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 15 of 21.
Rajesh said:
1 decade ago
1.91-43 = 48, 183-91 = 92, 183-43 = 140.
2.92-48 = 44, 140-92 = 48, 140-48 = 92.
For every time we eliminating uncommon Factors.
2.92-48 = 44, 140-92 = 48, 140-48 = 92.
For every time we eliminating uncommon Factors.
Vandana said:
1 decade ago
But there is that that same remainder in each case.
NARESH said:
1 decade ago
Please explain again simply way.
Somanath said:
1 decade ago
Find the highest number which divides 25 and 29 equally leaving 5 in each case.
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:
10 years 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:
10 years ago
I want the explanation of the logic.
Vimal kumar said:
10 years ago
If the sum of two numbers are 348 and their HCF is 48 then what is the difference of the numbers?
Mamang said:
10 years ago
Please tell me the most simple formula.
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