Question

A number in the range of 40 to 70 can be divided by 5 and have a remainder of 2. It can also be divided by 6 and also have a remainder of 2. What's the number?

Answer 1

62

Explanation:

One way we can do this is to list out the possibilities:

For a number divided by 5 with remainder of 2, we'll have a number that ends in either 2 or 7:

`42, 47, 52, 57, 62, 67`

For a number divided by 6 with remainder of 2, we'll have:

`44, 50, 56, 62, 68`

And so by observation it's 62.

Answer 2

`62`

Explanation:

To find the number, it must be the Least Common Multiple of both 5 and 6, times a multiplier, if necessary to get into the desired range, plus 2. Because 5 is prime, the LCM is simply `5 xx 6 = 30`

That is not in our desires 40 < n < 70 range. Doubling it puts it into the range, as 60. Tripling would again put it outside of the range.

Therefore, the number evenly-divisible by 5 and 6 in that range is 60. To have a remainder of 2 in either case means that is must be `60 + 2 = 62`

Check:
`62/5 = 12 + 2` and `62/6 = 10 + 2`

Algebraically,
`(n-2)/5 = A` and `(n-2)/6 = B`

`(n-2) = 5A = 6B` ; `A = 6/5B`

`A/B = 6/5` ; `A = 6` , `B = 5`

`5A = (n-2)` ; `5 xx 6 = n - 2` ; `30 + 2 = n`
`6B = (n-2)` ; `6 xx 5 = n - 2` ; `30 + 2 = n`

This is where the multiple comes in. To get 40 < n < 70 we multiply it by 2. `30 xx 2 = 60`. `n + 2 = 62`

That could also be done with an inequality equation, but this seems simpler.