There are three types of questions that are based on LCM with remainders. They are as follows.
When the remainders are same for all the divisors
In this case the required number will be the LCM × N + remainder, where N is any natural number
N=1, will give smallest such number
N=2, will give second smallest such number and so on
When the remainders are different for different divisors but the respective difference between the
divisors and the remainders remain constant.
In this case the required number will be LCM × N +difference of any (divisor - remainder)
refer to example 2.
When neither the divisors are same nor the respective difference between the divisors and the remaniders
remain constant.
In this case solve the question by forming equations and solving them.
Question 1 What is the least possible number which when divided by 4,5,6 leaves the remainder as 3,4,5 respectively.
59
60
61
119
A
Since the difference is same in all i.e (4-3)=(5-4)=(6-5)=1
∴ the required number = LCM (4,5,6) -1
= 60 - 1 = 59
Question 2 Find the least possible 5 digit number which when divided by 2,4,6,8 it leaves the remainder 1,3,5,7 respectively.
10006
10007
10008
None of these
B
i) LCM of 2,4,6,8 is 24
∴ all possible values = 24N - 1 , where N is a natural number
Least possible 5 digit number will be for N=417, 24 × 417 -1 = 10007
Question 3 What is the least possible number which when divided by 10,12,14 leaves the remainder as 2. How many such numbers are there between 5000 and 6000.
1
2
3
4
C
i) LCM of 10,12,14 is 420
∴ the least possible number that leaves the remainder as 2 when divided by 10,12,14 = 420 × 1 + 2 = 422
ii) Number of such numbers between 5000 and 6000 can be found by putting different values of N,
for N=12, 420N +2 = 5042
for N=14, 420N +2 = 5882
∴ there are 3 such numbers between 5000 and 6000.(for N=12,13,14 420N+2 lies between 5000and 6000)
Question 4 What is the least possible number which when divided by 11 leaves the remainder 3 and when divided by 5 leaves the remainder as 2.
46
47
48
49
B
Let the required number be N,
Since N divided by 5 gives remainder as 2 ∴ N = 5 × m + 2
Since N divided by 11 gives remainder as 3 ∴ N = 11 × n + 3
∴ 5m+2=11n+3
$m=\frac{11n+1}{5}$m=11n+15
For n = 4, m = 9 that is the smallest possible value of m for any n.
∴ the smallest number is 5×9+2=47
Next such number will be LCM of (11 and 5) + 47
Question 5 What will be the least possible number which when divided by 4,5,6 always leaves the remainder 3. Find the following such numbers which satisfy the given condition.
i) Smallest
ii) Second Smallest
iii) Greatest number smaller than 1000
The smallest number which when divided by 4,5,6 leaves the remainder 3 in each case is LCM (4,5,6) + 3 = 63
To find all the set of such numbers which gives remainder as 3 when divided by 4,5,6 will be
60N + 3, where N is a natural number.
∴ Second largest such number will be 60 × 2 + 3 = 123
To find the largest number of such form smaller than 1000, we will have to find the largest number of form 60N + 3 , that is smaller than 1000.
Taking N=16, 60N +3 = 963, that is the largest such number
LCM with Remainders
Reviewed by Ekansh Rastogi
on
01:18
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