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 11 of 21.
Pinku said:
6 years ago
43/4 = 3(remainder).
91/4 = 3(remainder).
183/4 = 3(remainder).
91/4 = 3(remainder).
183/4 = 3(remainder).
Saumya tyagi said:
1 decade ago
Well I also cannot understand the correct logic of solving this question please explain.
Arsh said:
8 years ago
Here, the remainder is unknown that is why we have only chance to find their difference.
Vamsi said:
5 years ago
48 = 4 * 12
92 = 4 * 23
140 = 4 * 35.
1,2,3,5 prime numbers so 4 is the correct answer.
92 = 4 * 23
140 = 4 * 35.
1,2,3,5 prime numbers so 4 is the correct answer.
(1)
Anomie said:
4 years ago
P is called the divisor & not dividend.
(Divisor x quotient) + remainder= dividend.
(Divisor x quotient) + remainder= dividend.
(1)
Anirudh rai said:
2 decades ago
Is this the only way to this type of problem?
If yes then would you please explain it.
If yes then would you please explain it.
Sowjanya said:
10 years ago
In @Prasanth answer, p is the divisor not dividend and the analysis of sum is correct.
Sendilnathan said:
1 decade ago
How to find highest number which divides some numbers and divide the same reminder?
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?
Sharmi said:
1 decade ago
Why you have to divide all the number with 4. What is the reason, please explain?
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