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. 4)
4.
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
Answer: Option
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Discussion:
157 comments Page 1 of 16.
Disha Mukherjee said:
2 months ago
I can't understand. Why should we subtract the numbers?
(8)
Chahat Mishra said:
2 months ago
@All. Here's my solution.
Num 1305, 4665, 6905.
Difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
Now,
3360/2240 = 1120,
5600/1120 = 0.
HCF 3360, 2240 & 5560 is 1120.
n = 1 + 1 + 2 + 0 = 4;
So, n = 4.
Num 1305, 4665, 6905.
Difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
Now,
3360/2240 = 1120,
5600/1120 = 0.
HCF 3360, 2240 & 5560 is 1120.
n = 1 + 1 + 2 + 0 = 4;
So, n = 4.
(1)
Arjun said:
4 months ago
Step 1: Differences between the numbers.
4665−1305 = 3360.
6905−4665 = 2240.
6905−1305 = 5600.
Step 2: Find the GCD of these differences
common prime factors are 2^5 . 5. 7 = 1120.
Step 3: Sum of the digits of 𝑁
1 + 1 + 2 + 0 = 4.
4665−1305 = 3360.
6905−4665 = 2240.
6905−1305 = 5600.
Step 2: Find the GCD of these differences
common prime factors are 2^5 . 5. 7 = 1120.
Step 3: Sum of the digits of 𝑁
1 + 1 + 2 + 0 = 4.
(1)
Vishnupriya Venkateshwaran said:
5 months ago
As explained in the question,
1305 = N(p) + r --->(1)
4665 = N(q) + r --->(2)
6905 = N(s) + r --->(3)
Now solve (2) - (1)=> 3360 = N (q-p) ---> (4)
and solve (3) - (2)=> 2240 = N (s-p) ---> (5)
Factorizing,
3360 = 1120 * 3 ---> (6)
2240 = 1120 * 2 ---> (7)
Comparing (4) and (5) with (6) and (7), we can clearly say that N=1120.
sum of digits in N = 1+1+2+0 = 4.
1305 = N(p) + r --->(1)
4665 = N(q) + r --->(2)
6905 = N(s) + r --->(3)
Now solve (2) - (1)=> 3360 = N (q-p) ---> (4)
and solve (3) - (2)=> 2240 = N (s-p) ---> (5)
Factorizing,
3360 = 1120 * 3 ---> (6)
2240 = 1120 * 2 ---> (7)
Comparing (4) and (5) with (6) and (7), we can clearly say that N=1120.
sum of digits in N = 1+1+2+0 = 4.
(1)
Nidhitiwari said:
9 months ago
Very good and helpful. Thanks all.
(2)
Sudha patil said:
9 months ago
I'm also not getting it please anyone can explain in detail?
The main thing is why should we subtract those numbers is there any trick?
The main thing is why should we subtract those numbers is there any trick?
(13)
Merlinissac said:
10 months ago
4 * 1304 + 1 = 1305
4 * 4664 + 1 = 4665
4 * 6904 + 1 = 6905.
4 * 4664 + 1 = 4665
4 * 6904 + 1 = 6905.
(3)
Shuvojoti said:
1 year ago
What is it exactly saying, when it says, "leaving the same remainder"?
We are ultimately finding the HCF, right? Then what is it defining by "Same remainder"?
Anyone, please explain me this in detail.
We are ultimately finding the HCF, right? Then what is it defining by "Same remainder"?
Anyone, please explain me this in detail.
(7)
Jai said:
1 year ago
Given numbers 1305, 4665, 6905.
Now take difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
now,
Divide 3360/2240=1120, and
Now divide 5600/1120=0.
so, the HCF of 3360, 2240 & 5560 is 1120.
n=1+1+2+0=4;
So, n=4
Now take difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.
now,
Divide 3360/2240=1120, and
Now divide 5600/1120=0.
so, the HCF of 3360, 2240 & 5560 is 1120.
n=1+1+2+0=4;
So, n=4
(48)
Surya said:
2 years ago
Why should we subtract the Numbers with the smallest number and so HCF for that?
(20)
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