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 10 of 21.
Mamang said:
10 years ago
Please tell me the most simple formula.
Manjunath said:
10 years ago
I want the explanation of the logic.
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.
Iegowg said:
1 decade ago
Just divide each option with all figures it is that simple.
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.
Somanath said:
1 decade ago
Find the highest number which divides 25 and 29 equally leaving 5 in each case.
NARESH said:
1 decade ago
Please explain again simply way.
Vandana said:
1 decade ago
But there is that that same remainder in each case.
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.
Phaaani said:
1 decade ago
What could be the value of x if HCF of 12, 24, x is as same as GCD of 18, 36 and x?
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