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?

GIVEN: N is a 2-digit integer. When the digits of N are reversed, the resulting number is M. -1 < N – M < 15 Let x = the tens digit of N

Let y = the units digit of N

So, the VALUE of N = 10x + y

When we reverse the digits, we get M = yx

So, the VALUE of M = 10y + x

So, N - M = (10x + y) - (10y + x)

= 9x - 9y

= 9(x - y)

In other words, N - M =

some multiple of 9We're told that -1 < N – M < 15

There are exactly two multiples of 9 between -1 and 15. They are 0 and 9.

So EITHER

N – M = 0 OR

N – M = 9Let's examine each case:

CASE A: If

N - M = 0, then 9(x - y) = 0, which means x - y = 0, which means

x = y CASE B: If

N - M = 9, then 9(x - y) = 9, which means x - y = 1, which means

x = y + 1 GIVEN: the sum of N’s digits is 15 In other words, x + y = 15

If x and y are INTEGERS, and if x + y = 15, then

x cannot equal yThis rules out CASE A, which means CASE B must be true. That is,

x = y + 1 We now have two equations:

x + y = 15

x = y + 1 When we solve the system, we get: x = 7 and y = 8

So, N = 78

Answer: 78

Cheers,

Brent

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Brent Hanneson – Creator of greenlighttestprep.com

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