If the greatest common factor (GCF) of two positive integers x and y i

Question

If the greatest common factor (GCF) of two positive integers x and y is greater than 1 and x + y + GCF(x, y) = 91, how many ordered pairs of (x, y) are possible ?

A. 3
B. 4
C. 5
D. 6
E. 8

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TestPrepUnlimited · Answer

Note both x and y contain GCF(x, y), and we can pull it out from  x + y + GCF(x, y) . So we can conclude, GCF(x, y) is a factor of 91. 
Then let's start breaking down the cases:
 GCF(x, y) = 1 . Already denied by the question. If it wasn't denied we would be counting nearly all cases with primes x's or y's such as 43 + 47, 41 + 49 etc.

GCF(x, y) = 7 . Then  x + y = 84  and  \frac{(x + y)}{7} = 12 . We have 12 multiples of seven to distribute among x and y, we can break them down into 1+11 or 5+7 multiples of 7 so that the GCF is 7. Since we have ordered pairs we will include 7 + 5 and 11 + 1 as well.
(e.g. If we break it down into 6+6, this means x = 6*7 = 42 and y = 6*7 = 42, then the GCF is not 7. We want to distribute the 7's among x and y so that the GCF is still 7. To do that the number of 7's distributed between x and y cannot have any common factor besides 1, hence 2 + 10 and 3 + 9 are also not viable).

GCF(x, y) = 13 . Then  x + y = 78  and  \frac{(x + y)}{13} = 6 . We have 6 multiples of thirteen to distribute among x and y, we can break them down into 1+5 only so that the GCF is 13. Include 5 + 1 since x and y are ordered.

GCF(x, y) = 91 . Then  x + y = 0  which isn't possible.

Finally, we only have 3 pairs of integers but they have to be ordered so it is 6 pairs in total. To list them they would be  (1*7, 11*7), (5*7, 7*7), (13, 5*13), (11*7, 7), (7*7, 5*7), (5*13, 13) .

Ans: D

amanjain9311 · Answer

IMO: D 
 
Here's how I solved it:*
x can be written as GCF(x,y)*n1, where n1 is any positive integer
Similarly, y can be written as GCF(x,y)*n2

substituting the above expressions in x+y+GCF(x,y)=91, we get,
GCF(x,y)*n1 + GCF(x,y)*n2 +GCF(x,y)=91
=> GCF(x,y)*(n1 + n2 +1) = 91

The GCD(x,y) should be a factor of 91 i.e 1,7,13 or 91. 
We reject "1" as It's given in the question that the GCD(x,y)>1
and we reject "91" as n1 and n2 can't be 0.

So the final 2 expressions and the ordered pairs we get for n1 and n2 are:

1. 7*(n1 + n2 + 1)=91
=> n1 + n2 =12
so (n1 ,n2) could be (1,11) , (2,10), (3,9) , (4,8) , (5,7) , (6,6), (7,5) , (8,4) , (9,3) , (10,2), (11,1)
i.e 11 values

2. 13(n1 + n2 +1)=91
=> n1 + n2=7
so (n1 ,n2) could be (1,5) , (2,4) , (3,3), (3,2) , (5,1)
i.e. 4 values

Now, since we are taking GCD(x,y) as 7 & 13 there should be no common factor between n1 and n2 i.e n1 and n2 should be coprime. 
The coprime ordered pair of n1 and n2 are:
(1,11), (5,7), (7,5), (11,1), (1,5), (5,1)
6 pairs. So the answer is D

amanjain9311 · Answer

IMO: D 
 
Here's how I solved it:*
x can be written as GCF(x,y)*n1, where n1 is any positive integer
Similarly, y can be written as GCF(x,y)*n2

substituting the above expressions in x+y+GCF(x,y)=91, we get,
GCF(x,y)*n1 + GCF(x,y)*n2 +GCF(x,y)=91
=> GCF(x,y)*(n1 + n2 +1) = 91

The GCD(x,y) should be a factor of 91 i.e 1,7,13 or 91. 
We reject "1" as It's given in the question that the GCD(x,y)>1
and we reject "91" as n1 and n2 can't be 0.

So the final 2 expressions and the ordered pairs we get for n1 and n2 are:

1. 7*(n1 + n2 + 1)=91
=> n1 + n2 =12
so (n1 ,n2) could be (1,11) , (2,10), (3,9) , (4,8) , (5,7) , (6,6), (7,5) , (8,4) , (9,3) , (10,2), (11,1)
i.e 11 values

2. 13(n1 + n2 +1)=91
=> n1 + n2=6
so (n1 ,n2) could be (1,5) , (2,4) , (3,3) , (4,2), (5,1)
i.e. 4 values

Now, since we are taking GCD(x,y) as 7 & 13 there should be no common factor between n1 and n2 i.e n1 and n2 should be coprime. 
The coprime ordered pair of n1 and n2 are:
(1,11), (5,7), (7,5), (11,1), (1,5), (5,1)
6 pair. So the answer is D

GaurSaini · Answer

We are given that GCD(x,y) > 1. Hence, x and y have integer greater than 1 in common.

