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In how many ways can $1500$ be resolved into two factors?

Is there a formula for that or a smart way because if I do that by listing all the divisors of $1500$ it will take a lot of time.

Vinyl_cape_jawa
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bulbasaur
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2 Answers2

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The divisors of $1500$ are as follows: $$ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 125, 150, 250, 300, 375, 500, 750, 1500 $$ Now, if you want to write $1500$ as a product of two numbers, you have to pick one number on that list, and the other number will be forced. For instance, you might pick $300$, and in that case you're forced to pick $5$. It can be shown that on this list, you have to pick exactly one number above $40$, and one number below $40$. That means that the number of ways to write $1500$ as a product of two numbers is exactly equal to the number of divisors it has below $40$, which is $12$.

There is some more theory behind it. For instance, it's not really $40$ that is the important number here, it's $\sqrt{1500} \approx 38.7$. Given a number $n$ that is non-square, exactly half of $n$'s divisors will be below $\sqrt{n}$ and half of them will be above. If $n$ is a square, then as we will see there are an odd number of divisors, and then half of the divisors that aren't $\sqrt{n}$ are below $\sqrt{n}$, and the rest are above.

Now, for the number of divisors, prime factorization is a very powerful tool. For instance, $1500 = 2^2 \cdot 3^1 \cdot 5^3$. A number that divides $1500$ cannot have other primes, and it cannot have any of the same primes to any highter power. That means that all divisors are of the form $$ 2^i\cdot 3^j \cdot 5^l $$ where $i \in \{0, 1, 2\}, j \in \{0,1\}$ and $l \in \{0, 1, 2, 3\}$. Within these limits, however, we are completely free, and that measn that there are $3$ possible values for $i$, there are $2$ possible values for $j$ and there are $4$ possible values for $l$. In total, there are $3\cdot 2 \cdot 4 = 24$ different divisors of $1500$.

In general, if we have a number $n$ with prime factorization $$ n = p_1^{a_1}\cdot p_2^{a_2}\cdots p_m^{a_m} $$ then any divisor of $n$ has no other primes in its factorization, and the number of factor $p_i$ cannot exceed $a_i$. Therefore, there are $a_i + 1$ different values to choose from. In total, this means that the number of divisors of $n$ is $$ (a_1 + 1)(a_2 + 1)\cdots (a_m + 1) $$ If $n$ is a square, that means that all the $a_i$ are even numbers, which means that this product is an odd number. Otherwise, there are an even number of divisors.

Arthur
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  • +1 though I would have repeated at the end your earlier comment, so halving your final expression (rounding up if necessary). Then you could confirm the particular example $1500=2^2\, 3^1 \,5^3$, the number of its divisor pairs is $\lceil(2+1)(1+1)(3+1)/2\rceil = \lceil24/2\rceil =\lceil12\rceil =12$ – Henry Jun 21 '15 at 21:22
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    +1, but that assumes (e.g.) 2 x 750 is the same 'way' of writing it as 750 x 2. If these are different, the answer is 24, not 12. – abligh Jun 21 '15 at 21:39
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    @abligh The fundamental theorem of arithmetic assumes just that. Since this is in the same ball park, it seems like a safe assumption to make. – Arthur Jun 21 '15 at 21:48
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$1500$ has prime factorisation

$$2^2 \cdot 3 \cdot 5^3 $$

and breaking it into two factors is equivalent to listing a factor $d$, since the matching other factor is evidently $15000/d$. In how many ways can you choose a factor $d$ then?

Write $d=2^a \cdot 3^b \cdot 5^c$, where $0 \leq a \leq 2,0 \leq b \leq 1,0\leq c \leq 3$, and don't forget that $d$ and $15000/d$ give the same pair of factors.

user1337
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