2
$\begingroup$

I would like to know how to solve the following question: Is 440 a factor of 72840?

I would have thought that this involves dividing 72840 by 440 and seeing if it produces an integer. However, the solution I have in front of me says "If 440 is a factor of 72840, then all prime factors of 440 should also be prime factors of 72840". So it asking us to find the prime factors of both numbers.

Now, this seems strange because it is a GRE question and therefore should be quite simple. However, 607 is a factor of 72840. I can't imagine having to check if 607 is a prime (which it is) on a GRE test.

Shouldn't it suffice to divide 72840 by 440 to see if it produces an integer? Or am I wrong in thinking that?

$\endgroup$
0

4 Answers 4

8
$\begingroup$

Yes, it does suffice to divide $72840$ by $440$ to see whether you get an integer. However, $440$ has an obvious factor of $11$, and there’s a simple test for divisibility by $11$ that $72840$ fails, so if you know that test, you can answer the question very quickly and move on $-$ a good thing on a timed exam.

The test, in case you don’t know it, is to form the alternating sum of the digits; the original number is divisible by $11$ if and only if the alternating sum of the digits is. In this case you get

$$7-2+8-4=9\;,$$

which is not a multiple of $11$, hence neither is $72840$.

$\endgroup$
1
$\begingroup$

You can use the divisibility rule for $11$: a base-$10$ number $$(a_na_{n-1}\ldots a_1a_0)_{10}=a_0+10a_1+\cdots +10^{n-1}a_{n-1}+10^na_n$$ is divisible by $11$ if and only if $$a_0-a_1+\cdots+(-1)^{n-1}a_{n-1}+(-1)^na_n$$ is divisible by $11$. This follows from the fact that $10\equiv -1\bmod 11$.

$\endgroup$
1
$\begingroup$

First note that you can remove the trailing zero from both the numbers. Hence, all you need to check is if $44$ divides $7284$.

To check if $44$ divides a number, all you need to check is whether $4$ divides it and $11$ divides it.

To check for divisibility by $4$, you need to check if the number formed by the last two digits is divisible by $4$. The number formed by the last two digits of $7284$ is $84$, which is divisible by $4$. Hence, $4$ divides $7284$.

To check for divisibility by $11$, you need to check if the difference between the sum of the odd placed digits and the even placed digits is a multiple of $11$. In your case the sum of the odd placed digits is $7+8 = 15$ and the sum of the even placed digits is $2+4 = 6$. The difference is $9$, which is not divisible by $11$.

Hence, $44$ doesn't divide $7284$. Hence, $440$ doesn't divide $72840$.

$\endgroup$
0
$\begingroup$

You are correct, with numbers this small you should just do long division. If the result is zero then 440 divides 72840. If the result is non-zero then 440 does not divide 72840.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .