Question

Let $a$ and $b$ be distinct positive divisors of $80.$ What is the smallest possible value of $ab$ which is not a divisor of $80?$

Answer 1

Answer:

The canonical factorisation of 80 is 24×5 . Any factor of 80 will be of the form 2u5v,0≤u≤4 and 0≤v≤1 .

Let a=2u5v and b=2w5x and 0≤u,w≤4 and 0≤v,x≤1 .

Now, ab=2u+w5v+x . If ab is not a divisor of 80 , one of the two conditions holds.

Either u+w>4 or v+x>1 .

If v+x>1 holds the only solution possible is v=x=1 . In this case, the least value of u+w will be 1 because u+v=0⟹u=v=0 which would mean a=b=5 . This is not the case because a,b are distinct. If we take u=0,w=1 then a=5 and b=10 and ab=50 .

Let us calculate the least value when u+w>4 holds. The least value of u+w is 5 . In this case take v=x=0 . This is acceptable because u=1,v=4 would give us distinct values of a and b i.e, 2 and 16 . The value of ab in this case is 32 .

In the two cases, 32 is the least and hence 32 is the desired number.