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 2 of 6.
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)
Anamika said:
7 years ago
Thanks all for the answer.
(1)
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.
Mouni said:
4 years ago
HCF of 551 and 1073 = 29 + 19 + 27 = 85.
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.
Diksha said:
9 years ago
Why the 29 divided the term. It's middle term but what logic behind this?
Ajit said:
8 years ago
Let me explain (551-1071) = 522 then again least no minus (551-522) = 29.
Divyanshu Priyadarshi said:
8 years ago
Let the three numbers be x, y and z.
Hence, xy=551 and yz=1073.
Middle number =HCF of 551 and 1073 = 29.
First number =551/29= 19.
Third number =1073/29= 37.
Required numbers are 19,29 and 37.
Hence, xy=551 and yz=1073.
Middle number =HCF of 551 and 1073 = 29.
First number =551/29= 19.
Third number =1073/29= 37.
Required numbers are 19,29 and 37.
Surya Deep said:
8 years ago
Consider numbers as X, Y, Z respectively.
Now,
Consider product XY = 551 ------eq1
Consider product YZ = 1073 -----eq2
So, clearly common highest factor btw XY and YZ should be Y
i.e HCF (551,1073) = 29.
From eq 1, X=19.
from eq 2, Z=37.
Now, Add this numbers (X+Y+Z)
(19+29+37)=85 And.
Now,
Consider product XY = 551 ------eq1
Consider product YZ = 1073 -----eq2
So, clearly common highest factor btw XY and YZ should be Y
i.e HCF (551,1073) = 29.
From eq 1, X=19.
from eq 2, Z=37.
Now, Add this numbers (X+Y+Z)
(19+29+37)=85 And.
Asaduzzaman said:
8 years ago
Thanks.
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