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Sauron [17]
3 years ago
6

m and n are given positive integers. how many odd numbers can we get when we multiply k*m*n given that k is any positive one-dig

it number?
Mathematics
1 answer:
DerKrebs [107]3 years ago
5 0

Unfortunately, there isn't enough info about m and n.

If either m or n is even, then m*n is even. By extension, k*m*n is always even as well.

This is because even*even = even and even*odd = even.

Some examples:

2*3 = 6

4*8 = 32

So if either m or n (or both) are even, then k*m*n is never odd no matter what you pick for k.

-----------------------------

If both m and n are odd together, then m*n is odd

Example: 3*5 = 15

To ensure that k*m*n is odd, we need to pick odd numbers for k. If k is even, then we fall into the same scenario as the last section.

The odd single digit numbers are {1,3,5,7,9}. We see that there are 5 items in this list. So there are 5 values we can pick for k to have k*m*n be odd. Furthermore, it means there are 5 different odd numbers of the form k*m*n, with k being a single digit number.

Again, this all relies on the assumption that m and n are both odd.

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