How to Solve: Remainder Problems

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Hope it will be useful. This post is about how to solve Remainder problems

TheoryGiven 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: what-is-the-remainder-when-is-divided-by-6-if-is-a-positiv-142343.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: what-is-the-remainder-when-positive-integer-t-is-divided-by-142376.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: if-p-and-n-are-positive-integers-and-p-n-what-is-the-128002.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: if-n-is-a-positive-integer-and-r-is-the-remainder-when-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: if-x-is-an-integer-is-x-between-27-and-108153.htmlLooking for a Quant Tutor?

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starting-gmat-quant-classes-tutoring-bangalore-online-135537.htmlHope it helps!

Good Luck!

_________________

Ankit

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