(Q). How many 2 digit numbers exist whose number of factors is 8?
Solution:-
If N = aᵖbᑫcʳ…..
then number of factors = (p+1)(q+1)(r+1)………
we are given that number of factors = 8
∴ (p+1)(q+1)(r+1)……. = 8
we have following cases:-
case(i).
If we take 8 = 2×2×2 then number will be in the format a¹b¹c¹, the two digit numbers in this format are
case(ii):-
If we take 8 as 2×4 then number will be in the format a¹b³, the two digit numbers in this format are:
case(iii).
If we take 8 = 1×8, then number will be in the format a⁷ but NO two digit number exist that is in this format
Hence total number of required numbers = 5 + 5 = 10 Answer