How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?
Approach Solution - 1
The product of the digits of the three-digit numbers should be more than 2 and less than 7.
Therefore, the possible numbers are as follows.
Hence there are a total of 21 possibilities.
Approach Solution -2
Let the number be represented as \(abc\). Then, we have the condition that \(2<a×b×c<7\).
The product can take on the values \(3, 4, 5,\) or \(6\).
We can obtain each of these products using the combination 1,1, x, where x can be \(3, 4, 5,\) or \(6\). Each number can be arranged in \(3\) ways, and since we have \(4\) such numbers, there are a total of \(12\) numbers satisfying the given criteria.
We can factorize \(4\) as \(2×2\), and the combination \(2,2,1\) can be used to form \(3\) more distinct numbers.
Similarly, we can factorize \(6\) as \(2×3\), and the combination \(1,2,3\) can be used to form \(6\) additional distinct numbers.
Thus, total numbers formed \(= 12+3+6=21\)
So, the answer is \(21\).
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