How To Solve: Remainder Problems : Quantitative Questions

BrushMyQuant · 2012-11-17T03:05:25-08:00

How to Solve: Remainder Problems Check out the post on Theory of Remainders First https://www.youtube.com/watch?v=P0S6j2uTaj4 https://www.youtube.com/watch?v=h1HTEhmUb1w Theory Given that an integer "n" when divided by an integer "a" gives "r" as remainder then the integer "n" can be written as n = ak + r where k is a constant integer. Example1. What is the remainder when B is divided by 6 if B is a positive integer? (1) When B is divided by 18, the remainder is 3 (2) When B is divided by 12, the remainder is 9 Sol STAT1 : When B is divided by 18, the remainder is 3 So, we can write B as B = 18k + 3 Now, to check the remainder when B is divided by 6, we essentially need to check the remainder when 18k + 3 is divided by 6 18k goes with 6 so the remainder will 3 So, its sufficient STAT2 : When B is divided by 12, the remainder is 9 So, we can write B as B = 12k + 9 Now, to check the remainder when B is divided by 6, we essentially need to check the remainder when 12k + 9 is divided by 6 12k goes with 6 so the remainder will be the same as the remainder for 9 divided by 6 which is 3 So, remainder is 3 So, its sufficient Answer will be D Link to the problem: https://gmatclub.com/forum/what-is-the-r ... 42343.html Example2: What is the remainder when positive integer t is divided by 5? (1) When t is divided by 4, the remainder is 1 (2) When t is divided by 3, the remainder is 1 Sol: STAT1: When t is divided by 4, the remainder is 1 t = 4k +1 possible values of t are 1,5,9,13 Clearly we cannot find a unique remainder when t is divided by 5 as in some cases(t=1) we are getting the remainder as 1 and in some(t=5) we are getting the remainder as 0. So, INSUFFICIENT STAT2: When t is divided by 3, the remainder is 1 t = 3s + 1 possible values of t are 1,4,7,10,13,16,19 Clearly we cannot find a unique remainder when t is divided by 5 as in some cases(t=1) we are getting the remainder as 1 and in some(t=10) we are getting the remainder as 0. So, INSUFFICIENT taking both together now there are two approaches 1. write the values of t from stat1 and then from stat2 and then take the common values from STAT1 t = 1,5,9,13,17,21,25,29,33 from STAT2 t = 1,4,7,10,13,16,19,22,25,28,31,34 common values are t = 1,13,25, 2. equate t = 4k+1 to t=3s+1 we have 4k + 1 = 3s+1 k = 3s/4 since, k is an integer so only those values of s which are multiple of 4 will satisfy both STAT1 and STAT2 so, common values are given by t = 3s + 1 where s is multiple of 4 so t = 1,13,25 (for s=0,4,8 respectively) Clearly we cannot find a unique remainder when t is divided by 5 as in some cases(t=1) we are getting the remainder as 1 and in some(t=10) we are getting the remainder as 0. So, INSUFFICIENT So, answer will be E Link to the problem: https://gmatclub.com/forum/what-is-the-r ... 42376.html Example3:If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ? (1) The remainder when p + n is divided by 5 is 1. (2) The remainder when p - n is divided by 3 is 1 Sol: STAT1 : The remainder when p + n is divided by 5 is 1. p+n = 5k + 1 but we cannot say anything about p^2 - n^2 just from this information. So, INSUFFICIENT STAT2 : The remainder when p - n is divided by 3 is 1 p-n = 3s + 1 but we cannot say anything about p^2 - n^2 just from this information. So, INSUFFICIENT Taking both together p^2 - n^2 = (p+n) * (p-n) = (5k + 1) * (3s + 1) = 15ks + 5k + 3s + 1 The remainder of the above expression by 15 is same as the remainder of 5k + 3s + 1 with 15 as 15ks will go with 15. But we cannot say anything about the remainder as its value will change with the values of k and s. So INSUFFICIENT Hence answer will be E Link to the problem: https://gmatclub.com/forum/if-p-and-n-ar ... 28002.html Example 4: If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r? (1) n+1 is divisible by 3 (2) n>20. Sol: r is the remainder when 4 + 7n is divided by 3 7n + 4 can we written as 6n + n + 3+ 1 = 3(2n+1) + n +1 remainder of 7n+4 by 3 will be same as remainder of 3(2n+1) + n +1 by 3 3*(2n+1) will go by 3 so the remainder will be the same as the remainder of (n+1) by 3 STAT1: n+1 is divisible by 3 n+1 = 3k (where k is an integer) n+1 will give 0 as the remainder when divided by 3 so, 7n+4 will also give 0 as the remainder when its divided by 3 (as its remainder is same as the remainder for (n+1) when divided by 3) => r =0 So, SUFFICIENT STAT2 n>20. we cannot do anything by this information as there are many values of n so, INSUFFICIENT. Hence, answer will be A Link to the problem: https://gmatclub.com/forum/if-n-is-a-pos ... 93364.html Example 5: If x is an integer, is x between 27 and 54? (1) The remainder when x is divided by 7 is 2. (2) The remainder when x is divided by 3 is 2. Sol: STAT1: The remainder when x is divided by 7 is 2. x = 7k + 2 Possible values of x are 2,9,16,...,51,... we cannot say anything about the values of x so, INSUFFICIENT STAT2: The remainder when x is divided by 3 is 2. x = 3s + 2 Possible values of x are 2,5,8,11,...,53,... we cannot say anything about the values of x so, INSUFFICIENT Taking both together now there are two approaches 1. write the values of t from stat1 and then from stat2 and then take the common values from STAT1 x = 2,9,16,23,30,37,44,51,58,...,65,... from STAT2 x = 2,5,8,...,23,...,44,...,59,65,... common values are x = 2,23,44,65,... 2. equate x = 7k+2 to x=3s+2 we have 7k + 2 = 3s+2 k = 3s/7 since, k is an integer so only those values of s which are multiple of 7 will satisfy both STAT1 and STAT2 so, common values are given by x = 3s + 2 where s is multiple of 7 so x = 2,23,44,65 (for s=0,7,14,21 respectively) Clearly there are values of x which are between 27 and 54 (i.e. 44) and those which are not (2,23,65) So, both together also INSUFFICIENT So, answer will be E Link to the problem: https://gmatclub.com/forum/if-x-is-an-in ... 08153.html Looking for a Quant Tutor? Check out my post for the same https://gmatclub.com/forum/starting-gmat ... 35537.html Hope it helps! Good Luck!

