SOLUTION: the tens' digit of a number is 3 less than the units' digits. if the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3. what is the original numb
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Question 990801: the tens' digit of a number is 3 less than the units' digits. 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?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
let a = the 10's digit
let b = the units
then
10a + b = the number
:
write an equation for each statement
:
the tens' digit of a number is 3 less than the units' digits.
a = b - 3
if the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3.
= 4
10a + b - 3 = 4(a+b)
10a + b - 3 = 4a + 4b
10a - 4a = 4b - b + 3
6a = 3b + 3
simplify, divide by 3
2a = b + 1
Replace a with (b-3) from the first statement
2(b-3) = b + 1
2b - 6 = b + 1
2b - b = 1 + 6
b = 7
then
a = 7 - 3
a = 4
:
what is the original number? 47
:
:
See if that checks out in the 2nd statement
=
= 4
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