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 16 of 16.
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)
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)
Nidhitiwari said:
9 months ago
Very good and helpful. Thanks all.
(2)
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)
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)
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)
Disha Mukherjee said:
2 months ago
I can't understand. Why should we subtract the numbers?
(8)
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