Number System

Let's call the first number ab and the second cd. Let's also assume a > c.

Any two-digit number ab can be written as 10a + b, so we have

(10a + b) - (10c + d) = x, where x is the original difference.

After exchanging the tens digits, we have

(10a + d) - (10c + b) = x - 4

Subtracting the second equation from the first gives us

10(a - c) + (b - d) - (10(a - c) + (d - b)) = 4

or (b - d) = 2

So the units digits of the original numbers must differ by 2.

To make the tens digits as different as possible, we could use a = 9 and c = 1. b and d must be such that b = d + 2, so we could choose b = 9 and d = 7. This gives us a difference of 99 - 17, or 82, and we're done. (We could also note that a difference ≥ 90 is impossible anyway, since we're dealing with two two-digit integers.)

Great question!

Hi Matt,

Please help understand the following -

Any two-digit number ab can be written as 10a + b, so we have

(10a + b) - (10c + d) = x, where x is the original difference.

After exchanging the tens digits, we have

(10a + d) - (10c + b) = x - 4

Doesn't exchanging the tens digits mean we interchange 'a' and 'c' in the original equation. I cannot understand why you have interchanged 'b' and 'd' digits.