# two-digit number

• Oct 10th 2009, 07:09 AM
ronaldj
two-digit number

The digit 3 is written at the right of a certain two-digit number thus forming a three-digit number. The new number is 372 more than the original two-digit number. What was the original two-digit number? my answer is 94?
• Oct 10th 2009, 07:16 AM
Defunkt
Quote:

Originally Posted by ronaldj

The digit 3 is written at the right of a certain two-digit number thus forming a three-digit number. The new number is 372 more than the original two-digit number. What was the original two-digit number? my answer is 94?

Let x be the original two-digit number. Then, the new number is:

$\displaystyle 10x + 3$. Now, we know that $\displaystyle 10x + 3 = x + 372$

Solving this gives us:

$\displaystyle 9x = 369 \Rightarrow x = 41$

How did you arrive at 94? You should have easily noted that if you modify the number as stated you get 943, which is obviously not 372 more than 94.
• Oct 10th 2009, 07:19 AM
earboth
Quote:

Originally Posted by ronaldj

The digit 3 is written at the right of a certain two-digit number thus forming a three-digit number. The new number is 372 more than the original two-digit number. What was the original two-digit number? my answer is 94?

Let the original number be 10x +y with $\displaystyle x\in\{1, 2, ..., 9\}$ and $\displaystyle y\in\{0, 1, 2, ..., 9\}$

Then the 3-digit-number is 100x+10y+3.

According to the wording of the question you get:

$\displaystyle 100x+10y+3-372=10x+y$

$\displaystyle 10x+y=41$

The only possible value for x = 4 and y = 1. Thus the original number is 41.
• Oct 10th 2009, 07:28 AM
ronaldj
Quote:

Originally Posted by Defunkt
Let x be the original two-digit number. Then, the new number is:

$\displaystyle 10x + 3$. Now, we know that $\displaystyle 10x + 3 = x + 372$

Solving this gives us:

$\displaystyle 9x = 369 \Rightarrow x = 41$

How did you arrive at 94? You should have easily noted that if you modify the number as stated you get 943, which is obviously not 372 more than 94.

Mea culpa. You are right - thanks!
• Oct 10th 2009, 07:35 AM
ronaldj
Quote:

Originally Posted by earboth
Let the original number be 10x +y with $\displaystyle x\in\{1, 2, ..., 9\}$ and $\displaystyle y\in\{0, 1, 2, ..., 9\}$

Then the 3-digit-number is 100x+10y+3.

According to the wording of the question you get:

$\displaystyle 100x+10y+3-372=10x+y$

$\displaystyle 10x+y=41$

The only possible value for x = 4 and y = 1. Thus the original number is 41.

QUESTION - my kid wants to know if it could also be 21 since 21 + 372 = 393/
• Oct 10th 2009, 07:53 AM
Defunkt
Quote:

Originally Posted by ronaldj
QUESTION - my kid wants to know if it could also be 21 since 21 + 372 = 393/

If you add 3 to the right of 21, do you get 393?
• Oct 10th 2009, 07:59 AM
ronaldj
you are sooh right!
My head hurts but you are right.

Tx for pointing out our mistake!