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How To Solve: Remainder Problems

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How To Solve: Remainder Problems [#permalink] How To Solve: Remainder Problems   Updated on: 13 Jul 2020, 23:36
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How to Solve: Remainder Problems


Check out the post on Theory of Remainders First







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!

Originally posted by BrushMyQuant on 17 Nov 2012, 04:05.
Last edited by BrushMyQuant on 13 Jul 2020, 23:36, edited 2 times in total.
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How to Solve: Remainders (Basics and Advanced) [#permalink] How To Solve: Remainder Problems   Updated on: 25 Jun 2024, 03:23
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How to Solve: Remainders (Basics and Advanced)

Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Hi All,

I have recently uploaded two Videos on YouTube to discuss Remainders Basics and Advanced problems on remainder:






Following is covered in the video:

Video 1

• Definition: What is "Remainder"?
• Remainder Notation
• How to find Remainders : Long Division Method
• Remainder Range
• PT1: Basic Problems on Finding Remainders
• PT2: Find divisor when remainder is given
• PT3: Find dividend when one divisor and remainder are given
• PT4: Find dividend when two divisors and remainders are given
• Remainder of sum of two numbers by a number
• Remainder of difference of two numbers by a number
• Remainder of product of two numbers by a number

Video 2

• Remainder of numbers when divided by 2 , 3, 5, 9, 10
• Remainder of numbers when divided by other numbers
• Binomial Theorem
• Application of Binomial Theorem in finding remainders


Remainders Basics

• Definition: What is "Remainder"?

Remainder is the integer which is left over in a division, when the divisor cannot evenly divide the dividend

Dividend = Divisor * Quotient + Remainder

Ex: 13 when divided by 3 gives 1 remainder. (13 = 3*4 + 1)
13 -> Dividend
3 -> Divisor
4 -> Quotient
1 -> Remainder

• Remainder Notation

1. A number when divided by 5 gives 3 as remainder.
Let the number be n
using Dividend = Divisor * Quotient + Remainder
n = 5k + 3 [where k is quotient and is an integer]

2. A number when divided by 13 gives 5 as remainder.
Let the number be n
using Dividend = Divisor * Quotient + Remainder
n = 13k + 5 [where k is quotient and is an integer]

3. A number when divided by 16 gives 3 as remainder.
Let the number be n
using Dividend = Divisor * Quotient + Remainder
n = 16k + 3 [where k is quotient and is an integer]


• How to find Remainders : Long Division Method -> Check the video link above

• Remainder Range
"A number when divided by a number k can give remainder from 0 to k-1"

Ex: If we are trying to divide a number by 6 then possible values of remainder are from 0-5

Q1. x when divided by "a" gives 3 as remainder. y when divided by "b" gives 4 as remainder. Find min(a+b)?

Show Spoiler
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

• PT1: Basic Problems on Finding Remainders -> Check out the video above for examples on long division method

Q1. Find the remainder when 200 is divided by 3.
Q2. Find the remainder when 80 is divided by 7.
Q3. Find the remainder when 100 is divided by 9.

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

• PT2: Find divisor when remainder is given

Q1. 20 when divided by which number will give 4 as remainder?

Show Spoiler
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

Q2. 50 when divided by which number will give 7 as remainder?

Show Spoiler
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

• PT3: Find dividend when one divisor and remainder are given

Q1. n when divided by 7 gives 4 as remainder. Find the possible values of the number.

Show Spoiler
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

Q2. n when divided by 5 gives 2 as remainder. Find the possible values of n.

Show Spoiler
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

• PT4: Find dividend when two divisors and remainders are given

Q1. n when divided by 7 gives 3 remainder and when divided by 5 gives 3 remainder. Find first 2 non-negative values of n

Show Spoiler
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

Q2. n when divided by 6 gives 4 remainder and when divided by 4 gives 2 remainder. Find first 2 non-negative values of n

Show Spoiler
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

Note: When the numerator is smaller than the denominator then the remainder is the numerator itself
Ex: If 2 is divided by 3 then remainder is 2

• Remainder of sum of two numbers by a number
13 when divided by 3 gives us 1 remainder

When we split 13 as 8 and 5 and divide 8 and 5 individually by 3 then we still get the same remainder
13/3 = (8+5)/3 = 8/3 + 5/3
8/3 will give 2 remainder
5/3 will give 2 remainder

Total remainder is 2+2 = 4, but remainder cant be greater than 3 so remainder will be 4-3 = 1 which is same as the remainder for 13/3

Q1. “A” when divided by 12 gives 3 as remainder. What is the remainder when A is divided by 4?

