kabirchaudhry92 wrote:

EgmatQuantExpert wrote:

Solution:

We are given:

\(F(n)= 1×2×3×4×5×……….×n\)

We need to find the number of odd factors of \(f(10)\).

Thus, we need to find the value of \(f(10).\) Then, we need to write \(f(10)\) in its prime factorized form. After which we can calculate the value of odd factors of\(f(10)\).

Number of Odd factors \(f(10)\)

\(f(10)= 1×2×3×4×5×6×7×8×9×10\)

\(f(10)= 1×2×3×(2×2)×5×(2*3)×7×(2*2*2)×(3*3)×(2*5)\)

\(f(10)= 2^8 * 3^4 * 5^2 *7\)

Odd factors of \(f(10)\)= (power of 3+1)*(power of 5+1)* (power of 7+1)

Odd factors of \(f(10)=5*3*2\)

Odd factors of \(f(10)=30\)

Hello!

I have a doubt here.. Why is 1 not included in the odd factors ?

Sorry for such a silly question but isn't that supposed to be counted too ?

It would be really helpful if you could provide your insights on the same.

Thanks in advance

Hi

kabirchaudhry92,

This is a good question. Let me explain with some small examples.

Q1. Find the number of factors of 25.

Step 1: Do the prime factorization. 25 = 5*5 = \(5^2\)

Step 2: Number of factors = (2+1) = 3.

Why have we added 1? By adding "1" we ensure that we count 1 as one of the factors.

Factors of 25: \(5^0 = 1, 5^1 = 5,\) and \(5^2 = 25.\) Hence, the answer is 3.

Q2. Find the number of factors of 75.

Step 1: Prime factorization = 3*5*5 = \(3^1 * 5^2\)

Step 2: Number of factors = (1+1) * (2+1) = 6.

Factors of 75:

- \(5^0 * 3^0 = 1\)
- \(5^0 * 3^1 = 3\)
- \(5^1 * 3^0 = 5\)
- \(5^1 * 3^1 = 15\)
- \(5^2 * 3^0 = 25\)
- \(5^2 * 3^1 = 75\)

.

Hence, 1 is already included in the given solution.

I hope this helps.

Thank you.