A 2-digit number, when divided by 4, has a remainder of 1. When the same number, when divided by 6, has a remainder of 3. What is the largest possible number?
Lowest common multiple of 4 and 6 = 12

Since it is a 2-digit number, to be the largest possible number, it has to be as close to 100 as possible.

100 ÷ 12 = 8 r 4

Quotient to be multiplied by 12 to produce the largest 2-digit number = 8

Largest 2-digit number that is divisible by 12
= 8 x 12
= 96

Since the number has a remainder of 1 when divided by 4 and it has to be the largest 2-digit possible number, we can find the difference between the divisor and the reminder.  

Number to be subtracted from the largest 2-digit number that is divisible by 12
= 4 - 1
= 3

Then we use the largest 2-digit number that is divisble by 12 and subtract it by the difference to find the answer.

Largest 2-digit number so that there is a remainder of 1 when divided by 4 and a remainder of 3 when divided by 6
= 96 - 3
= 93

Answer(s): 93