Let's say x = ma and y = mb
GCD(ma,mb) = m
and a and b are co-prime numbers.

Now, x+y+GCD(x,y) = 91
i.e. ma+mb+m = 91
 m(a+b+1) = 91
 
91 can be factorized as following, 
 1 * 91, 7*13

Now m>1 amd a+b+1 > 1
Hence, only possibility is 7*13

Therefore, 
 m(a+b+1) = 7*13 OR 13*7

when m = 7, a+b+1=13 => a+b = 12 
Now a and b are 'CO-PRIME' and statement has mentioned 'ORDERED pairs', Hence values possible for a+b = 1+11, 5+7, 7+5, 11+1
Hence, (1,11), (5,7), (7,5), (11,1)

when m = 13, a+b+1=7 => a+b=6
Values possible for (a,b) = (1,5), (5,1)

So, in total 6 ordered pairs are possible.

KarishmaB · Answer

Say the GCF of the two numbers is g. Then the two numbers are ga and gb (where a and b are co-prime i.e. they can have no common factors).

ga + gb + g = 91
g (a + b + 1) = 7*13

Hence, g can be 7 or 13.

If g = 7, a+b = 12. Then a = 1, b = 11; a = 5, b = 7; a = 7, b = 5; a = 11, b = 1. This gives us 4 ordered pairs.
If g = 13, a + b = 6. Then a = 1, b = 5; a = 5, b = 1. This gives us 2 ordered pairs.

Answer (D)

hangthanhtran0911 · Answer

thanks for your clear and helpful explanation. I have a doubt about how could you finish this sentence for approx 2m. when trying this for 2nd time, I need about 3-4m to solve this, so I want to hear your advice. thank you

Kinshook · Answer

If the greatest common factor (GCF) of two positive integers x and y is greater than 1 and x + y + GCF(x, y) = 91, how many ordered pairs of (x, y) are possible ?

x + y + GCF(x,y) = 91
x + y + 1 < 91 & x>0; y>0
0< x + y < 90

Case 1: GCF(x,y) = 2
x + y = 89; Not feasible since both x& y are not even

Case 2: GCF(x,y) = 3
x + y = 88
3(k + m) = 88; Not feasible since 88 is not a multiple of 3

Case 3: GCF(x,y) = 5
x + y = 86
5(k + m) = 86; Not feasible since 86 is not a multiple of 5

Case 4: GCF(x,y) = 7
x + y = 84
7(k + m) = 84; k + m = 12
(k,m) = {(1,11),(4,7),(5,6)}
(x,y) = {(7,77),(28,49),(35,42)}

Case 5: GCF(x,y) = 11
x + y = 80
11(k + m) = 80; Not feasible since 80 is not a multiple of 11

Case 6: GCF(x,y) = 13
x + y = 78
13(k + m) = 78;
k + m = 6
(k, m) = (1,5)
(x,y) = (13,65)

Case 7: GCF(x,y) = 17
x + y = 74
17(k + m) = 74; Not feasible since 74 is not a multiple of 17

Case 8: GCF(x,y) = 19
x + y = 72
19(k + m) = 72; Not feasible since 72 is not a multiple of 19

Case 9: GCF(x,y) = 23
x + y = 68
23(k + m) = 68; Not feasible since 68 is not a multiple of 23

Case 10: GCF(x,y) = 29
x + y = 62
29(k + m) = 62; Not feasible since 62 is not a multiple of 29

Case 11: GCF(x,y) = 31
x + y = 60
31(k + m) = 60; Not feasible since 60 is not a multiple of 31

(x,y) = {(7,77),(28,49),(35,42),(13,65)} : 4 possibilities

IMO B

KarishmaB · Answer

This is a challenging question. One gets to see challenging questions when one is doing well. Usually such people solve the easy/medium questions within a minute each and hence have some extra time on hand. Also, getting a challenging question wrong doesn't lead to much penalty. What it does do sometimes is derail the rest of your exam if you are not mindful of the time. GMAT tests your time management skills, your ability to keep the big picture in mind, not just aptitude.

Hence, never worry too much about hard questions. If the software gives one to you, give it your best in 2-3 mins. If unable to solve, bookmark and move on. If you get time later, come back to it. Else ignore.

If the greatest common factor (GCF) of two positive integers x and y i

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