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:
213 comments Page 16 of 22.
Manjunath said:
1 decade ago
I want the explanation of the logic.
Vimal kumar said:
1 decade ago
If the sum of two numbers are 348 and their HCF is 48 then what is the difference of the numbers?
Mamang said:
1 decade ago
Please tell me the most simple formula.
Trishana said:
1 decade ago
Here in this question greatest number is 7 and it gives the same remainder that is 1 when divided by all. So how is the answer 4?
Sowjanya said:
1 decade ago
In @Prasanth answer, p is the divisor not dividend and the analysis of sum is correct.
Dileep gowda said:
1 decade ago
Yes your right @Sowjanya. But @Prasanth is almost right I think you need more information.
Chandu said:
1 decade ago
1st divide 48 with 4 then remainder is 0.
2nd divide 92 with 4 then remainder is 0.
3rd divide 140 with 4 then remainder is 0.
Hence so 4 is a perfect number.
2nd divide 92 with 4 then remainder is 0.
3rd divide 140 with 4 then remainder is 0.
Hence so 4 is a perfect number.
Chirag said:
1 decade ago
Easy calculation, 7 and 13 are directly eliminated.
4 and 9 are the required. So divide it by the 9, no remainder will be same. So 4 is the option.
4 and 9 are the required. So divide it by the 9, no remainder will be same. So 4 is the option.
Umar said:
10 years ago
How to solve the same question if remainder is not same in each case?
Daniel said:
10 years ago
I can't understand the question.
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