Math Question?

Question

I got this question from my homework and I thought that instead of just doing trial and error technique, I might use some mathematical technique.

Would someone please tell me how to do "The sum of m and 6 is a multiple of 3. If the product of m and 6 is greater than 20, but less than 60, what is one possible value of m?"

How would you explain how to do this problem without trial and error?

I'm not sure, but I think this might require modular arithmetic. I'm completely new to that subject so please tell me in a very detailed explanation.

Answer 1

Here's what you're given:
m + 6 = 3z, where z is an integer
20 < m * 6 < 60

Take the equation:
m + 6 = 3z
and subtract 6 from both sides:
m = 3z - 6

Plug that new m into the inequality:
20 < (3z - 6) * 6 < 60
Divide through by 6:
20/6 < (3z - 6) < 10
Add 6 to each side:
20/6 + 6 < 3z < 10 + 6
Divide through by 3:
20/18 + 2 < z < 10/3 + 2
One fraction on each side:
56/18 < z < 16/3
3.111 < z < 5.333

Since z is an integer, we know that z = 4 or 5

Plug those values back into the original equation and simplify:
m + 6 = (3 * 4) or m + 6 = (3 * 5)
m + 6 = 12 or m + 6 = 12

So m = 6 or 9

Answer 2

6 is a multiple of 3, so "m+6 is a multiple of 3" is the same as "m is a multiple of 3," so you can make a variable "n" such that n=m/3. Then, 20 < 18n < 60. You just divide each part of the inequality by 18 and pick a random integer, then multiply by 3. You don't really need modular arithmetic.

Answer 3

I assume m is an integer.
Let x be an integer.

If m + 6 is a multiple of 3, then it is divisible by 3. Since 6 is also divisible by 3, then m is divisible by 3 as well.

20 < 6m < 60
10/3 < m < 10

Since m is an integer:

3 < m < 10

Since m is divisible by 3, the possible choices are 6 and 9.

Answer 4

m + 6=3x
20 < 6m >60

thats how i'm seeing it. im not sure you can do this without trial and error though. try to find a number that when you add 6 to it, its a multiple of 3 that is greater than 20 but less than 60.