Show Spoiler
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


Q2. “B” when divided by 15 gives 6 as remainder. What is the remainder when B is divided by 5?

Show Spoiler
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

• Remainder of difference of two numbers by a number
13 when divided by 3 gives us 1 remainder

When we split 13 as 15 - 2 and divide 15 and 2 individually by 3 then we still get the same remainder
13/3 = (15-2)/3 = 15/3 - 2/3
15/3 will give 0 remainder
2/3 will give 2 remainder
Total remainder = 0-2 = -2
But remainder cannot be negative so we are going to keep on adding 3 to -2 till the time the sum comes in the range of 0 and 2 [3-1]
=> -2 + 3 = 1
So, remainder is 1

• Remainder of product of two numbers by a number
21 when divided by 5 gives us 1 remainder. Now if we break 21 into product of two numbers and find the remainder of these individual numbers by 5 and multiply the remainders then we are going to get the same remainder as we got when we divided 21 by 5

Let's write 21 = 3*7 and divided 3 and 7 by 5
3/5 * 7/5 will give us 3 * 2 remainder respectively = 6
But we divided set of numbers by 5 so remainder cannot be more than 5, so we divide 6 again by 5 to get final remainder as 1

Q1. Find the remainder of 136 * 148 * 298 by 5.

Show Spoiler
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

Q2. Find the remainder of 1205 * 1208 * 2404 by 12.

Show Spoiler
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
​​​​​​​

More Examples of sample problems and their solution in this post

In next section I have just listed down the theory. For problems and Binomial theorem please refer thevideo

Check above video for examples
• Remainder of numbers when divided by 2

When we try to find remainder of a number by 2 then we just need to know if the number is odd or even.
If the number is odd then remainder is 1
If the number is even then the remainder is 0

• Remainder of numbers when divided by 3

Remainder of a number by 3 is same as the remainder of sum of digits of the number by 3

• Remainder of numbers when divided by 5

Remainder of a number by 5 is same as the remainder of the unit's digit of the number by 5

• Remainder of numbers when divided by 9

Remainder of a number by 9 is same as the remainder of sum of digits of the number by 9

• Remainder of numbers when divided by 10

Remainder of a number by 10 is same as the unit's digit of the number

• Remainder of numbers when divided by other numbers
• Binomial Theorem
• Application of Binomial Theorem in finding remainders -> Please check Video for this

Next Check out this post for more Remainder Problems!

Hope it helps!­

Originally posted by BrushMyQuant on 06 Jun 2020, 11:59.
Last edited by BrushMyQuant on 25 Jun 2024, 03:23, edited 2 times in total.
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How to Solve: LCM, GCD and Properties of LCM and GCD [#permalink] How To Solve: Remainder Problems   31 May 2020, 09:11
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How to Solve: LCM, GCD and Properties of LCM and GCD


Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Hi All,

I have recently uploaded a Video on YouTube to discuss LCM and GCD in detail.





Following is covered in the video:

What is LCM and how to find LCM of 2 or more numbers using three methods.
What is GCD and how to find GCD of 2 or more numbers using three methods.
• Listing Out Multiples
• Common Division Method
• Prime Factorization Method

How to find LCM and GCD of Exponents and Fractions.

