When the three digits of a three-digit number N are multiplied together, the result is a two digit number obtained by removing the middle digit of N. Find all such numbers N

thanks you very much

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- Jun 7th 2005, 12:28 AM #1

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- Jun 7th 2005, 05:55 AM #2Originally Posted by
**Chuck_3000**

The two digit number made from a and c is 10a + c, so the condition you are interested in is

$\displaystyle

a\cdot b \cdot c = 10a + c

$

Solve for a:

$\displaystyle

a = \frac{c}{bc-10}

$

Now c = 1,2,3,...,9, so for b = 1, a would have to be negative. Thus b=1 is impossible.

The idea then is to look at all larger values of b and try values of c such that this fraction is an integer between 1 and 9.

For b>1, we must have

$\displaystyle

a = \frac{c}{bc-10} \ge 1,

$

which leads to

$\displaystyle

c \le \frac{10}{b-1}

$

So for b = 4, e.g., we only need to look at c = 1, 2,3.

Using this information and trying all cases, you can find the solutions is reasonable time. I found four solutions.

- Jun 16th 2005, 04:17 PM #3

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