How many divisors of N ended by 5 I must know how to find how many divisor of N ended by 5 ? In my exercise, I have $\ N=63'000 = 2^3*3^2*5^3*7 $ and I can found the number of divisors of N using $\ (3+1)*(2+1)*(3+1)*(1+1)=96$ 
Among these 96 divisors, how many ended by 5 ? How can calculate this ?
Thank you so much

*

*Take out the $2$s, because when multiplied by $5$, the result will end with $0$.

*Find the number of divisors of $3^2\cdot7^1$, which is $(2+1)\cdot(1+1)=6$:


*

*$3^0\cdot7^0$

*$3^1\cdot7^0$

*$3^2\cdot7^0$

*$3^0\cdot7^1$

*$3^1\cdot7^1$

*$3^2\cdot7^1$


*Multiply each divisor by each one of the following $3$ powers of $5$:


*

*$5^1$

*$5^2$

*$5^3$


*Hence you have $6\cdot3=18$ divisors which end with $5$:


*

*$3^0\cdot7^0\cdot5^1$

*$3^1\cdot7^0\cdot5^1$

*$3^2\cdot7^0\cdot5^1$

*$3^0\cdot7^1\cdot5^1$

*$3^1\cdot7^1\cdot5^1$

*$3^2\cdot7^1\cdot5^1$

*$3^0\cdot7^0\cdot5^2$

*$3^1\cdot7^0\cdot5^2$

*$3^2\cdot7^0\cdot5^2$

*$3^0\cdot7^1\cdot5^2$

*$3^1\cdot7^1\cdot5^2$

*$3^2\cdot7^1\cdot5^2$

*$3^0\cdot7^0\cdot5^3$

*$3^1\cdot7^0\cdot5^3$

*$3^2\cdot7^0\cdot5^3$

*$3^0\cdot7^1\cdot5^3$

*$3^1\cdot7^1\cdot5^3$

*$3^2\cdot7^1\cdot5^3$


