The sum of two positive integers is 156 and their highest common facto

Question

The sum of two positive integers is 156 and their highest common factor is 13. How many pairs of such numbers are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

GMATinsight · Accepted Answer

First number = 13a
Second number = 13b  
 where and b are co-prime (relatively Prime) i.e. HCF of a and b = 1

13a+13b = 156
i.e. a+b = 12

i.e (a,b) =  (1, 11)   (2, 10)   (3, 9)   (4, 8)   (5, 7)   (6, 6)

i.e. only two such pairs

Answer: Option B

chipsy · Answer

65 = 5*13
 91 = 7*13

143= 13*11
13 = 1*13

HCF in both cases = 13
Sum : 65+91 = 143+13 = 156
Only 2 pairs are possible

Option : B

Nipungupta9081 · Answer

65 = 5*13
91 = 7*13

HCF = 13
Sum : 65+91 = 156
Only 1 pair is possible

Option : A

GMATinsight · Answer

Copying someone else's explanation is not a very good practice. I will encourage you to write your own explanation which will improve your understanding, accuracy and content richness of the GC community.

uchihaitachi · Answer

Two numbers have the common factor 13, which means that there is no other common factor between the two numbers other than 13.
So let the two numbers be 13A, 13B

13(A+B) = 156 
 A+B = 12

Now, as explained above A and B will be  co-primes (have no common factor other than 1)
So, the possible values for A and B are (1,11) and (5,7).

The number of pairs possible = 2

OA, B

rajatchopra1994 · Answer

Explanation:
As , 13 is the highest common factor of two number and sum of 2 is 156,
Therefore, 13x+13y = 156
13(x+y) = 156 ------ 1
x+y = 12 ( where x & y has to be co-primes)
 
Now we have to check values for x & y.
if x = 1, then y = 11 ; put in 1, so sum is 156.
If x= 5 , then y = 7 ; put in 1, so sum is 156.

So only 2 pairs are possible.

IMO-B

HouseStark · Answer

156=13*12

13+13*11= hcf(13) 
13*2+13*10=hcf(26)
13*3+13*9=hcf(39)
13*4+13*8=hcf(52)
 13*5+13*7=hcf(13) 
13*6+13*6=hcf(78)

Only those bold written has 13 hcf

Therefore only 2 pairs possible 
 Answer B

Posted from my mobile device

Kinshook · Answer

Given: The sum of two positive integers is 156 and their highest common factor is 13. 
Asked: How many pairs of such numbers are possible?

Let the two positive integers be x & y.

x + y = 156
HCF(x,y) = 13

x = 13k
y = 13m
where k & m are co-prime.

x + y = 13(k+m) = 156
k + m = 12

(k,m) = {(1,11),(5,7)}

Only 2 pairs (13, 143) & (65,91) are possible.

IMO B

boat32674 · Answer

How many pairs of such numbers are possible?

why does the answer not 4 ?

(a,b) = (1, 11) (5, 7) (7, 5) (11, 1)

Bunuel · Answer

Because (1, 11) and (11, 1) give the same pair, and (5, 7) and (7, 5) also give the same pair.

The sum of two positive integers is 156 and their highest common facto

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