If among the products a1b1,a2b2,..........,a100b100 if we get two products aibi and ajbj such that the difference in between them is divisible by 100 then the remainders after dividing aibi and ajbj with 100 would be same.
Now, consider the numbers {0,1,2,......,99} as holes and the pigeons be the products {a1b1,a2b2,....,a100b100}
Now, we are putting the product aibi in the hole r if it leaves the remainder r when divided by 100.
If we get that a hole contains more than one product then we are done.
Assume that all the 100 holes contain one product (aibi ) each.
Then there should be a product in hole 2 . Let the product be akbk.
Hence akbk = 100 q + 2. = (4 * 25) q + 2 = 4p + 2.
So, akbk leaves remainder 2 when divided by 4.
Now, since the holes contain one product each there are 50 products in the odd numbered holes.
Let the products in the odd numbered holes be albl . Then albl leaves odd remainder when divided by 100.
So, albl is an odd number and odd number is formed only when both al and bl are odd numbers.
Now, since albl is an odd number the remainder when divided by 4 can be either 1 or 3.
Now, the rest 50 products ahbh (say) are even and both ah and bh are even. So, the products ahbh are divisible by 4.
So, among the 100 products we are not getting any product which leaves remainder 2 when divided by 4.
But we know that akbk leaves remainder 2 when divided by 4. Hence contradiction.
So, the holes cannot contain 1 product each. So, there must be more than one products in at least one hole.
So, among the given products there must exists two products such that the difference between any two is divisible by 100.