how many positive dividers that aren't multiple of 2 are there in the number 52920?

Question

i need to know how many positive dividers that aren't multiple of 2 are there in the number 52920.

How do i eliminate multiples of 2?

Answer 1

Every odd factor of $52920$ is also a factor of the largest odd factor of $52920$, that is $6615 = 52920/2^3.$

Can you take it from there?

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Edit to add: As the OP has outlined in a comment below, the number of divisors of a number depends on the number of different ways it can be broken up into prime factors. Here we have $52920 = 2^3\cdot 3^3\cdot 5^1\cdot 7^2$, and in general we can independetly select from zero to maximum exponent of each prime for each divisor. In this case we want to know the number of divisors where the prime $2$ has exponent zero, so $1\cdot (1+3)(1+1)(1+2)=24$ options.

Answer 2

Just keep dividing by $2$ until it is no longer even:

$52920=2 \cdot 26460$

$2646- = 2 \cdot 13230$

$13230 = 2 \cdot 6615$

OK, so find out the dividers of $6615$.

Now, since it ends in $5$ it's divisible by $5$, so divide by $5$:

$6615 = 5 \cdot 1323$

Also, any number is divisible by $3$ iff the sum of its digits is divisible by $3$, so this number is divisible by $3$ since $1+3+2+3=9$:

$1323 = 3 \cdot 441$

$4+4+1=9$, so still divisible by $3$:

$441=3 \cdot 147$

$1+4+7=12$ so still divisible by $3$:

$147=3 \cdot 49$

OK, and we know $49 = 7 \cdot 7$

OK, so we know $6615=3^3 \cdot 5 \cdot 7^2$

So, any positive divisor will be of the form $3^i \cdot 5^j \cdot 7^k$ with $0 \le i \le 3$, $0\le j \le 1$, and $0 \le k \le 2$

So, final question left for you: in how many ways can I pick $i,j,k$?

Answer 3

Lemma: If $N = \prod p_i^{k_i}$ is the unique prime factor $N$ then $N$ has $\prod (k_i+1)$ factors.

Why. The only prime factors of $N$ are $\{p_i\}$ so all factors will be of the form $\prod p_i^{m_i}$ (if we allow an $m_i$ to be $0$). No $m_i$ can be greater then $k_i$ so each $m_i$ is such $0\le m_i \le k_i$. All numbers of the form $\prod p_i^{m_i}; 0\le m_i\le k_i$ is a factor and only numbers of that form are factors. And as each $m_i$ may be anything from $0$ to $k_i$ inclusive there are $k_i + 1$ possible values for each $m_i$. ANd so there are $\prod(k_i + 1)$ possible combinations and so $\prod(k_i + 1)$ factors.

$52920= 2^3*3^3*5*7^2$ so $N$ has $(3+1)(3+1)(1+1)(2+1)$ factors all of the form $2^a3^b5^c7^d$. But we don't want any of those that are even; we only want those of the form $2^0*3^b5^c7^d= 3^b5^c7^d$. and there are $(3+1)(1+1)(2+1) = 24$.

We can list them all and it might be informative to do so:

$3^05^07^0-3^35^07^0:$ $1=3^05^07^0;3=3^15^07^0;9=3^25^07^0;27=3^35^07^0$

$3^05^17^0-3^35^17^0:$ $5=3^05^17^0; 15=3^15^17^0;45=3^25^17^0;135==3^35^17^0$

$3^05^07^1-3^35^17^1:$ $7=3^05^07^1; 21;63;189; 35;105;315;945$

$3^05^07^2-3^35^17^2:$ $49=3^05^07^2; 147;441;1323; 245;735;2205;6615$

Hopefully it is clear how I listed them in that order and hopefully it is clear why there can not be any other factors.