# find the number

• May 19th 2008, 01:12 PM
sp0rtskid55
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.
• May 19th 2008, 01:16 PM
angel.white
Quote:

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"
• May 19th 2008, 02:22 PM
Soroban
Hello, sp0rtskid55!

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

The number is: . $10t + u$

Its reversal is: . $10u + t$

Quote:

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

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

Find the original number.

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

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

Therefore, the original number is 72.