The sum of the factors of 9!

Question

The sum of the factors of 9! which are odd and of the form 3m+2(m is a natural number) is equal to

$(A)40\hspace{1 cm} (B)42\hspace{1 cm}(C)46\hspace{1cm}(D)52$

I could not identify factors,i think they will be lot many.Can someone assist me in solving this question?

Answer 1

The possible numbers are $5,11,17,23,29,35$. So, we have at least $5+35=40$. From the given options, the answer is $(A)$.

Answer 2

Notice, $$9!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9$$$$=362880=2^7\times 3^4\times5\times 7$$ total number of odd factors $$=1\times5\times 2\times 2=40$$

But, the odd factors of form $(3m+2)$ are only two which can be found by putting $m=1$ & $m=11$ which are $3\times 1+2=5$ & $3\times 11+2=35$

Hence the sum of the odd factors of form $(3m+2)$ $$=5+35=40$$ Option (A) is correct.

Answer 3

$9!$ = $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9$. Of all the odd factors, only $5$ is of the form $3m + 2$. We can have $5$, and the only possible number to combine it is $7$ to give $35$ which is also of the form $3m + 2$. So these are the only 2 factors and their sum is $40$

Answer 4

The factors that you are looking for are not divisible by 2 (because they are odd) and not divisible by 3 (because they are divisible by 3).

We have 9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9.

If we remove the factors 2 and 3, then all that is left is 5 x 7, and the only factors that are odd and not divisible by 3 are 1, 5, 7 and 35. Of these, only 5 and 7 are of the form 3n + 2. So the answer is 5 + 35 = 40.

If we asked about 10!, the same argument would leave us with 5 x 7 x 5, and the six factors, 1, 5, 7, 35, 25, 175. The additional factors 25 and 175 are both not of the form 3n + 2, so the final answer is again 40. Same for 7! and 8!.