George's imagined numbers

Question

George pictured 4 natural numbers. He multiplied each of those numbers by three and wrote all four results on a blackboard. He also calculated all possible products of the pairs of the written numbers, he then wrote all 6 products on the blackboard. Prove that (of the ten numbers written on the blackboard) there are surely two numbers which end with the same digit.

Answer 1

HINT: Mod 10, all of the original four numbers are distinct (or else we're done immediately), and none can be $1$ or $3$. So we have a list of ten digits and none on the list can be $3$. That leaves only $9$ possible values for ten digits, so some pair must agree.

Answer 2

First, since we are only talking about the last digit of each number, we can think of this entire problem modulus ten.

So modulus ten, exactly one of the final numbers must be zero. If one of our initial numbers is zero, then more than one product will be zero. Otherwise one of our initial numbers must be five, in which case more than one of the products will be either zero or five.

Note that this solution doesn't use the fact that one of the initial numbers was three, so this is true for any five starting numbers.

Answer 3

Just to be ornery and different:

Each original number "influences" four of the ten results; once when multiplied by 3, and once when multiplied by each of the three other original numbers.

Given two original numbers each will have four results, but one of the results will be the common result of multiplying those two numbers together. So a pair of two original number "influences" seven of the ten results; The two where each are multiplied by 3, the four where each is multiplied by each of the other two original numbers, and the one where they are multiplied together.

Remember: Even times anything is even. Odd times odd is odd. To have an even result, there must be an even original. If there is an even original it will only have even results. So....

If there are two or more even original numbers they will influence 7 even results. They can not be distinct as there are only 5 even digits.

If there is exactly one even original number, it will influence 4 even results. All the remaining 6 results will be odd. As there are only 5 odd digits these will not be distinct.

Finally if there are no even original numbers, there will be no even results and the ten results can not be distinct.

So the results can not be distinct. (Indeed, they will have 4, 7, 9, or 10 even digits and 6,3,1, or 0 odd digits.)

....

But for simplicity, I prefer Mike Pierce's answer of noting $0$ as an original number results in repeated results. (In my terminology it would only have $1$ distinct result: $0$). $5$ does as well (as it would only have two distinct results: $5$ and $0$). Without $0$ or $5$ as original numbers, results of $0$ and $5$ are impossible ($a*b = 0$ implies one or the other is zero or one or the other is five and the other is even; $a*b = 5$ implies one is $5$ and the other is odd.) and without $5$ or $0$ as results there are only $8$, not $10$, possible distinct results.