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Question as in title,

Find the number of common, odd positive divisors of $27, 900$ & $20, 700$.

explaining the steps in your method would be much appreciated.

I prime factorized and got that the common factor is $2^2\cdot3^2\cdot5^2$, then my notes from the relevant lecture simply go on to a solution implied from that step that the number of common odd positive divisors = $(2+1)(2+1)=9$. How the one leads to the other is where I've fallen down.

So how can you tell the number of common, odd positive divisors of $2^2\cdot3^2\cdot5^2$?

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    $\begingroup$ What are your efforts? $\endgroup$ – Jaideep Khare May 2 '17 at 23:01
  • $\begingroup$ I prime factorized and got that the common factor is 2^2*3^2*5^2, then my notes from the relevant lecture simply go on to a solution implied from that step that the number of common odd positive divisors = (2+1)(2+1)=9. How the one leads to the other is where I've fallen down. $\endgroup$ – Asad3000 May 2 '17 at 23:04
  • $\begingroup$ So would the following question better reflect what you don't understand? "Why does the common factor of $2^3\cdot3^2\cdot5^2$ lead to there being $(2+1)(2+1)=9$ common, odd positive divisors?" $\endgroup$ – Ian Miller May 2 '17 at 23:08
  • $\begingroup$ Perhaps, or rather "How can you tell the number of common, odd positive divisors of 2^2 * 3^2 * 5^2" may be more indicative of my level of understanding of/difficulty with the original question. Any hints as to that would be super. $\endgroup$ – Asad3000 May 2 '17 at 23:17
  • $\begingroup$ @IanMiller Comment do not always last what does that mean? $\endgroup$ – Jaideep Khare May 2 '17 at 23:48
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Factorize $27{,}900$ and $20{,}700$ :

$$27{,}900=2^2 \times 3^2 \times 5^2 \times 31 ~;~ 20{,}700=2^2 \times 3^2 \times 5^2 \times 23$$

For their common odd divisors ; you need total number of divisors of $3^2 \times 5^2$.

For counting number of divisors of $3^2 \times 5^2$, you can have $0 , 1$ or $2$ power of $3$ from $3^2$. Hence three ways to chose exponent of $3$ in the common divisor. Similarly for exponent of $5$.

Therefore we have : $3 \times 3 = 9$ total number of common, odd positive divisors of $27{,}900$ & $20{,}700$.

In general, for exponent $\alpha$ of prime $p$ in a number, you'll have $\alpha+1$ ways to select an exponent of $p$ in the divisor.

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Alternative hint (without factoring the two numbers):

  • determine $\gcd(27900,20700)=900$ using the Euclidean algorithm for example;

  • drop the factors of $2$ from $900$ by halving until you get an odd number, in this case $225\,$.

The common, odd positive divisors of the two original numbers will be precisely the odd, positive divisors of $225=2^2 \cdot 3^2\,$, and there is $(2+1)\cdot(2+1)=9$ of them, as pointed out already.

The advantage of this approach is that you only have to factor the (small) number $225\,$, rather than the two (big) numbers originally given.

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