Question 42054
If any number has 'N' has 'r' prime number factors and can be expressed as
{{{N = (p[1])^(q[1])*(p[2])^(q[2])*(p[3])^(q[3])*....}}}{{{(p[r])^(q[r])}}}
where {{{p[1]}}}, {{{p[2]}}}, {{{p[3]}}}, ... are the prime factors and {{{q[1]}}}, {{{q[2]}}}, {{{q[3]}}}, .... are their respective indices.


For such a number N, the total number of factors (including 1 and itself) is given by {{{(q[1]+1)*(q[2]+1)*(q[3]+1)*....}}}{{{(q[r]+1)}}}.


Here total no. of factors = 10.
Hence, {{{(q[1]+1)*(q[2]+1)*(q[3]+1)*....}}}{{{(q[r]+1)=10}}}


As 10 itself has prime factors 2 and 5.
So the reqd. number must have only two prime factors such that the index of one of them is (2-1=) 1 and that of the other is (5-1=) 4.


So, {{{N = (p[1])^1*(p[2])^4}}}


As N has to be largest of all 3-digited number possible.


First, let us try with {{{p[2]}}}
Let, {{{p[2]}}} = 7. Then {{{p[2]^4=2401}}} > 1000.
Let, {{{p[2]}}} = 5. Then {{{p[2]^4=625}}} < 1000 but {{{p[1]}}} has to be atleast 2. So then N > 1000.
Let, {{{p[2]}}} = 3. Then {{{p[2]^4=81}}}. Under this condition {{{p[1]}}}(max) = 11 so N(max) = 891 < 1000. OK
Let, {{{p[2]}}} = 2. Then {{{p[2]^4=16}}}. Under this condition {{{p[1]}}}(max) = 61 so N(max) = 976 < 1000. OK


Hence 976 is the largest 3-digit number with exactly 10 factors including (1 and itself). 
These factors are: 1, 2, 4, 8, 16, 61, 122, 244, 488, 976.