REMAINDERS This post is a part of GMAT MATH BOOK ] created by: Bunuel edited by: bb , Bunuel -------------------------------------------------------- Definition If x and y are positive integers , there exist unique integers q and r , called the quotient and remainder , respectively, such that y =divisor*quotient+remainder= xq + r and 0\leq{r}<x . For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since 15 = 6*2 + 3 . Notice that 0\leq{r}<x means that remainder is a non-negative integer and always less than divisor. This formula can also be written as \frac{y}{x} = q + \frac{r}{x} . Properties When y is divided by x the remainder is 0 if y is a multiple of x . For example, 12 divided by 3 yields the remainder of 0 since 12 is a multiple of 3 and 12=3*4+0 . When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer. For example, 7 divided by 11 has the quotient 0 and the remainder 7 since 7=11*0+7 The possible remainders when positive integer y is divided by positive integer x can range from 0 to x-1 . For example, possible remainders when positive integer y is divided by 5 can range from 0 (when y is a multiple of 5) to 4 (when y is one less than a multiple of 5). If a number is divided by 10, its remainder is the last digit of that number. If it is divided by 100 then the remainder is the last two digits and so on. For example, 12 3 divided by 10 has the remainder 3 and 1 23 divided by 100 has the remainder of 23. Example #1 (easy) If the remainder is 7 when positive integer n is divided by 18, what is the remainder when n is divided by 6? A. 0 B. 1 C. 2 D. 3 E. 4 When positive integer n is dived by 18 the remainder is 7: n=18q+7 . Now, since the first term (18q) is divisible by 6, then the remainder will only be from the second term, which is 7. 7 divided by 6 yields the remainder of 1. Answer: B. Discuss this question HERE . Example #2 (easy) If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12 ? A. 0 B. 1 C. 2 D. 3 E. 5 There are several algebraic ways to solve this question, but the easiest way is as follows: since we cannot have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12. If n=5 , then n^2=25 . The remainder upon division 25 by 12 is 1. Answer: B. Discuss this question HERE . Example #3 (easy) What is the tens digit of positive integer x ? (1) x divided by 100 has a remainder of 30. (2) x divided by 110 has a remainder of 30. (1) x divided by 100 has a remainder of 30. We have that x=100q+30 : 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient. (2) x divided by 110 has a remainder of 30. We have that x=110p+30 : 30, 140, 250, 360, ... so, there are more than 1 value of the tens digit possible. Not sufficient. Answer: A. Discuss this question HERE . Example #4 (easy) What is the remainder when the positive integer n is divided by 6? (1) n is multiple of 5 (2) n is a multiple of 12 (1) n is multiple of 5. If n=5, then n yields the remainder of 5 when divided by 6 but if n=10, then n yields the remainder of 4 when divided by 6. We already have two different answers, which means that this statement is not sufficient. (2) n is a multiple of 12. Every multiple of 12 is also a multiple of 6, thus n divided by 6 yields the remainder of 0. Sufficient. Answer: B. Discuss this question HERE . Example #5 (medium) If s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t ? A. 2 B. 4 C. 8 D. 20 E. 45 s divided by t yields the remainder of r can always be expressed as: \frac{s}{t}=q+\frac{r}{t} (which is the same as s=qt+r ), where q is the quotient and r is the remainder. Given that \frac{s}{t}=64.12=64\frac{12}{100}=64\frac{3}{25}=64+\frac{3}{25} , so according to the above \frac{r}{t}=\frac{3}{25} , which means that r must be a multiple of 3. Only option E offers answer which is a multiple of 3 Answer. E. Discuss this question HERE . Example #6 (medium) Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30? A. 3 B. 12 C. 18 D. 22 E. 28 Positive integer n leaves a remainder of 4 after division by 6: n=6p+4 . Thus n could be: 4, 10, 16, 22, 28 , ... Positive integer n leaves a remainder of 3 after division by 5: n=5q+3 . Thus n could be: 3, 8, 13, 18, 23, 28 , ... There is a way to derive general formula for n (of a type n=mx+r , where x is a divisor and r is a remainder) based on above two statements: Divisor x would be the least common multiple of above two divisors 5 and 6, hence x=30 . Remainder r would be the first common integer in above two patterns, hence r=28 . Therefore general formula based on both statements is n=30m+28 . Hence the remainder when positive integer n is divided by 30 is 28. Answer. E. Discuss this question HERE . Example #7 (medium) If x^3 - x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. x^3-x=x(x^2-1)=(x-1)x(x+1) , notice that we have the product of three consecutive integers. Now, notice that if x=odd , then x-1 and x+1 are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)x(x+1) will be divisible by 8*3=24). (1) When 3x is divided by 2, there is a remainder. This implies that 3x=odd , which means that x=odd . Therefore (x-1)x(x+1) is divisible by 8. Sufficient. (2) x = 4y + 1, where y is an integer. We have that x=even+odd=odd , thus (x-1)x(x+1) is divisible by 8. Sufficient. Answer: D. Discuss this question HERE . Example #8 (hard) When 51^25 is divided by 13, the remainder obtained is: A. 12 B. 10 C. 2 D. 1 E. 0 51^{25}=(52-1)^{25} , now if we expand this expression all terms but the last one will have 52=13*4 in them, thus will leave no remainder upon division by 13, the last term will be (-1)^{25}=-1 . Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: -1=13*(-1)+12 . Answer: A. Discuss this question HERE . Example #9 (hard) When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x - y? A. 12 B. 15 C. 20 D. 28 E. 35 When the positive integer x is divided by 5 and 7, the remainder is 3 and 4, respectively: x=5q+3 (x could be 3, 8, 13, 18 , 23, ...) and x=7p+4 (x could be 4, 11, 18 , 25, ...). We can derive general formula based on above two statements the same way as for the example above: Divisor will be the least common multiple of above two divisors 5 and 7, hence 35. Remainder will be the first common integer in above two patterns, hence 18. So, to satisfy both this conditions x must be of a type x=35m+18 (18, 53, 88, ...); The same for y (as the same info is given about y): y=35n+18 ; x-y=(35m+18)-(35n+18)=35(m-n) . Thus x-y must be a multiple of 35. Answer: E. Discuss this question HERE . Example #10 (hard) If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5 (2) x – y = 3 (1) When p is divided by 8, the remainder is 5. This implies that p=8q+5=x^2+y^2 . Since given that y=odd=2k+1 , then 8q+5=x^2+(2k+1)^2 --> x^2=8q+4-4k^2-4k=4(2q+1-k^2-k) . So, x^2=4(2q+1-k^2-k) . Now, if k=odd then 2q+1-k^2-k=even+odd-odd-odd=odd and if k=even then 2q+1-k^2-k=even+odd-even-even=odd , so in any case 2q+1-k^2-k=odd --> x^2=4*odd --> in order x to be multiple of 4 x^2 must be multiple of 16 but as we see it's not, so x is not multiple of 4. Sufficient. (2) x – y = 3 --> x-odd=3 --> x=even but not sufficient to say whether it's multiple of 4. Answer: A. Discuss this question HERE . Example #11 (hard) m and n are positive integers. Is the remainder of \frac{10^m + n}{3} bigger than the remainder of \frac{10^n + m}{3} ? (1) m \gt n . (2) The remainder of \frac{n}{3} is 2 First of all any positive integer can yield only three remainders upon division by 3: 0, 1, or 2. Since, the sum of the digits of 10^m and 10^n is always 1 then the remainders of \frac{10^m + n}{3} and \frac{10^n + m}{3} are only dependent on the value of the number added to 10^m and 10^n . There are 3 cases: If the number added to them is: 0, 3, 6, 9, ... then the remainder will be 1 (as the sum of the digits of 10^m and 10^n will be 1 more than a multiple of 3); If the number added to them is: 1, 4, 7, 10, ... then the remainder will be 2 (as the sum of the digits of 10^m and 10^n will be 2 more than a multiple of 3); If the number added to them is: 2, 5, 8, 11, ... then the remainder will be 0 (as the sum of the digits of 10^m and 10^n will be a multiple of 3). (1) m \gt n . Not sufficient. (2) The remainder of \frac{n}{3} is 2 --> n is: 2, 5, 8, 11, ... so we have the third case. Which means that the remainder of \frac{10^m + n}{3} is 0. Now, the question asks whether the remainder of \frac{10^m + n}{3} , which is 0 , greater than the reminder of \frac{10^n + m}{3} , which is 0, 1, or 2 . Obviously it cannot be greater, it can be less than or equal to. So, the answer to the question is NO. Sufficient. Answer: B. Discuss this question HERE . ______________________________________________ Check more DS questions on remainders HERE . Check more PS questions on remainders HERE . For more on REMAINDERS check HERE .

