- Thread starter
- #1

- Thread starter jacks
- Start date

- Thread starter
- #1

- Admin
- #2

- Mar 5, 2012

- 9,485

Hi jacks!(1) The number of divisers of the form $2^2.3^3.5^3.7^5$ which are is in the form of $4n+1$ where $n\in\mathbb{N}$

(2) Calculate Total no. of positive Divisers of $7!$ which are is in the form of $3t+1\;,$ where $t\in \mathbb{N}$

Where are you stuck?

Did you try anything?