# Aptitude - Problems on H.C.F and L.C.M - Discussion

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 10)

10.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

Answer: Option

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90*k* + 4, which is multiple of 7.

Least value of *k* for which (90*k* + 4) is divisible by 7 is *k* = 4.

Required number = (90 x 4) + 4 = 364.

Discussion:

86 comments Page 1 of 9.
Amita said:
1 decade ago

First read the q. when we divide the all options with 6 then we get the reminder 4 in three option's 1st is 94 and 2nd 184 and 3rd is 364. when we divide the 94 and 184 with 7 we will get it is not divisible by 7, but when we divide the 364 with 7 then we get this is divisible by 7 so it will be answer.

After this only 364 is divisible by 7

7)364(52

-35

------

014

- 14

-------

00

-------

Another way is,

Divide all 4 option with 7. you will easily get which no. is divisible by 7.

And you will get only 364 is divisible by 7 so answer is 364.

6)74(12 6)94(15 6)364(6 6)184(3

-6 -6 -36 -18

----- ----- ------ ------

14 34 004 004

-12 -30

----- -----

02 04

After this only 364 is divisible by 7

7)364(52

-35

------

014

- 14

-------

00

-------

Another way is,

Divide all 4 option with 7. you will easily get which no. is divisible by 7.

And you will get only 364 is divisible by 7 so answer is 364.

Rahul said:
1 decade ago

The question is a bit confusing but once you get it, it is ridiculously easy to solve. We are basically looking for a number X that has the following properties:

X/7 - no remained

x/6 - remainder of 4

X/9 - remainder of 4

X/15 - remainder of 4

X/18 - remainder of 4

Ideally, one would try and find the LCM of 6, 9, 15 and 18 but we already know the number involved is a multiple of 7 so we can quickly eliminate by dividing everything by 7. Turns out only one option remains. Check if they all divide by 6, 9, 15 and 18 to give a remainder of 4 and this is one of the easiest questions that can be asked.

X/7 - no remained

x/6 - remainder of 4

X/9 - remainder of 4

X/15 - remainder of 4

X/18 - remainder of 4

Ideally, one would try and find the LCM of 6, 9, 15 and 18 but we already know the number involved is a multiple of 7 so we can quickly eliminate by dividing everything by 7. Turns out only one option remains. Check if they all divide by 6, 9, 15 and 18 to give a remainder of 4 and this is one of the easiest questions that can be asked.

Vikas said:
8 years ago

Hi, friends it's very simple trick.

LCM of 6, 9, 15, 18 is 90.

We can simply write the answer 90 + 4 but in question, it should be multiple of 7.

So (90k + 4) / 7 now we have to find the value of 'k' to complete the answer.

The first method is to substitute the value of k = 1, 2, 3. And divide it by 7 and sees the result. It actually time taking.

Now the trick:

(90k + 4) / 7 = 90k / 7 + 4 / 7,

The remainder of 4 / 7 is 4,

Remainder 90k / 7 is 6k.

Now add both remainders 6k+4 it should be divided by 7.

Now substitute the value of k it comes out be 4.

Put 90 * 4 + 4 = 364.

LCM of 6, 9, 15, 18 is 90.

We can simply write the answer 90 + 4 but in question, it should be multiple of 7.

So (90k + 4) / 7 now we have to find the value of 'k' to complete the answer.

The first method is to substitute the value of k = 1, 2, 3. And divide it by 7 and sees the result. It actually time taking.

Now the trick:

(90k + 4) / 7 = 90k / 7 + 4 / 7,

The remainder of 4 / 7 is 4,

Remainder 90k / 7 is 6k.

Now add both remainders 6k+4 it should be divided by 7.

Now substitute the value of k it comes out be 4.

Put 90 * 4 + 4 = 364.

Dipak lal said:
1 decade ago

L.C.M. of 6, 9, 15 and 18 is 90.

Take k =1,2,3,... till reminder is 00.

Let required number be (90 k + 4 )/7.

if k=1 then 90(1)+4/7 = 94/7 =13.42 = reminder is not 00.

if k=2 then 90(2)+4/7 = 184/7 =26.28 = reminder is not 00.

if k=3 then 90(3)+4/7 = 274/7 =39.14 = reminder is not 00.

if k=4 then 90(4)+4/7 = 364/7 =52.00 = reminder is 00.

OK now you take k = 4 because remainder 0.

Solution is 90k + 4 = 90*4 + 4 = 364.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

Take k =1,2,3,... till reminder is 00.

