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:
158 comments Page 10 of 16.
Bidya sagar behera said:
1 decade ago
I can't understand because of taking H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305).
Swaminathan Vembu said:
1 decade ago
I calculated the HCF for the 3 numbers and got an ans of 40.
This would also mean that the sum of the digits is 4 = correct answer.
1120 = 2*2*2*2*2*2*5*7.
3360 = 2*2*2*2*2*3*5*7.
5600 = 2*2*2*3*3*3*5*5.
This method does not warrant/validate the condition that the GCD needs to be calculated via successive difference.
This would also mean that the sum of the digits is 4 = correct answer.
1120 = 2*2*2*2*2*2*5*7.
3360 = 2*2*2*2*2*3*5*7.
5600 = 2*2*2*3*3*3*5*5.
This method does not warrant/validate the condition that the GCD needs to be calculated via successive difference.
Swaminathan said:
1 decade ago
Impossible to calculate the GCD without successive difference:
a. 6905 - 4665.
b. 4665 - 1305.
c. 6905 - 1305.
Interestingly when you replace the values used in the above method to calculate the successive difference as illustrated below :
a. 6905 - 1305 (c above).
b. 6905 - 4665 (a above).
and
c. 4665 - 1305 (above it would replace the statement => 4665 - 1305).
It will convincingly still give rise to a G.C.D = 1120.
a. 6905 - 4665.
b. 4665 - 1305.
c. 6905 - 1305.
Interestingly when you replace the values used in the above method to calculate the successive difference as illustrated below :
a. 6905 - 1305 (c above).
b. 6905 - 4665 (a above).
and
c. 4665 - 1305 (above it would replace the statement => 4665 - 1305).
It will convincingly still give rise to a G.C.D = 1120.
Geetika said:
1 decade ago
Please anyone tell me the concept of 1120?
Nandhini said:
1 decade ago
Please can you explain how to calculate HCF and explain?
Pallavi said:
1 decade ago
Please can you explain how to calculate HCF ?
Sasank said:
1 decade ago
Here taking difference is mentioned because we need to find h.c.f.
Here 1305=pq1+r;
4665=pq2+r;
6905=pq3+r;.here p is divident, q=quotient, r=reminder.
We want to find p so,
Totally 3 equations so,
P(q2-q1) = 3360.
p(q3-q2) = 2240.
p(q3-q1) = 5600 now there are 3 p's we need p that is h.c.f of (3360, 2240, 5600) here finding hcf by taking difference it will make easy now p = 1120 and (1+1+2+0) = 4.
Hope you all understood.
Here 1305=pq1+r;
4665=pq2+r;
6905=pq3+r;.here p is divident, q=quotient, r=reminder.
We want to find p so,
Totally 3 equations so,
P(q2-q1) = 3360.
p(q3-q2) = 2240.
p(q3-q1) = 5600 now there are 3 p's we need p that is h.c.f of (3360, 2240, 5600) here finding hcf by taking difference it will make easy now p = 1120 and (1+1+2+0) = 4.
Hope you all understood.
Rohit said:
1 decade ago
How to find HCF fast in simple method?
Vikram said:
1 decade ago
Why do we need to take the difference and then calculate HCF here? As per my understanding in the question to find HCF for given three numbers and the reminder should be same for all three numbers.
In that case why can't I divide all the numbers by 5 so that the reminder will be zero and I will get the HCF also. Where my idea gets slips. Please help.
In that case why can't I divide all the numbers by 5 so that the reminder will be zero and I will get the HCF also. Where my idea gets slips. Please help.
J.manisha said:
1 decade ago
Sorry I can't understand this process. Please kindly give correct explanation. How will take this "N" value N = 1+1+2+0 = 4.
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