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?

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.

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.