Factors and Multiples

The answer is C. As noted in the Definitions section, both divisor and factor refer to a number that divides evenly into another number. That means, in turn, that the greatest common divisor (identical to the greatest common factor) must be the largest number that divides into two or more numbers. Choice A doesn’t work here because, although $2$ is a divisor of both $8$ and $12$, it is not the largest divisor. Choice B is not right because $3$ divides evenly into $12$ but not into $8$. Similarly, choice D gives us a number that divides into $12$ but not into $8$. Choice C proposes the largest number that is a divisor of both $8$ and $12$.

The answer is B. Choice A doesn’t work because 15 is a factor of 90 but not a multiple of 6. Choice C has the same problem. Meanwhile, Choice D features the opposite conditions: 60 is a multiple of 6 but not a factor of 90.

The answer is A. We can approach this problem by recognizing that the prime factors of $6$ are $2$ and $3$. This means that any multiple of $6$ must also be multiples of $2$ and $3$. In one step, we’ve confirmed that statements I and II are true. What about statement III? For a number to be a multiple of $4$, it must contain two $2$’s as factors. But we only know for sure that this number contains one $2$. So statement III goes too far for us to know for certain.

If you prefer to plug in numbers, you could try $12$, $18$, and $24$. All three of those numbers are multiples of $2$ and $3$, so this exercise appears to confirm statements I and II. Regarding statement III, although $12$ and $24$ and multiples of $4$, $18$ is not, so we have proved that statement III does not work.

The answer is C. Let’s begin by removing the even numbers and leaving ourselves with only $31$, $33$, $35$, $37$, and $39$. Which of these appear to be composite? Certainly $35$, since its last digits tells us it’s divisible by $5$. Let’s use our divisibility trick for $3$ with the other numbers; which of them have digits that add up to a multiple of $3$? It looks like $33$ and $39$ fit the bill while $31$ and $37$ do not.

A handy way to speed up your process with primes is to memorize the perfect squares through at least $10$ (you should definitely do this for its own benefit!). If a number is below a perfect square and is not prime, it must be divisible by the prime numbers below the square root of that perfect square. For example, if a number is less than $7^2$, which is $49$, if it’s not prime it must be divisible by the prime numbers less than $7$: $2$, $3$, or $5$. Using this strategy, we can confidently say that $31$ and $37$ are prime because they are divisible by none of these factors.

It looks like we have three composite, odd numbers. But can we be sure that $31$ and $37$ are composite? A good thing to remember is that the smallest composite, odd number not divisible by $3$ or $5$ is $49$. So with smaller numbers, if an odd number is not divisible by $3$ or $5$, you can be confident that it is prime.

The answer is B. This is a difficult problem to approach without a calculator. Remember the UnCLES Method: look at the answers! The way the question is phrased, three of the answers must evenly divide into $1,001$, so we can begin anywhere. Why not start with $11$ since $10\times 11=1,100$, which is reasonably close to $1,001$? Since $1,100-1,001=99$ and $99$ is a multiple of $11$, we can see that $11$ works. But let’s use that to our advantage by finding the actual quotient; if $11$ goes into $1,100$ 100 times and we then subtracted a total of nine $11$’s, then the quotient must be $100-9-91$.

(There is also a cool divisibility trick for $11$, but this will apply only rarely, so feel free to skip to the next paragraph if you’d like to save time. If the sum of the odd-ordered (first, third, fifth, etc.) digits of a number is equal to the sum of the even-ordered digits of the number, or if the sums differ by $11$ or a multiple of $11$, then $11$ is a factor of that number. We can see at a glance that the year $2024$ is divisible by $11$, since $(2+2)=(0+4)$, and that $1969$ is also divisible by $11$ since $(1+6)$ is exactly $11$ less than $(9+9)$. One can see at a glance that 1,001 must be divisible by $11$.)

We now have a much smaller number to work with, but the number is still difficult. Does $91$ have any factors besides $1$ and itself? It might not seem so at first because, since its digits don’t add up to a multiple of $3$, $91$ is not divisible by $3$. It clearly doesn’t divide by $5$, so let’s try $7$. If you realize that $7$ goes into both $70$ and $21$, then $7$ must go evenly into $70+21$, which s $91$. (There is also a cool trick involving two numbers equidistant from a multiple of $10$: in this case $7\times 13=(10-3)(10+3)=10^2-3^2$. This always works with numbers on opposite sides of $10$-multiples: so, $98\times 102=100^2-2^2=9996)$. And $7$ goes in $10+3=13$ times, so have our other two factors of the original number: $7$ and $13$!

Before we leave this question, we would be remiss not to mention that there is a faster way to arrive at the solution here. You are looking for the answer choice that is NOT a factor of $1,001$. As a reminder, the “adding the digits” trick works not just with $3$ but with $9$ also. If a number’s digits don’t add to a multiple of $9$, then the number itself is not divisible by $9$. We can test this with $1,001$ and see right away that it’s not divisible by $9$ … which gives us the answer right away!