3, 5, 7 are the only three consecutive odd integers that are each a prime.

Digits

In the integer 5,432, 5 is the thousands digit, 4 is the hundreds digit, 3 is the tens digit, and 2 is the units digit.

In a multiple-digit number, there are 10 possible units digits (0 – 9). In the first digit, i.e. the digit farthest to the left, there are only nine possible digits. A zero as the first digit would not be valid. For instance, 0,453 is not a four-digit number. It’s – if anything – the three-digit number 453.

Divisibility Rules

3: Add up the digits in any number. If they are multiple of 3, then the number is divisible by 3. For example, 258 is divisible by 3 because 2 + 5 + 8 is 15. And 15 is a multiple of 3.

4: If the tens and units digits of integer X is a multiple of 4, then integer X is divisible by 4. For example, 364 is divisible by 4, because 64 is a multiple of 4.

5: If a number ends in 5 or 0, it is divisible by 5.

6: If the rules to divisibility to 2 and 3 are fulfilled, then a number is divisible by 6. One hundred and eleven is not divisible by 6. Two hundred and forty six, because it is divisible by both 2 and 3, is divisible by 6.

9: If the sum of the digits of a number is divisible by 9, then that number is divisible by 9. For example, adding up the digits in 4,536, we get 4+5+3+6 = 18. Because 18 is divisible by 9, the number 4,536 is also divisible by 9.

Odds (so to speak) and Ends

Zero is an even integer

Integers can be negative numbers (-3, -6, -17, etc.)

0! = 1

x^0 = 1

The problems below can be solved by referencing the above. Good luck!

Practice Problems

1. What is the smallest integer that can be divided by the product of a prime number and 7 while yielding a prime number?

(A) 7

(B) 14

(C) 24

(D) 28

(E) 35

2. The sum of five consecutive even integers is 20. What is the product of the median of the series and the smallest integer in the series?

(A) 12

(B) 10

(C) 6

(D) 4

(E) 0

3. Which of the following integers is NOT divisible by 3?

I did not know exactly which part of the blog is most suitable for putting this comment. I did not come across this method in the lesson videos so far so I am sharing it here:

There is another very useful and quick method to find GCF without prime factorization. For 1200 and 720,for example the method would be: 1200/720 gives us 1 as quotient and 480 as remainder, now divide 720 by 480 to get 1 as quotient and 240 as remainder, divide 480 by 240. We get 0 as remainder so 240 is the GCF of the two numbers. The idea is to divide till remainder is 0. Students often make mistakes while writing the exponents. In this method we don’t need that and the quotients are usually very small so it is easy.

I took the GRE last week and I still remember a question that bothers me so much because I don´t remember to review it during my preparation. I want now to take the GRE again because I´m not happy with my score. The problem was about a unit digit K, it was given an integer e.g. 567k89 and the question asked which of the following numbers (which I can´t remember) is a possible value of k. Could you please tell me in which category I can find problems like this or how can I prepare for this specific questions? Thanks!

Clemmonsdogpark Test Prep ExpertApril 5, 2016 at 11:29 am#

I’d have to see the full problem to be sure, but it sounds like you’re dealing with a “number properties” problem. Problems that deal with the characteristics of integers and other numbers are somewhat rare on the GRE, but you can expect to see at least five of these questions across the test, sometimes more. Because there are just a few of these problems and they can appear in any question type (quantitative comparison, data analysis, multiple answer, etc…), it’s hard to really find a full practice session that just focuses on this math skill.

However, you can find problems like the one you described alongside other related and relevant problems by seeking out arithmetic and integer property GRE math problems. Clemmonsdogpark GRE subscribers can view video lessons and customized practice sessions for these types of problems. You can also find materials and practice questions on arithmetic and integer property math in the GRE Official Guide. To seek out additioasl questions of this variety online, do searches for things like GRE arithmetic, GRE integer properties, GRE number theory, and so on. I’ve also found that searching for the word “GRE” along with word “integer” and the exact phrase “a possible value of” .