Let required number be (90 k + 4 )/7.

if k=1 then 90(1)+4/7 = 94/7 =13.42 = reminder is not 00.

if k=2 then 90(2)+4/7 = 184/7 =26.28 = reminder is not 00.

if k=3 then 90(3)+4/7 = 274/7 =39.14 = reminder is not 00.

if k=4 then 90(4)+4/7 = 364/7 =52.00 = reminder is 00.

OK now you take k = 4 because remainder 0.

Solution is 90k + 4 = 90*4 + 4 = 364.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

Sushil said:
7 years ago

Here is my simple explanation.

I think everyone is understanding how did they get ab = 12,

now if first no. is 13a,

and second no. is 13b,

factors of 12 can be (1,12) (3,4) (6,2),

Now, see if you include any common factor in a and b both then that will increase the H.C.F as HCF if highest common factor like if we take factor 6 and 2 then there is common factor between 6 and 2 which is 2 so HCF will change now to 26 that's why we can't take 6 and 2.

So only 2 pairs (1,12) (3,4).

I think everyone is understanding how did they get ab = 12,

now if first no. is 13a,

and second no. is 13b,

factors of 12 can be (1,12) (3,4) (6,2),

Now, see if you include any common factor in a and b both then that will increase the H.C.F as HCF if highest common factor like if we take factor 6 and 2 then there is common factor between 6 and 2 which is 2 so HCF will change now to 26 that's why we can't take 6 and 2.

So only 2 pairs (1,12) (3,4).

Tushar said:
5 years ago

L.C.M. of 6, 9, 15 and 18 is 90.

Let the required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

90K+4 is the required no.

How?

If K=1

90*1+4=94 , it is not multiple of 7.

90*2+4=184, it is not multiple of 7.

90*3+4=274, it is not multiple of 7.

90*4+4=364, it is multiple of 7.

So the least value will be 364.

Let the required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

90K+4 is the required no.

How?

If K=1

90*1+4=94 , it is not multiple of 7.

90*2+4=184, it is not multiple of 7.

90*3+4=274, it is not multiple of 7.

90*4+4=364, it is multiple of 7.

So the least value will be 364.

(1)

SRI said:
10 years ago

You will have 90k+4.

Now we know 90/7 gives - 12.85 so, we can write 90 as 84k+6k.

So 90k+4 can be written as (84k)+6k+4. The first part 84k is divisible by 7. So no need to check. The second part i.e. 6k+4 has to be divisible by 7.

K = 1, 6(1)+4 = 10 X.

K = 2, 6(2)+4 = 16 X.

K = 3, 6(3)+4 = 22 X.

K = 4, 6(4)+4 = 28.

So this is divisible by 7 hence your answer is 84(4)+6(4)+4.

Hope this clarifies.

Now we know 90/7 gives - 12.85 so, we can write 90 as 84k+6k.

So 90k+4 can be written as (84k)+6k+4. The first part 84k is divisible by 7. So no need to check. The second part i.e. 6k+4 has to be divisible by 7.

K = 1, 6(1)+4 = 10 X.

K = 2, 6(2)+4 = 16 X.

K = 3, 6(3)+4 = 22 X.

K = 4, 6(4)+4 = 28.

So this is divisible by 7 hence your answer is 84(4)+6(4)+4.

Hope this clarifies.

Jignesh rajput said:
1 decade ago

Here, (90k+4) put each of value starting from 1,2,3,.... than after divisible by 7.when in the case of put 4th value we have not get the any remainder of (90k+4)divisible by 7..dats y take k=4.

i.e (90(1)+4)/7= remainder 3

(90(2)+4)/7= remainder 3

(90(3)+4)/7= remainder 2

(90(4)+4)/7= remainder 0

thats y take k=4.

thanks u

i.e (90(1)+4)/7= remainder 3

(90(2)+4)/7= remainder 3

(90(3)+4)/7= remainder 2

(90(4)+4)/7= remainder 0

thats y take k=4.

thanks u

Shalini said:
5 years ago

LCM of 6, 9, 15, 18 is 90.

Such that there is one formula these type of questions ie "LCM*k+remainder "should be divisible by a multiple.

So we want to do "trail and error" method by giving value from the options to the 'k' so that the number is divisible, so we substitute k=4.

Such that there is one formula these type of questions ie "LCM*k+remainder "should be divisible by a multiple.

So we want to do "trail and error" method by giving value from the options to the 'k' so that the number is divisible, so we substitute k=4.

Sundar said:
1 decade ago

Short cut method: [ Eliminating the options method ]

From this "The least multiple of 7" - we can tell the the result should be divisible by 7.

From the given 4 options, 364 is the only number divisible by 7.

Therefore, 364 is the correct answer.

From this "The least multiple of 7" - we can tell the the result should be divisible by 7.

From the given 4 options, 364 is the only number divisible by 7.

Therefore, 364 is the correct answer.

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