Mathematics question

Question

Let a and b be positive integers such that

(i) both a and b have at least two digits;

(ii) a + b is divisible by 10;

(iii) a can be changed into b by changing its last digit.

Prove that the hundreds digit of the product ab is even.

Answer 1

a=73, b=37, a+b=110, ab=2701

73 and 37, switch the digits and a becomes b.

sum to 110 which is divisible by 10

the product 73(37)= 2701, hundreds digit=7 which is Not even

one counterexample disproves the claim that the hundreds digit is even

there is no proof possible that the hundreds digit is even

64 and 46 sum to 110 divisible by 10

64 = 46 when the digitis are reversed

64(46) = 2944 where the hundreds digit = 9 which is Not even

82 and 28 sum to 110, divisible by 10

82 =28 when digits are reversed

82(28) = 2296 where 2=the hundreds digit and is also even

91 and 19 sum to 110

91(19) = 1729 with 7 = hundreds digit which is Not even

55 and 55 sum to 110

reverse the digits they're the same

55(55)= 3025 with 0 the hundreds digit which is even

twice the hundreds digit is even

thrice it's Not even