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:
216 comments Page 16 of 22.
M sai teja said:
1 decade ago
I think 43 and 91 are prime numbers.
Prime numbers don't have factors, to find H.C.F we need factors, for that we are subtracting numbers to get non-prime numbers.
Prime numbers don't have factors, to find H.C.F we need factors, for that we are subtracting numbers to get non-prime numbers.
Iegowg said:
1 decade ago
Just divide each option with all figures it is that simple.
Anukriti srivastava said:
1 decade ago
Is there any quick and efficient method to find HCF and LCM of big numbers such as HCF of 38094, 43762, 84632 for competitive exam if yes then please help me out waiting.
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.
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