Properties of LCM and GCD

Prop 1:
LCM of 0 and any number does not exist
• LCM ( 0 , a ) = Does not exist

Prop 2:
GCD of 0 and any number is equal to the number itself. (Because all whole numbers are factors of 0. So, a will be also be a factor of 0.)
• GCD ( 0 , a ) = a

Prop 3:
LCM of 1 and any number is equal to the number itself
• LCM ( 1 , a ) = a

Prop 4:
GCD of 1 and any number is equal to 1
• GCD ( 1 , a ) = 1

Prop 5:
Product of two numbers = Product of their LCM and GCD
• a*b = LCM(a,b) * GCD(a,b)

Prop 6:
LCM of two numbers is always a multiple of their GCD
• LCM(a,b) = GCD(a,b) * k [where k is an integer]

Prop 7:
LCM of two numbers always lies between the larger of the two numbers and the product of those two numbers
• Larger (a,b) <= LCM (a,b) <= a*b

Prop 8:
GCD of two numbers is always smaller than or equal to the smaller of those two numbers
• GCD (a,b) <= Smaller (a,b)

Prop 9:
LCM of two numbers is equal to the larger of the two numbers when one number is multiple of other
• LCM(a,b) = Larger(a,b) => either a is a multiple of b or b is a multiple of a

Prop 10:
LCM of two numbers is equal to the product of those two numbers when the numbers are co-prime
• LCM(a,b) = a*b => a and b are co-prime numbers
Co-prime numbers are numbers which have only one factor in common (i.e. 1)

Prop 11:
GCD of two numbers is equal to the smaller of the two numbers when one number is multiple of other
• GCD(a,b) = Smaller(a,b) => either a is a multiple of b or b is a multiple of a

Prop 12:
GCD of two numbers is 1 when the numbers are co-prime
• GCD(a,b)=1 => a and b are co-prime numbers

Prop 13:
All numbers can be written as a multiple of their GCD
• a and b can be written as a multiple of their GCD
• a = GCD(a,b) * k [where k is an integer]
• b = GCD(a,b) * t [where t is an integer]

Prop 14:
All multiples of LCM are divisible by the numbers whose LCM we have calculated
• Multiples of LCM(a,b) will be divisible by both a and b

Prop 15:
a + b and a -b will always be divisible by GCD(a,b)
• \(\frac{a+b}{GCD(a,b) }=k \) [where k is an integer]
• \(\frac{a-b}{GCD(a,b) }=t \) [where t is an integer]

Sample Word Problems:

Prob 1: Joe has 12 A-grade projects and Chris has 18 B-grade projects. They have to divide up the projects into small teams with equal number of projects in each team; each team can have either A-grade project or B-grade project only. If there is no remainder, find the largest possible number of projects in each team.

Prob 2: Joe and Chris were each given a piece of wood of equal length. Joe cuts his wood piece into equal lengths of 12 cm, while Chris cuts his wood piece into equal lengths of 18 cm. If there was no remainder in both cases, find the shortest possible length of wood given to them.

Prob 3: Two steel pieces of length 340 cm and 408 cm are to be cut into pieces of all the same length without remainder. Find the greatest possible length of the pieces.

Solutions:
Prob 1: GCD (12,18) =6

Prob 2: LCM(12,18) = 36

Prob 3: GCD(340,408) = 68

Hope it helps!
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How to Solve: Divisibility Rules [#permalink] How To Solve: Remainder Problems   10 Apr 2021, 05:35
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How to Solve: Divisibility Rules


Hi All,

I have posted a video on YouTube to discuss Divisibility Rules



Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Following is Covered in the Video

Theory
    What are Divisibility Rules and why are they useful?
    Divisibility Rule for divisibility by 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
    Similarity in Divisibility rule for 2, 4 and 8
    Similarity in Divisibility rule for 3 and 9
    Example Problems


What are Divisibility Rule and why are they useful?

    Find Factors: Divisibility Rules help us in quickly identifying if a number is a factor of another number or not.
    Save Time: If we use divisibility rules then we do not have to go for Long Division method to find out the factors of the number.

Divisibility Rule for divisibility by 2

There are multiple ways of checking if a number is divisible by 2 or not, 3 of them are listed below
    Number should be even
    Last digit of the number should be divisible by 2
    Units digit should be 0,2,4,6,8


Q1: Check if 360 is divisible by 2 or not.