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Sol: As a and b are positive numbers so min(a+b) = min(a) + min(b)
x when divided by "a" gives 3 as remainder -> Since a is giving 3 remainder that means that a >= 4. So, min a = 4
y when divided by "b" gives 4 as remainder -> Since b is giving 4 remainder that means that b >= 5. So, min b = 5
=> min(a+b) = min(a) + min(b) = 4+ 5 =9

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Ans:
Q1. 2, Q2 3, Q3. 1

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Sol: 20 = nk + 4
=> nk= 20-4 = 16
=> n = 16/k
Now, n is giving 4 remainder that means that n>=5
Let's start putting values of k and get values of n
k=1 => n=16
k=2 => n=16/2 = 8
k=3 => n not integer
k=4 => n= 16/4 not possible as n>=5
So, possible values of the number (n) are 8 and 16

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Sol: 50 = nk + 7
=> nk = 50-7 = 43 [note 43 is prime so it has only two factors 1 and 43 itself]
=> n = 43/k
k= 1 => n = 43
k = 43 not possible as n becomes 1 [ but we know that n>=8 as n is giving 7 remainder]
So, Answer is 43

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Sol: n when divided by 7 gives 4 as remainder.
n = 7k + 4

We will start taking values of k starting from k=0 and find values of n correspondingly
k=0 n=7*0 + 4 = 4 k=5 n=7*5 + 4 = 39
k=1 n=7*1 + 4 = 11 k=6 n=7*6 + 4 = 46
k=2 n=7*2 + 4 = 18 k=7 n=7*7 + 4 = 53
k=3 n=7*3 + 4 = 25 k=8 n=7*8 + 4 = 60
k=4 n=7*4 + 4 = 32 k=9 n=7*9 + 4 = 67

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Sol: n when divided by 5 gives 2 as remainder.
n = 5k + 2

We will start taking values of k starting from k=0 and find values of n correspondingly
k=0 n=5*0 + 2 = 2 k=5 n=5*5 + 2 = 27
k=1 n=5*1 + 2 = 7 k=6 n=5*6 + 2 = 32
k=2 n=5*2 + 2 = 12 k=7 n=5*7 + 2 = 37
k=3 n=5*3 + 2 = 17 k=8 n=5*8 + 2 = 42
k=4 n=5*4 + 2 = 22 k=9 n=5*9 + 2 = 47

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Sol: Method -1
3. n when divided by 7 gives 3 remainder
n = 7k + 3
We will start taking values of k starting from k=0 and find values of n correspondingly
k = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 3 , 10, 17, 24, 31, 38, 45, 52, 59, 66

n when divided by 5 gives 3 remainder
n = 5t + 3
We will start taking values of t starting from t=0 and find values of n correspondingly
t = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 3 , 8, 13, 18, 23, 28, 33, 38, 43, 48

First two common values are 3 and 38

Sol: Method-2
n when divided by 7 gives 3 remainder
n = 7k + 3

n when divided by 5 gives 3 remainder
n = 5t + 3

7k+3 = 5t+3 => t = 7k/5
So, only those values of k will give us common values of n for which t is integer too.
k = 0 => t=7*0/5 = 0
k = 5 => t=7*5/5 = 7
So, k = 0 => n = 7*0 + 3 = 3
k = 5 => n = 7*5 + 3 = 38

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Sol: Method-1
n when divided by 6 gives 4 remainder
n = 6k + 4
We will start taking values of k starting from k=0 and find values of n correspondingly
k = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58

n when divided by 4 gives 2 remainder
n = 4t + 2
We will start taking values of t starting from t=0 and find values of n correspondingly
t = 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9
n = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38

First two common values are 10 and 22

Sol: Method-2
n when divided by 6 gives 4 remainder
n = 6k + 4

n when divided by 4 gives 2 remainder
n = 4t + 2

6k+4 = 4t+2 => t = (6k+2)/4 = (3k+1)/2
So, only those values of k will give us common values of n for which t is integer too.
k = 1 => t=(6*1 + 2)/4 = 2
k = 3 => t=(6*3 + 2)/4 = 5
So, k = 1 => n = 6*1 + 4 = 10
k = 3 => n = 6*3 + 4 = 22

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Sol: A = 12k + 3
When A is divided by 4 then 12k will give 0 remainder and 3 will give 3 remainder. Total remainder is 3

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Sol: B = 15k + 6
When B is divided by 5 then 15k will give 0 remainder and 6 will give 1 remainder. Total remainder is 1

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Sol: 136/5 remainder is 1
148/5 remainder is 3
298/5 remainder is 3
Total remainder = 1*3*3 = 9 [>=5]
=> Final remainder = remainder of 9/5 = 4

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Sol: 1205/12 remainder is 5
1208/12 remainder is 8
2404/12 remainder is 4
Total remainder = 5*8*4 = 160 [>=12]
=> Final remainder = remainder of 160/12 = 4

How To Solve: Remainder Problems

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