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.

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- May 19th 2008, 01:12 PMsp0rtskid55find 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 PMangel.white
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 PMSoroban
Hello, sp0rtskid55!

Let $\displaystyle t$ = tens digit

Let $\displaystyle u$ = ones digit

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

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

Quote:

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**.