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REMAINDERS

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New post Updated on: 26 Feb 2019, 05:26
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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, 123 divided by 10 has the remainder 3 and 123 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 (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 #9 (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 #10 (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 #11 (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 #12 (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.


Edit: For more on REMAINDERS check HERE.

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Originally posted by Bunuel on 24 Dec 2012, 08:48.
Last edited by Bunuel on 26 Feb 2019, 05:26, edited 1 time in total.
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New post 18 Jun 2014, 15:54
I have reviewed this theory but am still having a hard time understanding a couple of things. I will be using the following 2 examples:

Example #1 & 6.

From Example 1: "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." I don't understand why it matters if 18q is divisible by 6 or not and how do you determine if the remainder comes from the second term or not.

From Example 2: Why are we trying to derive an LCM of 5 & 6?

Greatly appreciate the assistance.
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New post 21 Jun 2014, 07:33
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farhanabad wrote:
I have reviewed this theory but am still having a hard time understanding a couple of things. I will be using the following 2 examples:

Example #1 & 6.

From Example 1: "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." I don't understand why it matters if 18q is divisible by 6 or not and how do you determine if the remainder comes from the second term or not.

From Example 2: Why are we trying to derive an LCM of 5 & 6?

Greatly appreciate the assistance.


18q can be divided equally into 6 groups, so it will leave the remainder of 0. So, the remainder when 18q+1 is divided by 6 will come only from the second term, which is 1.

For example:
18 + 1 = 19 divided by 6 gives the remainder of 1;
18*2 + 1 = 37 divided by 6 gives the remainder of 1;
18*3 + 1 = 55 divided by 6 gives the remainder of 1;
...

Hope it's clear.
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New post 27 Aug 2014, 05:52
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I would like to add the following shortcuts:

1)
Remainder of (ax + 1)^n / a = 1

Example: 46^3578 / 9 will have a remainder of 1.

2)
Similarly, remainder of (ax - 1)^n / a = 1 (if n is even) or -1 (if n is odd)

Example: 27^79 / 14 will have a remainder of -1 or 13.
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New post 27 Aug 2014, 09:48
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Another tip which may be handy.

(Fermat's theorem)

If ‘p’ is a prime number and ‘a’ and ‘p’ are co-primes:

1)
Remainder of (a^p)/p = a
(a^p – a) will be divisible by p.

Example: remainder of (10^13)/13 = 13

2)
Remainder of [a^(p-1)]/p = 1

Example: remainder of (10^12)/13 = 1
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New post 14 Dec 2015, 04:56
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Alternative solution to #6

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


We have -
1) n - 4 = 6t and
2) n - 3 = 5k

We need to find
3) n - R = 30x

Multiply (1) by 5 and (2) by 6
1) 5n - 20 = 30t
2) 6n - 18 = 30k

(2) - (1)
n + 2 = 30(t - k)

t and k are positive integers. So we can assume them as x fro m(3)
Thus R = -2 or 30-2 = 28.
Answer E.

+Kudos, if this helped :lol:
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New post 25 Aug 2016, 00:02
Hi Bunuel,

the second point under properties violates example 9, i.e. when a small integer is divided by a larger one, the quotient is 0 and the remainder is the smaller number.

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New post 04 Jul 2017, 06:01
Hey, what will be the reminder when any positive integer is divided by 0?
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New post 04 Jul 2017, 06:10
Hey sorry, wrong question what will be the reminder when 0 is divided by any positive integer?

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New post 21 Sep 2018, 22:24
How to find reminder using cyclic method :


Q1. Find the reminder of 14^77/11
step 1:
14/11 gives us a reminder of 3 so basically the question becomes : 3^77/11
Step 2:
3/11 gives us a reminder of 3
3^2/11 gives us a reminder of 9
3^3/11 gives us a reminder of 5
3^4/11 gives us a reminder of 4 ( easy way to calculate reminder : 3^4/11 = 3^3/11*3 = 5*3/11 =4 )
3^5 /11 gives us a reminder of 1 ( similarly 3^5/11 = 4*3/11 =12/11=1)
Whenever we get a reminder of 1 , we get our cyclic value. Here in this case the cyclicity is 5
Step 3
Divide the power with the cyclic value
77/5 which gives us a reminder of 2. Thus the reminder of the expression 14^77/11 is the second term of cycle. Here it is 9

Q2. Find the reminder of the expression 4^98/25
here since the numerator is small than denominator we go directly to cycle
4^1/25 gives us a reminder of 4
4^2/25 gives us a reminder of 16
4^3/25 gives us a reminder of 14 (16*4/25 =14)
4^4/25 gives us a reminder of 6 (14*4/25 =6)
4^5/25 gives us a reminder of 24 or -1 (6*4/25=24 or -1)
Whenever we have a reminder of -1 , then the cycle would be double of its position. Here at 5th position the reminder becomes -1, hence the cyclicity would be 10)
divide the power with the cycle number
98/10 , this gives a reminder of 8
hence 8th term of the cycle is the reminder for the expression.
(to find the 8th term : 4^4*4^4/25 = 6*6/25 =11)
Hence 11 is the answer.
Another method for peculiar cases

Q3. 2^300/9 find the reminder
2^300/9 can be written as (2^3)^100/9 or 8^100/9 or (-1)^100/9 = 1/9 or the reminder is 1

Q4. 12^175/13. Find the reminder
we can write this as (-1)^175/13 or -1/13 . Hence the reminder would be 12

Q5. find the reminder of 14^21^39/11
take 21^39 as N, the expression becomes 14^N/11
14/11 gives a reminder of 3 , hence the expression becomes 3^N/11
so continuing with cycle
3/11 gives us a reminder of 3
3^2/11 gives us a reminder of 9
3^3/11 gives us a reminder of 5
3^4/11 gives us a reminder of 4
3^5/11 gives us a reminder of 1
Hence the cyclicity is 5
so we need to divide the power by cycle
21^39/5
21/5 gives a reminder of 1
so the21^39/5 expression becomes 1^39/5 . the reminder is 1
hence the reminder of 14^21^39/11 is the first term of the cycle or 3
Hence the reminder is 3

Q6. Find the reminder of 17^100/51
if we look closely we will find 51 is divisible by 17.
So take one 17 out and keep outside. The expression becomes
17 * REM[17^99/3]
or 17 * REM[ (-1)^99/3]
or 17*2 = 34 (answer)

Q7. Find the reminder of 30^40/17
30/17 gives us a reminder of 13
hence the expression becomes
13^40/17
13^1/17 gives us a reminder of 13
13^2/17 gives us a reminder of -1 or 16.
Hence the cycle must be 4
Divide the power with cycle
40/4 gives us a reminder of 0
Hence it should be the last term of the cycle
The 4th term can be found out by : 13^2*13^2/17 or (-1)*(-1)/17 or 1/17 or 1
Hence the answer should be 1

Q8. find the reminder of 73^382/100
we can write the above expression as (73^2)^191/100
or (5329)^191/100
In other words we are asked to find the last 2 digit
the expression becomes 29^191/100 (we are only concerned with last 2 digit .)
or (-1+30)^191/100
Using binomial expression
191C0*(-1)^191*30^0+ 191C1*(-1)^190*30^1 +......
= -1+ 191*30
= -1+ 5730
= 5729
or the last 2 digit/ reminder is 29


May be some cases are unwarranted for GMAT and i would recommend what @bunnel advised as the best method to go for reminders.

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New post 26 Feb 2019, 05:24

5. Divisibility/Multiples/Factors




6. Remainders



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Re: REMAINDERS   [#permalink] 26 Feb 2019, 05:24
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