Question

Let a, b,c, and d be four integers show that
(a-b)(a-c)(a-d)(b-c)(b-d)(c-d) is
divisible by 12

Answer 1

Hello ma'am, I think you might know this as you are a retired professor in math. Can you tell me how to prove this stuff, as I have no idea of how to do all this proving.

Answer 2

Let us wait, may be some other participants can do this first.

Answer 3

Okay.

Answer 4

@eliassaab is a professor o.O no wonder she's good..i cant believe i didnt see it coming *facepalm*

Answer 5

eliassaab is a he

Answer 6

haha

Answer 7

but @ParthKohli said ma'am

Answer 8

Oops. Your name suits more with females. Sorry, sir :)

Answer 9

*facepalm*

Answer 10

Hint for this problem show that
(a-b)
(a-c)
(b-c)
Cannot be all odd.

Answer 11

well i think i got it

Answer 12

if i recall correctly @SmoothMath loves puzzles

Answer 13

If we take all of the numbers in each (a - b),(a - c) and (b - c) as even, then they will be even.

Answer 14

Without reading the puzzle, I'll boldly agree with that statement.

Answer 15

consider division by 4.....it leaves nly 4 remainders...0,1,2,3

Answer 16

so i start by considering two cases.......ab,c,d all give diff remainders-case 1
nd atleast two have same remaider- case 2...
wen it's case 2....problem's solved....we have that the product is divisible by 4

Answer 17

what is this i don't even

Answer 18

To prove the hint
If the first two are odd
(a-b)
(a-c)
(b-c)

then
(b-c)= (a-c) -(a-b) is even.

Answer 19

Using the hint, we will be able to find 2 terms which are divisible by 2. So the product of the six terms will be divisible by 4. Now all that remains is to find a term divisible by 3. By the pigeon hole principle, since there are four numbers, and only 3 possible remainders when dividing by 3, at least 2 of the numbers will have the same remainder. Since every possible combination of 2 numbers is present, the difference of the two numbers is one of the terms, and their difference is divisible by 3.

Answer 20

I do believe @joemath314159's way does work.

Answer 21

pigeonhole principle ftw :)

Answer 22

You could also use the pigeonhole principle (or something very similar) for showing that 4 divides the expression as well.

Answer 23

ah thats true!