I have my exam in 2 days and i have a few doubts, hope you can help me out:

1) What is the root of 0? does it exist?is it 0?

2) Is 0 an even number? many question explanations on Clemmonsdogpark use 0 as an even number to test for substitution when the question requires an even number.

In another example where there was a QC question asking to compare the product of all even integers from say -42 to 8. Here we just count the number of integers and try to find the sign but shouldn’t the product also include a 0?

Clemmonsdogpark Test Prep ExpertJune 28, 2018 at 9:09 pm#

Hi Kinnar,

Before we get started, my apologies to Yogesh. I read through the problem a little too fast. Column A is actually NOT the higher quantity,because it’s a negative number. Column B is 0, which is NOT negative, and thus would be the higher (or “greater”) quantity.

Kinnar, you are correct that taking out the 0 would keep the answer the same. However, the answer would still be Quantity B, not Quantity A. Instead, Quantity A would be the larger negative number, still. With 0 removed from Quantity B, B is 21 negative numbers, with the largest negative being -42, multiplied by each other. But A is 27 negative numbers multiplied by each other, with the largest negative number starting at 57. So Quantity A, when compared to quantity B, will yield the larger negative number, which is thus “lower” because it’s farther below 0.

So the square root of 0 is simply 0. (I don’t see this coming up much in test questions).

Next, 0 is definitely an even integer. This comes up from time to time, so if a question is asking for the product of even integers and one of those even integers is 0, then the product will equal ‘0’.

Hello, what is the quickest/most efficient way to solve question 2? Is there a formula or strategy for solving a question like this for much larger number? Thanks!

So the most efficient way may not necessarily seem that efficient, and I think a lot of people expect some kind of formula. But really what the GRE is testing is your ability to think on the spot and test out solutions. It intentionally makes many problems “formula proof”–so you can’t just solve them by merely “plugging and chugging”.

So the GRE would never create a problem in which the numbers are so big that you can’t just test out numbers to arrive at a solution.

With this problem, you know the numbers can’t be that large, since the sum has to be less than 20. 2 is a good place to start. I add up 2 + 4 + 6 + 8 + 10. Even without adding those all up I can see that is way to large. So I have to make the numbers smaller. The trick is remembering that zero is even. That way the next lowest five consecutive even numbers are 0 + 2 + 4 + 6 + 8, which equals 20. Anything times the least of these numbers (which is ‘0’) will give you zero. Therefore, (E).

It may seem that that approach is time-consuming. But we barely had to add up any numbers in the first case, and adding the numbers in the second case is a straightforward. Should be a 45 sec to 1 minute solution.

Clemmonsdogpark Test Prep ExpertMarch 25, 2016 at 12:09 pm#

Hi Tanvir,

I’m happy to help with this question. First, it’s important to understand what’s being asked. If we were to write out the question as an equation, we could say

x / (P_1*7) = P_2

where x is the correct answer choice, P_1 is a prime number and P_2 is a prime number. At this point, it’s good to note that P_1 and P_2 may or may not be equal–we don’t know based on the information given in the problem.

Solving the equation for x, we get the following:

x = P_1*P_2*7

From here, we can determine the correct answer by finding the prime factorization of each answer choice. The prime factorization of the correct answer will be in the form P_1*P_2*7.

A) 7 = 7 B) 14 = 2*7 C) 24 = 2*2*3 D) 28 = 2*2*7 E) 35 = 5*7

As we can see, the only answer choice that fits our criteria is D, 28. We can double check by comparing 28 to the first equation I mentioned:

x / (P_1*7) = P_2 28/(2*7) = 2

In this case, P_1 = P_2 = 2, and D is the correct answer 🙂

The first question–which is confusingly worded–is asking what is the smallest number that you can divide by a prime number and 7, and end up with a prime number.

