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 14 of 16.
Dilz said:
8 years ago
Subtraction of these numbers cancels remainder but how?
Sumit said:
8 years ago
For
4*2+3=11
4*8+3=35
4*11+3=47
Here you all can see that taking same remainder h.c.f seems 4.
H.C.F. OF (35-11),(47-11) and (47-35)=12.
So taking this method answers can differ.
4*2+3=11
4*8+3=35
4*11+3=47
Here you all can see that taking same remainder h.c.f seems 4.
H.C.F. OF (35-11),(47-11) and (47-35)=12.
So taking this method answers can differ.
Meher priyanka said:
8 years ago
N(q2-q1)=(4665-1305)=3360
N(q3-q2)=(6905-4665)=2240
N(q3-q1)=(6905-1305)=5600
N(q1+q2+q3)=3360+2240+5600=11200
N(1+1+2+0+0)=4.
N(q3-q2)=(6905-4665)=2240
N(q3-q1)=(6905-1305)=5600
N(q1+q2+q3)=3360+2240+5600=11200
N(1+1+2+0+0)=4.
Meghana said:
8 years ago
Indeed,
1. It is given in the question that the remainder is same and
2. we know that the greatest number is leaving the same number.
Thus we are subtracting two numbers to get value exclusive of remainder
Now, if A and B are divisible by say x them A-B is also divisible by x.
With this logic, we are first eliminating the common remainder and later finding the HCF.
1. It is given in the question that the remainder is same and
2. we know that the greatest number is leaving the same number.
Thus we are subtracting two numbers to get value exclusive of remainder
Now, if A and B are divisible by say x them A-B is also divisible by x.
With this logic, we are first eliminating the common remainder and later finding the HCF.
Guru prasad said:
8 years ago
Subtract the numbers such as,
4665-1305 = 3360 and 6905 - 4665 = 2240 then subtracting the obtained results.
3360-2240 = 1120. Therefore 1 + 1 + 2 + 0 = 4.
4665-1305 = 3360 and 6905 - 4665 = 2240 then subtracting the obtained results.
3360-2240 = 1120. Therefore 1 + 1 + 2 + 0 = 4.
Sibaram said:
8 years ago
What if the question is for finding smallest such number? Is it same like we have to find lcm of difference between numbers?
Samir Alvani said:
8 years ago
When you divide 1305, 4665 and 6905 by 1120 you will get the same remainder 185 so N is 1120 so the sum of the digits is 1+1+2+0=4.
Gopika said:
8 years ago
@Saraswati.
I can't understand why you are multiplying 10?
Please explain it.
I can't understand why you are multiplying 10?
Please explain it.
Aman Srivastava said:
8 years ago
Thanks @Yogesh.
Kashi said:
8 years ago
Thanks @Prakash.
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