A three-digit number has two properties. The tens-digit and the
ones-digit add up to 5. If the number is written with the digits in the
reverse order, and then subtracted from the original number, the result
is 792. Use a system of equations to find all of the 3-digit numbers
with these properties.
Define an unknown 3-digit number: abc
a = hundreds place b = tens place c = ones place
Represent each as a single digit with the corresponding multiplier
(so if a = 3, 100a = 300)
abc; 100a + 10b + 1c
We’re given 2 properties;
- b + c = 5 -> b = 5 - c
- abc - cba = 792
cba; 100c + 10b + 1a
abc - cba = 792
- (100a + 10b + 1c) - (100c + 10b + 1a) = 792
- 100a + 10b + 1c - 100c - 10b - 1a = 792
- 99a - 99c = 792
- a - c = 8
We’re left with the following sets;
Digits can only range from 0 - 9, this narrows a to be values of 8
and 9 (therefore c can only have values of 0 and 1)
Therefore b can have values of 5 or 4