# digits problem

• December 8th 2012, 05:48 PM
rcs
digits problem
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
he original number?

attempt:

let : x be unit digit
x - 3 be the tens digit
10(x-3) + x be the number

solve: [10(x-3) + x] / (x-3)+x = 4 + 3

please do check my representations and equation... and please correct my work..

thanks a lot

can anybody help me in representing the unknown please.
• December 8th 2012, 07:04 PM
Soroban
Re: digits problem
Hello, rcs!

Quote:

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 $u$ = units digit.
Let $t$ = tens digit.

The original number is: . $10t + u$
. . where: $t \:=\:u-3 \quad\Rightarrow\quad t - u \:=\:\text{-}3$ .[1]

Then: . $10t + u \:=\:4(t+u) + 3 \quad\Rightarrow\quad 2t - u \:=\:1$ .[2]

Subtract [1] from [2]: . $t \:=\:4$

Substitute into [1]: . $4 - u \:=\:\text{-}3 \quad\Rightarrow\quad u \:=\:7$

The original number is $47.$