Solution: There are multiple ways of checking it. 3 or them are given below:
    360 is even so 360 divisible by 3
    Last digit of 360 which is 0 is divisible by 2, so 360 divisible by 3
    Unit digit of 360 is 0, so 360 divisible by 3

Divisibility Rule for divisibility by 3

    Sum of all the digits of the number should be divisible by 3


Q2: Check if 360 is divisible by 3 or not.

Solution: Sum of all the digits of 360 = 3 + 6 + 0 =9
We know that 9 is divisible by 3 => 360 is divisible by 3

Divisibility Rule for divisibility by 4

    Number formed by last two digits should be divisible by 4


Q3: Check if 360 is divisible by 4 or not.

Solution: Number formed by last two digits of 360 is 60.
We know that 60 is divisible by 4 => 360 is divisible by 4

Divisibility Rule for divisibility by 5

    Number should end with 0 or 5


Q4: Check if 360 is divisible by 5 or not.

Solution: Since 360 ends with a 0 => 360 is divisible by 5

Divisibility Rule for divisibility by 6

    Number should be divisible by both 2 and 3


Q5: Check if 360 is divisible by 6 or not.

Solution: 360 is divisible by both 2 and 3 (Check above problems)
=> 360 is divisible by 6

Divisibility Rule for divisibility by 7

    Remove the last digit and double it and subtract it from the rest of the number.
    If the result is divisible by 7 then number is divisible by 7, else it is not


Q6: Check if 343 is divisible by 7 or not.

Solution: Yes 343 is divisible by 7. Check the video for detailed explanation.

Divisibility Rule for divisibility by 8

    Number formed by last three digits should be divisible by 8


Q7: Check if 1360 is divisible by 8 or not.

Solution: Number formed by last three digits of 1360 is 360.
We know that 360 is divisible by 8 => 1360 is divisible by 8

Divisibility Rule for divisibility by 9

    Sum of all the digits of the number should be divisible by 9


Q8: Check if 9360 is divisible by 9 or not.

Solution: Sum of all the digits of 9360 = 9 + 3 + 6 + 0 = 18
We know that 18 is divisible by 9 => 9360 is divisible by 9

Divisibility Rule for divisibility by 10

    Number should end with 0


Q9: Check if 360 is divisible by 10 or not.

Solution: Since 360 ends with a 0 => 360 is divisible by 10

Divisibility Rule for divisibility by 11

    If the difference of the sum of odd place digits and the sum of even place digits of the number is divisible by 11, then the number is divisible by 11, else it is not


Q10: Check if 1320 is divisible by 11 or not.

Solution: In 1320
Sum of odd places = 1 + 2 = 3
Sum of even places = 3 + 0 = 3

Sum of odd places - sum of even places = 3-3 =0
And 0 is divisible by all the numbers
=> 1320 is divisible by 11

Divisibility Rule for divisibility by 12

    Number should be divisible by both 3 and 4


Q11: Check if 360 is divisible by 12 or not.

Solution: 360 is divisible by both 3 and 4 (Check above problems)
=> 360 is divisible by 12


Similarity in Divisibility rule for 2, 4 and 8

    2 can be written as \(2^1\) -> Rule -> Number formed by last 1 digit(s) should be divisible by 2
    4 can be written as \(2^2\) -> Rule -> Number formed by last 2 digit(s) should be divisible by 4
    8 can be written as \(2^3\) -> Rule -> Number formed by last 3 digit(s) should be divisible by 8


Similarity in Divisibility rule for 3 and 9

    Divisibility Rule for 3: Sum of all the digits of the number should be divisible by 3
    Divisibility Rule for 9: Sum of all the digits of the number should be divisible by 9


Sample Problems

Q12: 336 is divisible by which of the following (multiple options correct)?
A. 2
B. 3
C. 6
D. 7
E. 9

Answer: A,B, C, D. [ Check Video for Explanation ]

Q13: 3773 is divisible by which of the following (multiple options correct)?
A. 2
B. 3
C. 7
D. 9
E. 11

Answer: C, E [ Check Video for Explanation ]

Hope it helps!
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How To Solve: Factors and Multiples [#permalink] How To Solve: Remainder Problems   21 Apr 2023, 05:21
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Hi All,

I have recently uploaded a video on YouTube to discuss Factors and Multiples in Detail:




Following is covered in the video

    ¤ What are Multiples?
    ¤ What are Factors?
    ¤ How to find the number whose two factors are given.
    ¤ Relationship between divisor, factor and multiple

What are Multiples?