Let’s call X the number we are looking for: X/(7)(prime number). Since we want to make X as small as possible, we want the prime number in the denominator to be the smallest possible prime, which is 2: X/(7)(2) = X/14. We need to make sure that when you divide X by 14 that you end up with a prime number. Again, we want to make X as small as possible. Therefore, the prime number we end up with should be as small as possible (which again is the number 2): X/14 = 2. Therefore, X = 28.

The question says divisible. 24/21 is not divisible, it leads to a fraction ( not sure if fractions can be considered in prime no.) While 28/14 is divisible and yield to a Prime no.

I have a question related to this post: Usually when a question says “the positive integers” I usually find in the answer, zero was not considered. In fact in ALL questions that say this, zero was not considered as a possibility.

I am not sure which questions you are talking about as none of the above mention “positive integers.” Also one of the above questions includes zero as an answer choice.

For problem 1, why is the answer not B = 14? 14 can be divided by 2, which yields 7 which is a prime number; and 14 can also be divided by 7, which yields 2 which is a prime number.

Yep, those are the integers and because ‘0’ is the smallest number anything you multiply by it will give you zero so you don’t even have to worry about finding the median.

Im a little confused to question two as well… My first thought to solve this problem is to write out: x, x+1, x+2, x+3, x+4=20 and solve by x which i get 2, then to write out the answers 2,3,4,5,6.. which shows the median of 4. Not too sure how to even think about getting 0 from that because I can’t just assume that the “even integers” start at 0..

With the algebraic equation you want to make sure to use x+2, x+4, etc. because the numbers are consecutive even numbers, meaning that they have to jump up by 2. That gives us the equation x+x+2+x+4+x+6+x+8 =20, x =0.

And that’s always a good one of those little math conventions to know: ‘0’ is an even number.

Aw, gotcha… that helps a lot.. because i’m still keeping that same formula that I finally mastered. So in essence is they were asking for the “odd consecutive integers” it would be as follows: x,x+1,x+3 etc.. x=1. Aside from this problem, I get a little confused with the procedure when they ask for the product of consecutive integers. Would you mind briefly going over a good strategy for that. Aside from this problem, I get a little confused with the procedure when they ask for the product of consecutive integers. Would you mind briefly going over a good strategy for that.

So, for consecutive odds you would still use the exact same formula, since odds are separated by 2. Using your formula (x, x+1, x+3…), if x = 3, then the series would be 3, 4, 6, etc.

If they ask for the product of consecutive integers that is just a factorial sign (assuming the number starts with 1). Typically, the GRE won’t just ask you to multiply a bunch of consecutive integers.

Clemmonsdogpark Test Prep ExpertApril 28, 2016 at 9:08 am#

Hi Avani,

You’re on the right track but your equation only includes 4 even consecutive integers, while the problem asks us to consider a situation with 5 even consecutive integers. If you include another integer, either x-4 or x+6, you should get the correct answer 🙂

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is zero a positive integer or non negative integer ?

Hi Hardik,

On the GRE, 0 is considered even but it is neither positive nor negative. This can be seen in the official GRE math conventions. I hope that helps! 🙂

But zero is technically an “integer” on the GRE, correct?

Yes! 0 is considered an integer.

I did not know exactly which part of the blog is most suitable for putting this comment. I did not come across this method in the lesson videos so far so I am sharing it here:

There is another very useful and quick method to find GCF without prime factorization. For 1200 and 720,for example the method would be:

1200/720 gives us 1 as quotient and 480 as remainder, now divide 720 by 480 to get 1 as quotient and 240 as remainder, divide 480 by 240. We get 0 as remainder so 240 is the GCF of the two numbers. The idea is to divide till remainder is 0. Students often make mistakes while writing the exponents. In this method we don’t need that and the quotients are usually very small so it is easy.

Hey Chris,

I took the GRE last week and I still remember a question that bothers me so much because I don´t remember to review it during my preparation. I want now to take the GRE again because I´m not happy with my score. The problem was about a unit digit K, it was given an integer e.g. 567k89 and the question asked which of the following numbers (which I can´t remember) is a possible value of k. Could you please tell me in which category I can find problems like this or how can I prepare for this specific questions? Thanks!

