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 = 8 and y = 7
So, N = 87
Answer: 87
Cheers,
Brent