We say that “a” is a multiple of “b” when we can express “a” as a product of “b” and any integer

    a = kb , where k is an integer

Example 1: 6 is a multiple of 2 because we can express 6 as a product of 2 and an integer (3). (6 = 2*3)

Example 2: “a number(n) is a multiple of 3”.
=> n = 3*k (where k is an integer)


What are Factors?

We say that a number “a” is a factor of a number “b” when we can write “b” as a multiple of “a”

    b = ka , where k is an integer

Example 1: 2 is a factor of 6 because we can write 6 as a multiple of 2 i.e. 6 = 2*3

Example 2: “a number(n) has 3 as one of its factor's”.
So, n is a multiple of 3.
=> n = 3*k (where k is an integer)

How to find the number whose two factors are given.

If a number has 2 and 3 as its factors then the number will be a multiple of LCM of these two factors

    n = LCM (2,3) k = 6k


Relationship between divisor, factor and multiple

If n is divisible by 3 then

    ¤ 3 is a factor of n
    ¤ n is a multiple of 3
    ¤ n can be written as, n = 3k, where k is an integer

Hope it helps!
Good Luck!
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How To Solve: Number and Sum of Factors of a number [#permalink] How To Solve: Remainder Problems   21 Apr 2023, 05:59
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Hi All,

I have recently uploaded a video on YouTube to discuss Number and Sum of Factors of a number in Detail:




Following is covered in the video

    ¤ How to find Number of Factors of a number.
    ¤ Solved Problems: Number of Factors
    ¤ How to find Sum of Factors of a number.
    ¤ Solved Problems: Sum of Factors

How to find Number of Factors of a number.

To find number of factors of a number we need to write the number as product of power of prime number and add one to the powers and multiply the powers.

Example : Find the number of factors of 72.

Show Spoiler
72 = \(2^3 * 3^2\)
=> Number of factors of 72 = (3+1) * (2+1) = 4 * 3 = 12

We can list down the factors as
1 * 72
2 * 36
3 * 24
4 * 18
6 * 12
8 * 9


Solved Problems: Number of Factors

Q1 : Find number of factors of 100 and list the factors?

Show Spoiler
Sol: 100 = \(2^2 * 5^2\)
=> Number of factors of 100 = (2+1) * (2+1) = 3 * 3 = 9

We can list down the factors as
1 * 100
2 * 50
4 * 25
5 * 20
10 * 10

Q2 : Find number of factors of 8?

Show Spoiler
Sol: 8 = \(2^3\)
=> Number of factors of 8 = (3+1) = 4

We can list down the factors as 1, 2, 4, 8


How to find Sum of Factors of a number

Sum of factors of an integer n, where n = \(p^a * q^b * r^c\) (p, q, r being prime number) is given by

\(\frac{(p^{a+1} – 1) * (q^{b+1} – 1) * (r^{c+1} – 1)}{ (p-1) * (q-1) * (r-1)}\)

Solved Problems: Sum of Factors

Q1 : Find the sum of Factors of 60.

Show Spoiler
Sol: 60 = \(2^2 * 3^1 * 5^1\)
Sum of factors of 60 = \(\frac{(2^{2+1} – 1) * (3^{1+1} – 1) * (5^{1+1} – 1)}{ (2-1) * (3-1) * (5-1)}\) = \(\frac{7 * 8 * 24}{ 1 * 2 * 4}\) = 168

Hope it helps!
Good Luck!
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Factors and Multiples Question Set (Easy-Medium) [#permalink] How To Solve: Remainder Problems   22 Apr 2023, 21:19
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Hi All,

I have recently uploaded a video on YouTube to discuss Factors and Multiples Question Set (Easy-Medium) in Detail:




Seven solved problems are covered in the video.

