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Question

Answers

(a) 40

(b) 42

(c) 46

(d) 52

Answer
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We have a number given as 9!. We need to find the sum of the factors of given numbers which are odd and in the form of 3m + 2.

We know that n! is defined as $ n.\left( n-1 \right).\left( n-2 \right).\left( n-3 \right).....1 $ .

So, we have got $ 9!=9.8.7.6.5.4.3.2.1 $ . We need odd factors which are in the form of 3m + 2.

So, we should not submit even natural numbers in place of ‘m’ as it gives even numbers.

So, We substitute odd natural numbers in place of ‘m’ i.e., 1, 3, 5, 7, 9, 11…….. .

Let us find the values of 3m + 2 on substituting these values.

On substituting the value m = 1, we get $ 3(1)+2=3+2=5 $ .

On substituting the value m = 3, we get $ 3(3)+2=9+2=11 $ .

On substituting the value m = 5, we get $ 3(5)+2=15+2=17 $ .

On substituting the value m = 7, we get $ 3(7)+2=21+2=23 $ .

On substituting the value m = 9, we get $ 3(9)+2=27+2=29 $ .

On substituting the value m = 11, we get $ 3(11)+2=33+2=35 $ .

On substituting the value m = 13, we get $ 3(13)+2=39+2=41 $ .

We got the values of 3m + 2 for m = 1, 3, 5, 7,…… are 5, 11, 17, 23, 29, 35, 41,…….. .

So, 3m + 2 cannot be divisible by 3. So, we need to find the factors of 9! that are not multiples of even numbers, 3 and 9. So, we get only two factors 5 and 35 that are in this from.

So, the factors that are odd and in the form of 3m + 2 for 9! are 5 and 35.

So, the sum of those factors is $ 5+35=40 $ .

∴ The sum of the factors of 9! that are odd and in the form of 3m + 2 is 40.