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. 1)
1.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Answer: Option
Explanation:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
Discussion:
210 comments Page 4 of 21.
Mozammil anwar said:
6 years ago
4/43 = remainder is 3.
4/183 =remainder is 3.
4/91 = remainder is 3.
So that' why the answer is 4.
4/183 =remainder is 3.
4/91 = remainder is 3.
So that' why the answer is 4.
(1)
Lokesh said:
6 years ago
Why we have to use hcf?
Bala said:
6 years ago
@All.
They asked to divide by a number also we need to have the same remainder in all cases.
So, dividing by 4, we get remainder 3 in all cases.
They asked to divide by a number also we need to have the same remainder in all cases.
So, dividing by 4, we get remainder 3 in all cases.
(1)
Pinku said:
6 years ago
43/4 = 3(remainder).
91/4 = 3(remainder).
183/4 = 3(remainder).
91/4 = 3(remainder).
183/4 = 3(remainder).
Pabitra Saha said:
6 years ago
Very good explanation, Thanks @Kuldeep Sharma.
Kuldeep Sharma said:
7 years ago
let the greatest number is H.
Now;
H*Quotient(Q1)+Reminder(R)= 43 -------> (1)
H*Quotient(Q2)+Reminder(R)= 91 -------> (2)
H*Quotient(Q3)+Reminder(R)= 183 -------> (3)
Now, (1)-(2), (3)-(2), (3)-(1)
H*(Q2-Q1)= 48,
H*(Q3-Q2)= 92,
H*(Q3-Q1)= 140.
So 48, 92 and 140 are completely divisible by H, that will be the HCF of these numbers.
So, the HCF of 48, 92 and 140 is 4.
Now;
H*Quotient(Q1)+Reminder(R)= 43 -------> (1)
H*Quotient(Q2)+Reminder(R)= 91 -------> (2)
H*Quotient(Q3)+Reminder(R)= 183 -------> (3)
Now, (1)-(2), (3)-(2), (3)-(1)
H*(Q2-Q1)= 48,
H*(Q3-Q2)= 92,
H*(Q3-Q1)= 140.
So 48, 92 and 140 are completely divisible by H, that will be the HCF of these numbers.
So, the HCF of 48, 92 and 140 is 4.
(3)
Davoas said:
7 years ago
What if we have a fixed remainder.
For example take 90, 114 and 230 with remainder 6 what will be the divisor?
Anyone explain me.
For example take 90, 114 and 230 with remainder 6 what will be the divisor?
Anyone explain me.
Magnus Udoka said:
7 years ago
Your explanation was superb, Thanks @Prashant.
Shreekanth said:
7 years ago
Keep on dividing all the three numbers by 2 so that reminder should be the same ie
2[43 91 183]
[ 21 45 91] remainder will be 1 for all the three.
Divide again by 2.
2[21 45 91]
10 22 45 remainders will be same ie 1.
So after this, if you try to divide all three again by 2 then you get a different value for the reminder.
And hence; 2 * 2 = 4.
2[43 91 183]
[ 21 45 91] remainder will be 1 for all the three.
Divide again by 2.
2[21 45 91]
10 22 45 remainders will be same ie 1.
So after this, if you try to divide all three again by 2 then you get a different value for the reminder.
And hence; 2 * 2 = 4.
(1)
Pabitra Bhusan Saha said:
7 years ago
Your explanation is very easy to understand and this is the best process, Thanks @Saroja and @Sourav.
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