Consider all positive integers a, b, and c such that

Question

Consider all positive integers a, b, and c such that 135 is a factor of each of the expressions 15a, 25b and 30c. What is the greatest common factor of all possible sums a + b + c ?

A. 3
B. 5
C. 9
D. 15
E. 27

Bunuel · Accepted Answer

Factor 135:  135 = 3^3 * 5

3^3 * 5  to be a factor of  15a = 3 * 5 * a ,  a  must be a multiple of  3^2 ;
 3^3 * 5  to be a factor of  25b = 5^2 * b ,  b  must be a multiple of  3^3 ;
 3^3 * 5  to be a factor of  30c = 2 * 3 * 5 * c ,  c  must be a multiple of  3^2 ;

Thus,  a + b + c = 3^2x + 3^3y + 3^2z = 3^2(x + 3y + z) . Since x + 3y + z can take any integer value greater than or equal to 5 (the least value of x + 3y + z is 5 when x = y = z = 1), the possible values of a + b + c are:

9 * 5; 
9 * 6; 
9 * 7;
...

Therefore, the greatest common factor of all possible sums a + b + c will be 9.

Answer: C.

Bunuel · Answer

Can you please post a screenshot of that question from GMAT Prep Focus? Thank you!

GMATinsight · Answer

135 is a factor of each of the expressions 15a, 25b and 30c

135 =  15*9 = 3^3*5

135 is a factor of 15a (3*5*a) i.e. a must have 3^2

135 is a factor of 25b (5^2*b) i.e. b must have 3^3

135 is a factor of 30c (3*5*2*c) i.e. c must have 3^2

a_{min} = 9 
 b_{min} = 27 
 c_{min} = 9

i.e.  (a+b+c)_{min} = 9+27+9 = 45 = 9*5

i.e. all values of a+b+c will be multiple of 9 as a, b and c ae all multiples of 9

hence GCD of all values of (a+b+c) will be 9

Answer: Option C

nbharti · Answer

15a, 25b and 30c , One quick question: How do we interpret 15a as 15*a and not An integer with 15 as the starting two digits?

Bunuel · Answer

If "15a" were a three-digit number, it would have been explicitly mentioned. Without such clarification, "15a" can only represent "15 * a", as the multiplication sign (*) is typically omitted.

KarishmaB · Answer

135 = 5 * 3^3

15a = 3*5a (a must be at least 9 and could have another factor k) 
 25b = 5*5b (b must be at least 27 and could have another factor m) 
 30c = 3*5*2c (c must be at least 9 and could have another factor n)

a + b + c = 9k + 27m + 9n = 9(k + 3m + n)

We need the GCD of all possible values of 9(k + 3m + n). We know that all possible values will have 9 as a factor.

Say k = 1, m = 1, n = 1 
  a + b + c = 9 * 5

Say k = 2, m = 1, n = 1
 a + b + c = 9 * 6 
 
Only 9 is common in these two instances and it will always be common. Hence GCD is 9.

Answer (C)

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