I’d have to see the full problem to be sure, but it sounds like you’re dealing with a “number properties” problem. Problems that deal with the characteristics of integers and other numbers are somewhat rare on the GRE, but you can expect to see at least five of these questions across the test, sometimes more. Because there are just a few of these problems and they can appear in any question type (quantitative comparison, data analysis, multiple answer, etc…), it’s hard to really find a full practice session that just focuses on this math skill.

However, you can find problems like the one you described alongside other related and relevant problems by seeking out arithmetic and integer property GRE math problems. Clemmonsdogpark GRE subscribers can view video lessons and customized practice sessions for these types of problems. You can also find materials and practice questions on arithmetic and integer property math in the GRE Official Guide. To seek out additioasl questions of this variety online, do searches for things like GRE arithmetic, GRE integer properties, GRE number theory, and so on. I’ve also found that searching for the word “GRE” along with word “integer” and the exact phrase “a possible value of” .

Hey Chris,

I have my exam in 2 days and i have a few doubts, hope you can help me out:

1)

What is the root of 0? does it exist?is it 0?

2)

Is 0 an even number? many question explanations on Clemmonsdogpark use 0 as an even number to test for substitution when the question requires an even number.

In another example where there was a QC question asking to compare the product of all even integers from say -42 to 8. Here we just count the number of integers and try to find the sign but shouldn’t the product also include a 0?

Hi Rahul,

The square rt of 0 is equal to 0, since ‘0’ is the only number that solves the equation: x^2 = 0.

Also, ‘0’ is an even number. Therefore, the product of that list of numbers would be ‘0’.

By the way, where was that question coming from, the one with -42 to 8?

Thanks for replying

There was this question under Quantative Comparison on Clemmonsdogpark which asked to compare e.g

Column A

product of all odd integers from say -53 to 7

Column B

product of all even integers from say -42 to 6

the solution to this only considered the number of negative integers….. but if 0 is even shouldn’t the second column be 0?

Hello Chris Sir,

Is the answer A?

Because in Column B, product of even integer is O. As 0 is even!

Hi Yogesh,

Great job! 🙂

Even if we don’t consider 0 in column B, then also it yields to answer A.

Hi Kinnar,

Before we get started, my apologies to Yogesh. I read through the problem a little too fast. Column A is actually NOT the higher quantity,because it’s a negative number. Column B is 0, which is NOT negative, and thus would be the higher (or “greater”) quantity.

Kinnar, you are correct that taking out the 0 would keep the answer the same. However, the answer would still be Quantity B, not Quantity A. Instead, Quantity A would be the larger negative number, still. With 0 removed from Quantity B, B is 21 negative numbers, with the largest negative being -42, multiplied by each other. But A is 27 negative numbers multiplied by each other, with the largest negative number starting at 57. So Quantity A, when compared to quantity B, will yield the larger negative number, which is thus “lower” because it’s farther below 0.

Thanks for replying .

that second question was actually my misunderstanding so ignore it 😛

just took my first power prep test and scored 170 in quant..

Clemmonsdogpark helped me a LOT!

Rahul,

Wow, good score! Sorry for any confusion from our question 🙂

gave my GRE scored a 330!!!

thank you magoosh! 😀

Wow, that’s awesome! Quite an amazing score 🙂

Hi Rahul,

So the square root of 0 is simply 0. (I don’t see this coming up much in test questions).

Next, 0 is definitely an even integer. This comes up from time to time, so if a question is asking for the product of even integers and one of those even integers is 0, then the product will equal ‘0’.

Hope that helps 🙂

Hello, what is the quickest/most efficient way to solve question 2? Is there a formula or strategy for solving a question like this for much larger number? Thanks!

