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

2. Originally Posted by Chuck_3000
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
Let a, b, c be the digits of the three digit number, e.g. a=1, b=2, c=3 for 123. None of the digits can be zero, since then the product would be zero.
The two digit number made from a and c is 10a + c, so the condition you are interested in is
$
a\cdot b \cdot c = 10a + c
$

Solve for a:
$
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
$
a = \frac{c}{bc-10} \ge 1,
$

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