How to Solve: Remainder Problems
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Theory of Remainders FirstTheoryGiven 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.htmlExample2:
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.htmlExample3: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.htmlExample 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.htmlExample 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.htmlLooking for a Quant Tutor?
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