Hi Shea,

So the most efficient way may not necessarily seem that efficient, and I think a lot of people expect some kind of formula. But really what the GRE is testing is your ability to think on the spot and test out solutions. It intentionally makes many problems “formula proof”–so you can’t just solve them by merely “plugging and chugging”.

So the GRE would never create a problem in which the numbers are so big that you can’t just test out numbers to arrive at a solution.

With this problem, you know the numbers can’t be that large, since the sum has to be less than 20. 2 is a good place to start. I add up 2 + 4 + 6 + 8 + 10. Even without adding those all up I can see that is way to large. So I have to make the numbers smaller. The trick is remembering that zero is even. That way the next lowest five consecutive even numbers are 0 + 2 + 4 + 6 + 8, which equals 20. Anything times the least of these numbers (which is ‘0’) will give you zero. Therefore, (E).

It may seem that that approach is time-consuming. But we barely had to add up any numbers in the first case, and adding the numbers in the second case is a straightforward. Should be a 45 sec to 1 minute solution.

Hope that helps!

I solved the problem below.

5x = 20

x = 4

So x is the median.

The smallest number is x-4 which equals 0.

0*4 = 0

Solved in < 30 seconds?

1. What is the smallest integer that can be divided by the product of a prime number and 7 while yielding a prime number?

(A) 7

(B) 14

(C) 24

(D) 28

(E) 35

I could not understand this question .plz help me

Hi Tanvir,

I’m happy to help with this question. First, it’s important to understand what’s being asked. If we were to write out the question as an equation, we could say

x / (P_1*7) = P_2

where x is the correct answer choice, P_1 is a prime number and P_2 is a prime number. At this point, it’s good to note that P_1 and P_2 may or may not be equal–we don’t know based on the information given in the problem.

Solving the equation for x, we get the following:

x = P_1*P_2*7

From here, we can determine the correct answer by finding the prime factorization of each answer choice. The prime factorization of the correct answer will be in the form P_1*P_2*7.

A) 7 = 7

B) 14 = 2*7

C) 24 = 2*2*3

D) 28 = 2*2*7

E) 35 = 5*7

As we can see, the only answer choice that fits our criteria is D, 28. We can double check by comparing 28 to the first equation I mentioned:

x / (P_1*7) = P_2

28/(2*7) = 2

In this case, P_1 = P_2 = 2, and D is the correct answer 🙂

I hope this helps!

“3, 5, 7 are the only odd consecutive sequence of three integers” – please clarify the meaning of the aforementioned statement.

Should not it be “3, 5, 7 are the only odd consecutive sequence of three prime numbers”.

correct me if I am wrong.

Oh my! Yes, that is clear typo on my part :). It should read “the only three consecutive odd integers that are each a prime.”

Thanks for catching that 🙂

For Q 1: I was wondering why 7 would not be the answer.

2 is the smallest prime number prod 2*7 (product of prime number and 7)

when divided by 7 yields 2 which is prime and 7 is least integer.

Please correct me if I am wrong

Hi Lisa,

Q1 asked for the least integer that can be divided by a prime (2) and 7 and still yield a prime.

The answer is 28 because: 28 divided by 2 and 7: 28/(2)(7) = 2, which is a prime number.

Hope that makes sense 🙂

Is 28 a prime number? If it is not a prime, what does the first question exactly mean?

Hi Aaron,

The first question–which is confusingly worded–is asking what is the smallest number that you can divide by a prime number and 7, and end up with a prime number.

Let’s call X the number we are looking for: X/(7)(prime number). Since we want to make X as small as possible, we want the prime number in the denominator to be the smallest possible prime, which is 2: X/(7)(2) = X/14. We need to make sure that when you divide X by 14 that you end up with a prime number. Again, we want to make X as small as possible. Therefore, the prime number we end up with should be as small as possible (which again is the number 2): X/14 = 2. Therefore, X = 28.

Hope that helps!

Hello Chris,

Why cant the answer be ‘C’ 24.???

Cos, the question is asking for the smallest interger, which when divided by a prod of 7 and prime number, yields a prime number.

