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 1 of 6.
Satyajit said:
5 years ago
To find hcf of big no:
1073/551=1,rem=522
Then 551/522=1,rem=29.
Then 522/29=18,rem=0.
So HCF=29.
1073/551=1,rem=522
Then 551/522=1,rem=29.
Then 522/29=18,rem=0.
So HCF=29.
(18)
Aparna said:
1 decade ago
Nothing to worry,
see in HCF we get a number which is common to both numbers.
So find the hcf of 551 and 1073, we get 29
which is one number among three nos.
Let other two numbers be a and b
Given
a*29=551
a=551/29
a=19
Now,
b*29=1073
b=37
Hence three numbers. are 29,19,37
see in HCF we get a number which is common to both numbers.
So find the hcf of 551 and 1073, we get 29
which is one number among three nos.
Let other two numbers be a and b
Given
a*29=551
a=551/29
a=19
Now,
b*29=1073
b=37
Hence three numbers. are 29,19,37
(16)
Nivi said:
5 years ago
Let me explain.
Co-prime : factors are 1 and itself.
Given: XY=551.
YZ=1073.
Since why is a common factor in both the equations and the as the no's are caprice then the other factor becomes the highest common factor.
So, Y is the HCF of XY and YZ.
Co-prime : factors are 1 and itself.
Given: XY=551.
YZ=1073.
Since why is a common factor in both the equations and the as the no's are caprice then the other factor becomes the highest common factor.
So, Y is the HCF of XY and YZ.
(7)
Muhammad safeer ahmed said:
7 years ago
a*b=551
b*c=1073
So, b is =1073-551=522,
522-551=29,
29 is middle and common term exist in both and b =29.
Put the value
a*b=551
a*29=551
a=551/29=19
a=19.
Put the value
b*c=1073
29*c=1073
C=1073/29=37.
So
a=19
b=29
c=37
19+29+37=85.
b*c=1073
So, b is =1073-551=522,
522-551=29,
29 is middle and common term exist in both and b =29.
Put the value
a*b=551
a*29=551
a=551/29=19
a=19.
Put the value
b*c=1073
29*c=1073
C=1073/29=37.
So
a=19
b=29
c=37
19+29+37=85.
(6)
Shreya said:
6 years ago
I don't understand why we take the HCF of 551 and 1073.
(6)
Anvi said:
4 years ago
Why we use HCF ? Explain clearly.
(3)
Shaurya said:
7 years ago
Let the three no be a,b,c.
a*b=551
a*b=19*29
Now
b*c=1073
b*c=29*37
So we get a=19,b=29 and c=37.
On adding a+b+c=19+29+37=85.
a*b=551
a*b=19*29
Now
b*c=1073
b*c=29*37
So we get a=19,b=29 and c=37.
On adding a+b+c=19+29+37=85.
(2)
Riteeka said:
4 years ago
Hey @Satyajit.
Your method is perfect. Helped a lot. Thank you.
Your method is perfect. Helped a lot. Thank you.
(2)
Priya said:
4 years ago
Really Awesome explanation, Thanks @Aparna.
(2)
Priya said:
4 years ago
Really Awesome explanation, Thanks @Aparna.
(2)
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