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 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.
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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