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. 22)
22.
Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
Answer: Option
Explanation:
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
| First number = |  |
551 |  |
= 19; Third number = | |
1073 | |
= 37. |
| 29 | 29 |
Required sum = (19 + 29 + 37) = 85.
Discussion:
53 comments Page 3 of 6.
Mahi said:
9 years ago
Good and clear explanation @Ashutosh.
Krishna said:
9 years ago
@Jyoti, Can you explain it clearly?
Rudrika said:
9 years ago
@Ashutosh,
Your explanation is very nice.
Your explanation is very nice.
Anant said:
9 years ago
Thanks @Ashutosh.
Ranjith said:
9 years ago
xy = 551 & yz = 1073; clearly y is a HCF.
Multiply xy * yz = 1073 * 551 we get xzy^2 = 29sqroot703;
So 29 is y; then it is easy to proceed .
Multiply xy * yz = 1073 * 551 we get xzy^2 = 29sqroot703;
So 29 is y; then it is easy to proceed .
(1)
Preethi said:
9 years ago
I can't understand how 29 comes?
Ram said:
9 years ago
29 is the HCF of 551 and 1073.
To find HCF of 551 and 1073 follow these steps.
551)1071(1
551
----------------
522)551(1
522
------------
29)522(18
29
---------
232
232
------------
XXX
So,HCF of 551 and 1073 is 29.
To find HCF of 551 and 1073 follow these steps.
551)1071(1
551
----------------
522)551(1
522
------------
29)522(18
29
---------
232
232
------------
XXX
So,HCF of 551 and 1073 is 29.
Anonymous said:
9 years ago
How will the solution change if the given numbers are not co-prime?
Tanvir said:
9 years ago
Which least number should be added to 2497 so that the sum is exactly divisible by 3, 4, 5, and 6?
Solve it and find the answer.
Solve it and find the answer.
Purva said:
9 years ago
@Tanvir.
2497/60 we get remainder as 37.
Now: LCM of 5, 6, 4, 3 is 60.
(60 - 37) = 23.
2497/60 we get remainder as 37.
Now: LCM of 5, 6, 4, 3 is 60.
(60 - 37) = 23.
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