no. of divisers

[lampat]

Total no. of divisors of \(\displaystyle 10!\) which is in the form of \(\displaystyle 5m+2\). where \(\displaystyle m\in \mathbb{N}\)

 

[pka]

Total no. of divisors of \(\displaystyle 10!\) which is in the form of \(\displaystyle 5m+2\). where \(\displaystyle m\in \mathbb{N}\)

Take a look at this webpage.

 

[lookagain]

It looks like you left out some needed words for clarification:

Total no. of divisors of \(\displaystyle 10! \ \), limited to each of those that are of the form of \(\displaystyle \ 5m+2\). where \(\displaystyle m\in \mathbb{N}\)

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[lampat]

\(\displaystyle 10! = 2^8 \times 3^4 \times 5^2 \times 7 \)

Now here we have to calculate divisers of the form \(\displaystyle 5m+2\) , where \(\displaystyle m\in \mathbb{N}\)

\(\displaystyle \bullet \)for \(\displaystyle 2\)

\(\displaystyle 2^0 = 1 = 5m+1 \in \mathbb{N}\)

\(\displaystyle 2^1 = 2 = 5m+2 \in \mathbb{N}\)

\(\displaystyle 2^2 = 4 = 5m+4 \in \mathbb{N}\)

\(\displaystyle 2^3 = 8 = 5m+3 \in \mathbb{N}\)

\(\displaystyle 2^4 = 16 = 5m+1 \in \mathbb{N}\)

\(\displaystyle 2^5 = 32 = 5m+2 \in \mathbb{N}\)

\(\displaystyle 2^6 = 64 = 5m+4 \in \mathbb{N}\)

\(\displaystyle 2^7 = 128 = 5m+3 \in \mathbb{N}\)

\(\displaystyle 2^8 = 256 = 5m+1 \in \mathbb{N}\)

\(\displaystyle \bullet \) Now for \(\displaystyle 3\)

\(\displaystyle 3^0 = 1 = 5m+1 \in \mathbb{N}\)

\(\displaystyle 3^1 = 3 = 5m+3 \in \mathbb{N}\)

\(\displaystyle 3^2 = 9 = 5m+4 \in \mathbb{N}\)

\(\displaystyle 3^3 = 27 = 5m+2 \in \mathbb{N}\)

\(\displaystyle 3^4 = 81 = 5m+1 \in \mathbb{N}\)

\(\displaystyle \bullet \) and for \(\displaystyle 5\) , we can not calculate for \(\displaystyle 5\) bcz all are in \(\displaystyle 5m\) form

\(\displaystyle \bullet \) So for \(\displaystyle 7\)

\(\displaystyle 7^0 = 1 = 5m+1 \in \mathbb{N}\)

\(\displaystyle 7^1 = 7 = 5m+2 \in \mathbb{N}\)

Now I did not understand how can i proceed after that

please help me , thanks

 

[JeffM]

How does your book define the natural numbers?