Given N= $2^2*6^4*5^6=A*B$ In how many ways N can be written as a product of two numbers such that one of them is even and the other is odd?

My answer $N= 2^6*3^4*5^6$ Number of odd factors= $5*7=35$ factors multiplied by total number of even factor that is $6$. $35*6=210$ But the answer is $35$. How?

  • 4
    $\begingroup$ You are right $N$ has 35 odd factors. And once you select the odd factor, the even factor is completely determined. Hence there are 35 such pairs. $\endgroup$ – symplectomorphic Aug 8 '17 at 19:08
  • $\begingroup$ Why do you think the number of ways to write a product with will be the number of each type of factor multiplied together? $\endgroup$ – fleablood Aug 8 '17 at 19:25
  • $\begingroup$ Think of this. How many ways are there to write 12 as a product of a prime and a composite. There are two primes factors. And there are three composite factors. So there are $12$ ways to write this??? How many ways are there to write 12 as a product of two numbers. There are 6 factors so there are $6^2$ ways to write this? Are there $6^3$ ways to write it as a product of $3$ factors? $\endgroup$ – fleablood Aug 8 '17 at 19:31
  • $\begingroup$ The number of even factors of $N$ is $70,$ not $6.$ If $A$ is any odd factor of $N$ then $2N$ and $4N$ are even factors of $N.$ $\endgroup$ – bof Aug 8 '17 at 21:49

Your logic simply does not follow. The odd factors are $3^i*5^j$ and there are $35$ of them. But the even factors are not the $6$ of the form $2^k; k \ge 1$. There are of the factors of the form $2^k3^i5^j; k > 1$ and there are $210$ of them.

But that doesn't matter. Because you don't multiply the number of even factors by the number of odd factors to get the answer. Why would you? Just because one term is even and another odd doesn't mean they will multiply out to $N$.

You need $a*b$ $a$ odd and be even and $a*b = 2^6*3^4*5^6$. So $a = 2^j3^k5^l; b = 2^{6-j}3^{4-k}5^{6-k}$. But as $a$ is odd, $j=0$ and $6-j= 6$.

So $a = 3^k5^l$ and $b=3^{4-k}5^{6-l}$. $k$ can be anything from $0$ to $4$ so there $5$ choices, and $l$ can be anything from $0$ to $6$ so there and $7$ choices for that.

So there are $35$ ways of doing this.

Another, and far easier way to do it is to note that for any value of $a$ then $b$ must equal $N/a$. So the for each odd factor, the other even factor must be fixed. SO there are as many ways to do this as there are odd factors.

===old answer ===

There are 35 odd factors. True.

There are 6 even factors. False.

There are $6$ even powers of $2$; $2^1, 2^2,.... , 2^6$. And each of those can be multiplied by an odd factor to get another even factor. Examep: $2^i*3$ or $2^i*3^3 *5^2$ etc. So there are $6*35 = 210$ even factors.

There are 210 even factors. True.

There are $210*35$ ways to combine these factors. True.

Evey way to combine these factors will equal $N$. !!!!!FALSE!!!!

Example. $3$ is an odd factor. $2$ is an even factor. $3*2 \ne N$ so we can not count this example. Taking all $210*35$ combinations does take that example.

If $a$ is an odd factor. And $a*b=N$ then $b = \frac Na$ and that is the only possible even factor that can be multiplied by $a$ to get $N$.

So for each of the $35$ odd factors, we do NOT have $210$ options for an even factor. We have ONE option. One only!. So we do not have $35*210$ ways of doing this.

We had $35*1$ ways of doing this.

  • $\begingroup$ Right. Or maybe $35*2=70$ ways if we count $N=81*1,000,000$ and $N=1,000,000*81$ as two different ways. $\endgroup$ – bof Aug 8 '17 at 21:52
  • $\begingroup$ @bof, good point. I thought of that ... and then I completely forgot it. I'd say most would say "a is the even, and b the odd" would be the same whether we multiply the odd first or the even first. $\endgroup$ – fleablood Aug 9 '17 at 1:30
  • $\begingroup$ Seem I misunderstood the OP. S/he thought there were only $6$ even factors. There are only $6$ even powers of $2$ but those can be multiplied by odd factors to get an even number of factors. $\endgroup$ – fleablood Aug 9 '17 at 1:53
  • $\begingroup$ Of course the ambiguity is resolved by the OP telling us that the answer is supposed to be $35.$ $\endgroup$ – bof Aug 9 '17 at 1:54
  • $\begingroup$ Oops, I just now noticed that the OP has both $N=2^2*6^4*5^6$ (twice counting the title) and $N=2^6*3*4*5*6$ once. $\endgroup$ – bof Aug 9 '17 at 1:58

Once you choose the odd factor (say, $A$), there's only one possibility for the other (it has to be $\frac{N}{A}$).

  • $\begingroup$ I did not understand @platty $\endgroup$ – Sakuzi Markel Aug 8 '17 at 19:03
  • $\begingroup$ We need $AB = N$. Let's say $A$ is the odd factor and $B$ is the even factor. You are correct in stating that there are 35 choices for $A$. But after you choose $A$, there's only one possible way to choose $B$ so that you get $AB = N$. $\endgroup$ – platty Aug 8 '17 at 19:04
  • $\begingroup$ How come there is just one possibility to choose B. I want the product such that one is even and the other is odd. How can you say this? @platty $\endgroup$ – Sakuzi Markel Aug 8 '17 at 19:06
  • $\begingroup$ Try a small example: $24 = 2^3 \times 3$. How many ways can you choose an odd $A$ and even $B$ such that $AB = 24$? Can you list them out? $\endgroup$ – platty Aug 8 '17 at 19:07
  • $\begingroup$ 8 ways. @platty $\endgroup$ – Sakuzi Markel Aug 8 '17 at 19:08


The 2 factors must be even or 1 factor odd another factor even. There is no possibility that both factors are odd as you have a 2 as one of the factors of $N$

No. of odd factors $=35$

For each odd factor there must an even factor to get product as $N$

So no. of ways $=35$


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