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 \(N\)?
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 \(N\)?
Updated On: Apr 28, 2024
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Approach Solution - 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.
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Approach Solution -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.
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