Q1. If 26 is a factor of n, then is 6 a factor of n?

Show Spoiler
Sol: 26 is a factor of n => n is a multiple of 26
=> n = 26k where k is an integer

For 6 to be a factor of n, 26k should be divisible by 6
We need a 2 and a 3 for 6, but we have only 2 in 26k (=2 * 13k)

So, 26k will be divisible by 6 only when k is divisible by 3, which might not be true always
=> We CANNOT say for sure if 6 is a factor of n.

Q2. If 90 is a factor of n, then is 20 a factor of n?

Show Spoiler
Sol: 90 is a factor of n => n is a multiple of 90
=> n = 90k where k is an integer

For 20 to be a factor of n, 90k should be divisible by 6
We need a 2 and a 10 for 20, but we have only 10 in 90k (=10 * 9k)

So, 90k will be divisible by 20 only when k is divisible by 2, which might not be true always
=> We CANNOT say for sure if 20 is a factor of n.

Q3. If 24 is a factor of n, then is n a multiple of 8?

Show Spoiler
Sol: 24 is a factor of n => n is a multiple of 24
=> n = 24k where k is an integer

n = 24k = 8 * 3k
=> Clearly, n is a multiple of 8 => TRUE.

Q4. If 4 and 6 are factors of x, then which of the following will be a factor of x for sure
A. 8
B. 12
C. 24
D. 36
E. 48

Show Spoiler
Sol: 4 and 6 are factors of n => n is a multiple of LCM(4,6)
=> n = LCM(4,6) * k where k is an integer
= 12k
=> n is a multiple of 12 for sure

=> So, Answer will be D.

Q5. Is 192 divisible by x?
A. x is divisible by 12
B. x is divisible by 16

Show Spoiler
Sol:
STAT A: x is divisible by 12
=> x will be a multiple of 12.
=> x can be 12, 24, 192 or even 192 *10.
When x is 12, 24, 192 then 192/x is an integer but when x is 1920 then 192/x is not an integer.
=> NOT SUFFICIENT

STAT B:x is divisible by 16.
=> x will be a multiple of 16.
=> x can be 16, 32, 192 or even 192 *10.
When x is 16, 32, 192 then 192/x is an integer but when x is 1920 then 192/x is not an integer.
=> NOT SUFFICIENT

Taking both the statements together we get

Combining both also gives us that x will be a multiple of LCM(12,16)= 48k.
=> x can be 48,..192, 1920.
When x is 48, 96, 192 then 192/x is an integer but when x is 1920 then 192/x is not an integer.
=> NOT SUFFICIENT

=> So, Answer will be E.

Q6. If 12 is a factor of x and 16 is a factor of y, then
6.1 Is 48 a factor of xy?
6.2 Is 96 a factor of xy?
6.3 Is 192 a factor of xy?

Show Spoiler
Sol: 12 is a factor of x
=> x is a multiple of 12 => x = 12a
16 is a factor of y
=> y is a multiple of 16 => y = 16b
xy = 12a * 16b = 192 ab

So, xy is a multiple of
48 because xy = 48*4*ab
96 because xy = 96*2 *ab
192 because xy= 192*1*ab

So, Answer will be TRUE for all three problems

Q7. In a competition, gold medal gives 7 points, Silver gives 5 points and bronze gives 3 points. Roger participated in the competition in various events and the product of all the points earned by him is 33075. Then find the number of gold, silver and bronze medal earned by Roger

Show Spoiler
Sol: Bronze = x, Silver = y, Gold = z
Product of all the points = 33075
=> 33075 = \(3^x * 5^y * 7^z\)
33075 = \(3^3 * 5^2 * 7^2\)
\(3^3 * 5^2 * 7^2\) = \(3^x * 5^y * 7^z\)

So, Answer will Roger won: Gold – 2, Silver – 2, Bronze - 3


Watch the following video to learn the Basics of Factors and Multiples




Hope it helps!
Good Luck!
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