Question 155014
The tens digit of a number is 3 less than the units digit. If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3. What is the original number?
:
Let x = the 10's digit; Let y = units digit
Then
10x+y = "the number"
:
"The tens digit of a number is 3 less than the units digit.",therefore
x = (y-3)
:
" If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3.
:
Subtract the remainder from the number and you have:
{{{((10x+y)-3)/((x+y))}}} = 4
:
multiply both sides by (x+y)
10x + y - 3 = 4(x+y)
:
10x + y - 3 = 4x + 4y
Arrange x on the left and y on the right with the remainder:
10x - 4x = 4y - y + 3
:
6x = 3y + 3
Substitute (y-3) for x:
6(y-3) = 3y + 3
:
6y - 18 = 3y + 3
:
6y - 3y = 3 + 18
:
3y = 21
y ={{{21/3}}}
y = 7
then
x = 7 -3
x = 4
:
47 = "the number"
:
Check solution in the statement:
"If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3."
{{{47/((4+7))}}} = 4, remainder of 3