# Thread: find the number

1. ## find the number

Heres the problem:
The tens digit of a two-digit number is 1 more than 3 times the ones digit. Subtracting 45 from the number reverses it. Find the original number.

2. Originally Posted by sp0rtskid55
Heres the problem:
The tens digit of a two-digit number is 1 more than 3 times the ones digit. Subtracting 45 from the number reverses it. Find the original number.
well,
If the 1's digit is 0, then 3*0+1 = 1, so the number is 10
If the 1's digit is 1, then 3*1+1 = 4, so the number is 41
If the 1's digit is 2, then 3*2+1 = 7, so the number is 72
If the 1's digit is 3, then 3*3+1 = 10, this is not a two digit number, so you are left with the first three cases.

Just check which one meets the criteria of "Subtracting 45 from the number reverses it"

3. Hello, sp0rtskid55!

Let $\displaystyle t$ = tens digit
Let $\displaystyle u$ = ones digit

The number is: .$\displaystyle 10t + u$

Its reversal is: .$\displaystyle 10u + t$

The tens digit of a two-digit number is 1 more than 3 times the ones digit.
$\displaystyle {\color{blue}t \:=\:3u+1}$

Subtracting 45 from the number reverses it.
$\displaystyle {\color{blue} 10t + u - 45 \:=\:10u + t}$

Find the original number.

The second equation is: .$\displaystyle 9t\:=\:9u + 45\quad\Rightarrow\quad t \:=\:u + 5$

Equate to the first equation: .$\displaystyle 3t + 1 \:=\:u + 5\quad\Rightarrow\quad u \:=\:2$
. . Then: .$\displaystyle t = 7$

Therefore, the original number is 72.