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.
Hanumant said:
9 years ago
Thank you all for explaining it.
Shirish said:
9 years ago
Thanks to all for the correct explanation.
Diya said:
8 years ago
Thanks @Prakash.
Amit yadav said:
8 years ago
Any no. Can be represented as of the form D*q+r right. so we want remainder same in each case. So
1305=N*q1+r
4665=N*q2+r
6905=N*q3+r
Now, now subtracting 2 eq from 1 then ..3 to 2 then 3 to 1 we get N*(q2-q1)=3360.
Similarly, we will get the rest ones. now here, we just want the greatest no. Which divides all this num3360 2240 nd 5600 ....so simply take the hcf off all. We will get N=1120.
I hope this helps you.
1305=N*q1+r
4665=N*q2+r
6905=N*q3+r
Now, now subtracting 2 eq from 1 then ..3 to 2 then 3 to 1 we get N*(q2-q1)=3360.
Similarly, we will get the rest ones. now here, we just want the greatest no. Which divides all this num3360 2240 nd 5600 ....so simply take the hcf off all. We will get N=1120.
I hope this helps you.
(1)
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.
Jayant Verma said:
8 years ago
8*163 = 1304 thus 1 remainder
583*8 = 4664 thus 1 remainder
863*8 = 6904 thus 1 remainder
So the answer should be 8.
583*8 = 4664 thus 1 remainder
863*8 = 6904 thus 1 remainder
So the answer should be 8.
(1)
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.
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.
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.
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