18, 19, 20, 21, 22\} $. $N = 8 + 18 = 26$, $N = 8 + 19 = 27$, $N = 8 + 20 = 28$, $N = 8 + 21 = 29$, $N = 8 + 22 = 30$.
For $x = 9$, minimum $y$ is 15, $N_{min} = 9 + 15 = 24$. To get $N > 25$, $y$ must be at least $25 - 9 + 1 = 17$. Valid $y$: $ \{17, 18, 19, 20, 21, 22\} $. $N = 9 + 17 = 26$, $N = 9 + 18 = 27$, $N = 9 + 19 = 28$, $N = 9 + 20 = 29$, $N = 9 + 21 = 30$, $N = 9 + 22 = 31$.

Step 3: List the distinct values of $N$ obtained
The set of distinct values for $N$ from the valid combinations is: $ \{26, 27, 28, 29, 30, 31\} $.
The total count of distinct values of $N$ is 6.

Answer: There are 6 distinct possible values for $N$. \[ \boxed{6} \]

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 \(N\)? [This Question was asked as TITA]

The Correct Option is D

Solution and Explanation

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