One of
the two digits of a
two digit number is
three times the
other digit. If you
interchange the
digits of this two-
digit number and add
the resulting number
to the original
number, you get 88.
What is the original
number?
let the tens digit be x
then the unit digit is 3x
so the number is 10x + 3x = 13x
the number reversed is 10(3x) + x = 31x
31x + 13x = 88
44x = 88
x = 2
3x = 6
The original number was 26
check:
is 26 + 62 = 88 ? YES
my answer is correct
To solve this problem, let's break it down step by step.
Let's assume the tens digit of the two-digit number is 'x' and the ones digit is 'y'. We know from the problem statement that one of the digits is three times the other digit. So we can write two equations based on the information given:
Equation 1: x = 3y (One digit is three times the other)
Equation 2: (10x + y) + (10y + x) = 88 (When we interchange the digits and add the resulting number to the original number, we get 88.)
Let's simplify Equation 2:
11x + 11y = 88
Divide both sides of the equation by 11:
x + y = 8
Now we have a system of two equations:
x = 3y
x + y = 8
We can solve this system of equations using substitution. Let's start by solving Equation 1 for 'x':
x = 3y
Now substitute this value of 'x' into Equation 2:
(3y) + y = 8
4y = 8
Divide both sides of the equation by 4:
y = 2
Now substitute the value of 'y' back into Equation 1 to find 'x':
x = 3(2)
x = 6
So, the original number is 62.