jhudson wrote:
I have the answer to the question. I do not understand how to determine all 5 values or know you have found all 2-digit numbers that make of the possible options.
You missed the closing bracket and that created confusion.
As for the explanation, I have given it above. Nevertheless, if you are looking for a more detailed solution, here it is:
Think of the divisibility rule of 5. Every number which ends with 5 or 0 is divisible by 5. What happens if a number ends with 1 or 6? The number previous to it which ends with 0 or 5 will be divisible by 5 so remainder will be 1. e.g. 151126 will give remainder 1 when divided by 5 because 151125 is divisible by 5.
To get a remainder of 2 on division by 5, the last digit of the number i.e. Q must be either 2 or 7 (2 more than a multiple of 5). Notice that Q cannot be anything other than 2 or 7 since the remainder will be different in that case. e.g. if Q = 4, remainder when you divide the number by 5 will be 4.
If Q is 7, 6PQ is 6P7 i.e. an odd number. Hence it is certainly not divisible by 4. Hence Q must be 2.
If Q is 2 and 6P2 is divisible by 4, P must be 1 or 3 or 5 or 7 or 9 i.e. it can take 5 values. Just think of the divisibility rule of 4. For a number to be divisible by 4, the last two digits must be divisible by 4. Hence if P2 is divisible by 4, so is 6P2.
What values can P take such that P2 is divisible by 4?
Can P be 0? No.
Can P be 1? Yes, 12 is divisible by 4.
Can P be 2? No, 22 is not divisible by 4.
Can P be 3? Yes, 32 is divisible by 4.
Notice the pattern. For every odd value of P, P2 is divisible by 4. There are 5 odd values so there will be 5 solutions. This is something you should quickly figure in your head.