I have seen 3 types of questions on LCM..
Type 1
Que 1: Which series will give all the +ve nos. Which when divided by 15, 18, 24, remainder is 1?
Soln. Of course this is an easy nut 2 crack.....You will immediately say…
The series will be represented by L.C.M.(15, 18, 24)K + 1
Where k = 0, 1, 2, 3..........
Now the 2nd type....
Type 2
Que 2: Which series will give all the +ve nos. which when divided by 15, 18, 24, remainders are 13, 16, 22 respectively?
Soln. this is the same question as above....watch it carefully..
Divisor remainder
15 13
18 16
24 22
Now in each case,remainder can be written as -2 (-ve remainder) also.This is what the question requires...but now since the remainders r same..We can opt 1st type..
So, it will be L.C.M. (15, 18, 24) K - 2
Where k = 1, 2, 3,4.........(neglecting zero because we want +ve nos)
Now comes d mind boggling 1....I mean 3rd type
Type 3
First of I’ll take a smaller example...
Que 3: Which series will give all the nos. which when divided by 9 and 8, remainders are 5 and 1 respectively???
Soln: I’ll obviously use Chinese Reminder theorem…but not algebraically..i have picked up d logic only…see
I’ll pick d bigger divisor which is 9 here (why???cz otherwise it’ll be more calculative).
My first condition is nos should leave remainder 5 when divided by 9.It’ll be of 9k + 5 type
K 9k + 5
0 5
1 14
2 23
3 32
and so on……..
Out of these nos.. I want 2 choose that no which leave remainder 1 when divided by 8…we’ll start checking…
K 9k + 5 remainder wn divided by 8
0 5 5
1 14 6....(increase of 1.same pattern)
2 23 7
3 32 8
We’ll check for only first 2 nos for the middle column....as we can see there is increase of 1 in the last column, so we’ll keep on increasing the remainders by 1 until we get 1. (the required remainder).....and that corresponds to k = 4…
So, answer is 9×4 +5 = 41.
Now come 2 the original question..
Answer will be L.C.M. (9, 8) k+41, where k = 0, 1, 2, 3………this series will give all d nos. of above said type.
Qus 4: What is the smallest +ve no. which when divided by 72 and 10; remainders are 41 and 9 respectively?
Soln. I’ll pick the bigger divisor which is 72 here.
My first condition is that the number should leave remainder 41 when divided by 72.It’ll be of 72k + 41 type.
Out of these nos.. I want 2 choose that no. which leave remainder 9 when divided by 10…we’ll start checking…
K 72k+41 remainder wn % by 10
0 +4 41 1
1 113 3....(increase of 2..so, +2 in each term)
2 __ 5
3 __ 7
4 __ 9 Answer
We’ll check for only first 2 nos for d middle column....as we can see there is increase of 2 in the last column, So we’ll keep on increasing the nos. by 2 until we get 9 (the required remainder).....and that corresponds to k = 4…
So, answer is 72*4+41 = 329.
Qus 5:
What is the smallest +ve no. which when divided by 8, 9 and 10; remainders are 1, 5 and 9 respectively?
Ans. This question consists of above two questions (3 and 4). First of all we will find for 8 and 9. By above question the required number will be 41.Now question becomes
What is the smallest +ve no. which when divided by 72 and 10; remainders are 41 and 9 respectively?
And the ans. to this question is 329.
Application of type 3 in Remainder theory
Qus 6: What is the remainder when N= 777……777 (41 times) is divided by 36?
Ans: 36 = 9*4 (We will Split the divisor in to 2 or 3 factors which are Co-Prime to each other)
Now, Remainder when N%9 = 7+7+7….. (41 Times)
= 7*41
= 7*5
= 35 =8
Remainder when N%4 = 1 (We will Check for last 2 places)
Now, we are looking for the smallest positive integer which leaves remainder 8 and 1 respectively, when divided by 9 and 4.
K 9k+8 remainder wn % by 4
0 8 0
1 17 1
Luckily, we got our ans. at the initial stage so our ans. Correspond to K=1 i.e. 17.