24/ (7*3) gives a reminder of 3, which is another prime number.

though 28 is a valid answer, wen comes to smallest integer, I feel 24 is the right answer. Please correct me if I’m wrong.

Hi,

Yes even I am confused. Shouldn’t it be 24 ?

Yes, wont the smallest integer be 24? Please advice.

The question says divisible.

24/21 is not divisible, it leads to a fraction ( not sure if fractions can be considered in prime no.)

While 28/14 is divisible and yield to a Prime no.

Hi Asma,

Thanks for the explanation here! Fractions cannot be considered prime numbers. Prime numbers by definition are whole numbers.

I have a question related to this post:

Usually when a question says “the positive integers” I usually find in the answer, zero was not considered. In fact in ALL questions that say this, zero was not considered as a possibility.

Hi Javeria,

I am not sure which questions you are talking about as none of the above mention “positive integers.” Also one of the above questions includes zero as an answer choice.

hi can u explain the 1st one i am little bit confused

I already purchased your practice material and I’m practicing now. It’s good.

Thanks, Jayanti! We’re glad you find it helpful.

For problem 1, why is the answer not B = 14? 14 can be divided by 2, which yields 7 which is a prime number; and 14 can also be divided by 7, which yields 2 which is a prime number.

Hi Arnita,

The problem says, “by a prime number and 7…” which I guess can be a little ambiguous. I’ll change it to “by the product of 7 and a prime number.”

Thanks!

changing the wording clears up my question.

Problem 2 Are the even consecutive integers 0, 2, 4, 6, 8?

Yep, those are the integers and because ‘0’ is the smallest number anything you multiply by it will give you zero so you don’t even have to worry about finding the median.

Im a little confused to question two as well… My first thought to solve this problem is to write out: x, x+1, x+2, x+3, x+4=20 and solve by x which i get 2, then to write out the answers 2,3,4,5,6.. which shows the median of 4. Not too sure how to even think about getting 0 from that because I can’t just assume that the “even integers” start at 0..

Hi April,

With the algebraic equation you want to make sure to use x+2, x+4, etc. because the numbers are consecutive even numbers, meaning that they have to jump up by 2. That gives us the equation x+x+2+x+4+x+6+x+8 =20, x =0.

And that’s always a good one of those little math conventions to know: ‘0’ is an even number.

Hope that helps!

Aw, gotcha… that helps a lot.. because i’m still keeping that same formula that I finally mastered. So in essence is they were asking for the “odd consecutive integers” it would be as follows: x,x+1,x+3 etc.. x=1.

Aside from this problem, I get a little confused with the procedure when they ask for the product of consecutive integers. Would you mind briefly going over a good strategy for that. Aside from this problem, I get a little confused with the procedure when they ask for the product of consecutive integers. Would you mind briefly going over a good strategy for that.

Hi April,

I’m happy that helped :).

So, for consecutive odds you would still use the exact same formula, since odds are separated by 2. Using your formula (x, x+1, x+3…), if x = 3, then the series would be 3, 4, 6, etc.

If they ask for the product of consecutive integers that is just a factorial sign (assuming the number starts with 1). Typically, the GRE won’t just ask you to multiply a bunch of consecutive integers.

For some more factorials, check out this post:

http://clemmonsdogpark.info/gre/2011/its-not-an-excited-number-its-a-factorial/

Hope that helps!

Can the equation be written as : (x-2)+x+(x+2)+(x+4)=20?

then x=4 and median will be 5. So answer could be (b) also. Please help, i’m confused

Hi Avani,

You’re on the right track but your equation only includes 4 even consecutive integers, while the problem asks us to consider a situation with 5 even consecutive integers. If you include another integer, either x-4 or x+6, you should get the correct answer 🙂

Hope that clears things up!

Problem 3 The digits add up to be 11 and 11 is not a multiple of 3. Why is this not the answer.

Hmm, that’s a typo – it should read 624. Thanks for catching that!