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Re: N is a 2-digit integer. When the digits of N are reversed, [#permalink]
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GreenlightTestPrep wrote:
N is a 2-digit integer. When the digits of N are reversed, the resulting number is M. If the sum of N’s digits is 15, and -1 < N - M < 15, what is the value of N?

Show: ::
87


So we know that n = ab where ab are integers representing a two digit integer. We also know that the sum of the digits for n equal 15 so that mean that a + b = 15. We need to find a value of n that when n - m is between -1 and 15.

Lets try \(a=9\) and \(b=6\) in which case \(n = 96\) and \(m = 69\):
\(96 - 69 = 27\) which is greater than 15 so no go

Lets try \(a=8\) and \(b=7\) in which case \(n = 87\) and \(m = 78\):
\(87 - 78 = 9\) which is greater than -1 and less than 15.

The answer is \(n = 87\)
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Re: N is a 2-digit integer. When the digits of N are reversed, [#permalink]
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N is a 2-digit integer. When the digits of N are reversed, [#permalink]
1
GreenlightTestPrep wrote:
N is a 2-digit integer. When the digits of N are reversed, the resulting number is M. If the sum of N’s digits is 15, and -1 < N - M < 15, what is the value of N?

Show: ::
87


Any 2 digit number \(xy\) can be written in the form of \(10x + y\)
For example: \(27 = 2(10) + 7\)

Let \(N = 10x + y\) and \(M = 10y + x\)

we know; \(-1 < N - M < 15\)
So, \(-1 < 10x + y - 10y - x < 15\)
i.e. \(-1 < 9(x - y) < 15\)

\(\frac{-1}{9} < x - y < \frac{15}{9}\)
\(-0.111 < x - y < 1.667\)

So as per the question, the difference of \(x\) and \(y\) could only be \(1\) and the numbers as \(7\) and \(8\)

Case I: N = 78, M = 87
Case II: N = 87, M = 78

Hence, N = 87
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Re: N is a 2-digit integer. When the digits of N are reversed, [#permalink]
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Re: N is a 2-digit integer. When the digits of N are reversed, [#permalink]
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