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:
75
81
85
89
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.

Gopi said:   1 decade ago
In-case of co primes, if 29 is the middle number then first should be 23 and last should be 31. Then the given values not match. Please clarify.

Prathyusha.j said:   1 decade ago
@Gopi.

If we take the numbers to be 23 and 31,

We get 23*29=667.. which is in contrast with the question.

So the numbers should be 19 and 37.

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.
(2)

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.

Ashutosh said:   1 decade ago
Hello friends its simple to get 29.

1 step:-1073-551 we will get 522.
Now,
2 step:- 551-522 we will get 29

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.
(18)

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.

Jyoti. said:   9 years ago
@Sujeeth

1*3 = 3
5*7 = 35
5*0 = 0
1.
= 3 - 35 - 0 - 1.
= 28 - 0 - 1.
= 28 + 1.
= 29.

Ajit said:   8 years ago
Let me explain (551-1071) = 522 then again least no minus (551-522) = 29.

Diksha said:   9 years ago
Why the 29 divided the term. It's middle term but what logic behind this?


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