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ReedArnoldMPREP

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I think it's best maybe to start with a simpler version of that formula: y = Qx + r

So, for instance, 25 = 2(11)+ 3

y=25, x=11, Q=2, and r = 3. 25/11 has remainder three.

So if we divide both sides by 11:

25/11 = 2 + 3/11

Take the general formula y = Qx + r and divide by x, and you'll get y/x = Q + r/x

We typically say the remainder is what's 'left over' in division, which isn't wrong, but I also think it sometimes easiest to think of it as 'how many more than the nearest (lower) multiple of the divisor?'

So 31 has remainder 3 when divided by 7 because the nearest multiple of 7 (less than 31) is 28. 31 is 3 more than that, so it has a remainder of 3.

31 = 4(7) + 3

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Problem 7:

how do I recognize that x(x^2−1) equals (x−1)x(x+1)?

Bunuel

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Schachfreizeit

Problem 7:

how do I recognize that x(x^2−1) equals (x−1)x(x+1)?

\(a^2 - b^2 = (a - b)(a + b)\), so \(x^2 - 1 = x^2 - 1^2 = (x - 1)(x + 1)\).

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rijain

Can someone please explain how we derive this formula - y/x=Q+r/x with an Example.

Thanks in advance !

Hi rijain

We already know that dividend (y) = divisor (x) * quotient (Q) + remainder (r) ie. y = xQ + r.

Now if you divide both sides of this expression by x you get. y/x = (xQ + r)/x => y/x = xQ/x + r/x => y/x = Q + r/x.

Hope it helps.!

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Hi Bunuel
How can i take [rint of this article or any other article that you have posted on GMAT club for quant for revision and all.
If i am doing ctrl+P and save as it coming as blank .

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