Textbook

*Two conditions* questions are unique to the CLT among college admissions tests. They provide a range of integers and ask the student to determine how many integers in the given range meet both, at least one, or neither of two criteria. Unfamiliarity may make them seem challenging at first, but those that appear in the first half of the CLT Quantitative Reasoning section need not be intimidating. Try the question below, then explore strategies for these types of questions before attempting examples at whichever difficulty levels are appropriate for your current abilities.

Note any key terms that require defining in order to answer the question below. Then make your best attempt.

How many integers between $100$ and $500$, inclusive, meet both of the following conditions given in the statements below?

1: The number is a perfect square.

2: The product of the number’s digits is not zero.

A) 0

B) 2

C) 9

D) 11

Among the 401 numbers in the range from $100$ to $500$ *inclusive*, we need to find how many meet both listed criteria. *Perfect squares* are those values that are equal to integer squared (the first few examples are $1$, $4$, $9$, and $16$). (We *highly* recommend you memorize all the perfect squares all the way up to $20_{2}$). We would do well to begin with this condition, because the other condition, having digits whose product is not zero, contains hundreds of examples in the range between 100 and 500 inclusive.

To list the perfect squares, we can start at the smallest number in the range, 100, noting that we have already found our first perfect square, since $100=10_{2}$. From there, we can list $11_{2}$, $12_{2}$, etc., until we get close to $500$. That gives us $100$, $121$, $144$, $169$, $196$, $225$, $256$, $289$, $324$, $361$, $400$, $441$, and $484$. As you may notice, the difference between any two consecutive numbers is growing as we go, so the next number after $484$ will be too large to stay less than or equal to $500$.

The list above contains 13 numbers. But we need to remove any number where the product of its digits is zero. That eliminates $100$ and $400$. Eleven numbers remain, so the answer is **D**.

Instead of asking for integers that meet both given conditions, the CLT may ask you to meet *at least one* or *neither* of the two conditions.

*Important reminder:*The CLT does not allow calculator usage! But as we can see from the approach question at the beginning of this module, the test still expects you to know your multiplication facts. We recommend that you*memorize the times tables up to $15×15$*. Using physical flashcards or a flashcard problem like Quizlet is highly recommended. It’s important to be automatic with your multiplication facts so you can spend your time with actual problem-solving.- In addition, even beyond $15$, you should
*memorize the perfect squares up to $20×20$*. You’ll be happy you did when the question calls for such calculation and you’re not stuck writing out long multiplication! - Always make definition flashcards for anything listed under “key terms” above that you are not already familiar with.

**Difficulty 1**

How many integers between 100 and 200, inclusive, meet both of the following conditions given in the statements below?

1: The number is divisible by 13.

2: The number is even.

A) 4

B) 6

C) 8

D) 10

(spoiler)

The answer is **A**. Following the strategy of starting with the more accessible statement, let’s begin with statement 1 because it will have fewer examples than statement 2 - better not to list all the even numbers from 100 to 200 if we don’t have to! If you take our advice and do your times tables up to $15$, you will be in good shape working with $13$ as a factor. But if you need help, consider that $13×10=130$. From $130$, we can count up and down by $13$ at a time. The lowest number will be $104$ and the highest will be $195$. Among this whole list, we now count the even integers: $104$, $130$, $156$, and $182$. There are four.

**Difficulty 2**

How many numbers between 10 and 50 (inclusive) meet both of the conditions given in the statements below?

1: The number is divisible by 4.

2: The last digit is a composite (non-prime), nonzero number.

A. 6

B. 7

C. 8

D. 9

(spoiler)

The answer is **A**. The simpler of the two tasks here is to list the multiples of $4$ in the relevant range: $12$, $16$, $20$, $24$, $28$, $32$, $36$, $40$, $44$, $48$. Which of these fits the second condition? We can rule out $20$ and $40$ because the last digit must be *nonzero*. But the last digit also must be composite, and we mustn’t forget that *2 is prime*. That takes out $12$ and $32$ as well, leaving only **6** examples.

**Difficulty 3**

How many integers between $1$ and $100$ (inclusive) meet both of the conditions below?

1: The integer is a prime number doubled.

2: The integer’s units digit is 2.

A. 3

B. 4

C. 5

D. 6

(spoiler)

The answer is **A**. The second statement is the simpler, and it doesn’t take long to list out $2$, $12$, $22$, etc. all the way to $92$. Which of these is a prime number doubled? Certainly not $2$, since $1$ isn’t prime. $12$ doesn’t work either since half of it is $6$. In fact, all the numbers that are double an even number don’t work, so we can take out $32$, $52$, $72$, and $92$. Of the remaining numbers, $22$ looks good; half of it is $11$, which is prime. $42$ might look good at first, but $21$ is not prime since it is divisible by $3$. The remaining numbers, $62$ and $82$, join $22$ as good candidates, so there are a total of **3**.

We can sum up by noting that if a number’s units digit is $2$, half that number must end in either $1$ or $6$. All the values that end in $6$, being even numbers, are immediately eliminated from prime number contention.

**Difficulty 4**

How many even integers between 2 and 200 (inclusive) meet at least one of the conditions below?

1: The integer is a perfect square.

2: The integer is a multiple of 98.

A. 8

B. 9

C. 14

D. 16

(spoiler)

The answer is **A**. From the second statement (and make sure you don’t confuse *multiple* with *factor* here) we find a very small number of fitting examples: just $98$ and $196$. So we will need to list the examples filling the first statement. But let’s keep two important words/phrases in mind. First, unlike most “Two Conditions” questions, this question is asking which number meet *at least one* condition (not both). Second, we dare not overlook the word “even” in the question, which narrows down our list considerably.

With these things in mind, we start listing the perfect squares starting with $2_{2}=4$. How high does the list go? This is another example of a problem showing why knowing multiplication facts up to $15$ is helpful: $14_{2}=196$, so that’s our last example. So the numbers satisfying the first condition are $4$, $16$, $36$, $64$, $100$, $144$, and $196$. Adding the two examples we got from the other statement, we get a total of 9 numbers. But watch out for double counting! $196$ is present in both lists but only gets counted once. Therefore the answer is A.

**Difficulty 5**

How many integers between 80 and 120 (inclusive) meet neither of the conditions below?

1: The integer is a multiple of 3.

2: The integer’s digits add up to an even number.

A. 10

B. 11

C. 12

D. 13

(spoiler)

The answer is **C**. Like the last question, this “Two Conditions” question is unusual, but this time it’s because the key word is “neither”. We must catalog all the numbers that don’t fulfill *either* condition.

To do this, we should list the *non-multiples* of 3 in the given range, even though the list is pretty long. (Try it now on your scratch paper.) Starting with $80$ and ending $119$, we have a total of 26 integers. Now let’s go through those integers and cross out the ones whose digits add up to an even number. So in the $80$’s we can cross out $80$, $82$, $86$, and $88$. Continuing the process, we’ll get rid of $91$, $95$, and $97$. The numbers greater than $100$ in the “cross-out” category are $101$, $103$, $107$, $109$, $112$, and $116$. There should be a total of 12 numbers left on your list.

Interestingly, if you scrutinize your list carefully, you’ll notice that exactly three numbers remain from each “decade” in our list: the 80’s, the 90’s, the 100’s, and the 110’s. At least we have some sort of pattern, so if you had little time on this question and had to make an educated guess, you might find three candidates in the 80’s and assume that each of the following decades has the same number of candidates. There are worse educated guesses out there, no doubt!

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