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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

Cyrill Gremaud
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    The ones that have at least one factor $5$, and at the same time no factor $2$ will end in $5$. – Arthur Jan 13 '15 at 12:17

1 Answers1

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

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

    • $3^0\cdot7^0$
    • $3^1\cdot7^0$
    • $3^2\cdot7^0$
    • $3^0\cdot7^1$
    • $3^1\cdot7^1$
    • $3^2\cdot7^1$

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

    • $5^1$
    • $5^2$
    • $5^3$

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

    • $3^0\cdot7^0\cdot5^1$
    • $3^1\cdot7^0\cdot5^1$
    • $3^2\cdot7^0\cdot5^1$
    • $3^0\cdot7^1\cdot5^1$
    • $3^1\cdot7^1\cdot5^1$
    • $3^2\cdot7^1\cdot5^1$
    • $3^0\cdot7^0\cdot5^2$
    • $3^1\cdot7^0\cdot5^2$
    • $3^2\cdot7^0\cdot5^2$
    • $3^0\cdot7^1\cdot5^2$
    • $3^1\cdot7^1\cdot5^2$
    • $3^2\cdot7^1\cdot5^2$
    • $3^0\cdot7^0\cdot5^3$
    • $3^1\cdot7^0\cdot5^3$
    • $3^2\cdot7^0\cdot5^3$
    • $3^0\cdot7^1\cdot5^3$
    • $3^1\cdot7^1\cdot5^3$
    • $3^2\cdot7^1\cdot5^3$
barak manos
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