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:
212 comments Page 10 of 22.
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?
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.
Manjunath said:
1 decade ago
I want the explanation of the logic.
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.
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.
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