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 3 of 22.
Narendra Singh said:
4 years ago
Thanks everyone for explaianing.
(4)
Anomie said:
4 years ago
P is called the divisor & not dividend.
(Divisor x quotient) + remainder= dividend.
(Divisor x quotient) + remainder= dividend.
(1)
Ishaq said:
4 years ago
Thanks everyone for explaining.
Darshil said:
5 years ago
Why they subtracted? Explain, please.
Aravind B said:
5 years ago
Thanks all for giving the clear explanation.
Sakthi said:
5 years ago
Why they took difference here? Anyone explain about it.
Nic said:
5 years ago
Same remainder means? Please anyone explain for me.
Tejveer said:
5 years ago
@Karthik V.
Great explanation, Thanks for solving this.
Great explanation, Thanks for solving this.
Karthick v said:
5 years ago
@All.
Here you need the highest common factor of three numbers those leave the same remainder
Let's do with the formula.
(p)q1 + r = 43.
(p)q2 + r = 91.
(p)q3 + r = 183.
we have used the same 'r' because all three leave the same remainder.
we need to find the 'p' value, which is the highest common factor.
to find the 'p' value, we need to remove 'r' first, ---------> (important point).
it seems that we can do it by making a difference between two equations.
Let's do it.
=> (pq2 +r) - (pq1 +r)
=> pq2+r-pq1-r
=> pq2-pq1 ----> now the reminder is taken out.
=> p(q2-q1) -----> need to find p value ----> (91-43)----> 1
Other numbers after diff will be like;
=> p(q3-q2) --------->(183-91)---------------> 2.
=> p(q3-q1) -------->(183-43) ---------------> 3.
Now we can find HCF of these three numbers to find p-value. Because 'p' is common in all numbers.
Hope you got it.
Here you need the highest common factor of three numbers those leave the same remainder
Let's do with the formula.
(p)q1 + r = 43.
(p)q2 + r = 91.
(p)q3 + r = 183.
we have used the same 'r' because all three leave the same remainder.
we need to find the 'p' value, which is the highest common factor.
to find the 'p' value, we need to remove 'r' first, ---------> (important point).
it seems that we can do it by making a difference between two equations.
Let's do it.
=> (pq2 +r) - (pq1 +r)
=> pq2+r-pq1-r
=> pq2-pq1 ----> now the reminder is taken out.
=> p(q2-q1) -----> need to find p value ----> (91-43)----> 1
Other numbers after diff will be like;
=> p(q3-q2) --------->(183-91)---------------> 2.
=> p(q3-q1) -------->(183-43) ---------------> 3.
Now we can find HCF of these three numbers to find p-value. Because 'p' is common in all numbers.
Hope you got it.
(6)
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)
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