Question

The addition of 7 distinct positive integers is 1740. What is the largest possible “greatest common divisor” of these 7 distinct positive integers?

• A. 42
• B. 60
• C. 74
• D. 140
• E. None of the above.

Answer 1

The Correct Option is B

To determine the largest possible greatest common divisor (GCD) of 7 distinct positive integers whose sum equals 1740, follow these steps:

1. Consider the sum of the integers \( a_1 + a_2 + ... + a_7 = 1740 \).
2. If these integers share a common divisor \( d \), then they can be expressed as \( a_1 = d \times b_1, a_2 = d \times b_2, ..., a_7 = d \times b_7 \), where \( b_1, b_2, ..., b_7 \) are distinct integers.
3. Consequently, the sum can be rewritten as: \(d \times (b_1 + b_2 + ... + b_7) = 1740\)
4. Thus, \( d \) must be a divisor of 1740. First, we calculate the divisors of 1740 by finding its prime factorization:
5. The prime factorization of 1740 is: \(1740 = 2^2 \times 3 \times 5 \times 29\)
6. From this, we find the possible divisors of 1740. To find the largest GCD, we need the largest divisor that allows the remaining conditions to hold.
   • Divisors of 1740 include: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 29, 30, 58, 60, 87, 116, 145, 174, 290, 348, 435, 580, 870, 1740.
7. Given the options: 42, 60, 74, 140, and None of the above, check to see which can be the GCD:
   • 42: Not a divisor of 1740 (sums of factors do not match).
   • 60: A divisor, and dividing 1740 by 60 gives the sum of distinct integers:
     • If we have \( b_1 + b_2 + ... + b_7 = \frac{1740}{60} = 29 \).
     • Possible distinct integer solution: (1, 2, 3, 4, 5, 6, 8), each multiplied by 60 gives distinct multiples equating to 1740.
   • 74 and 140 do not evenly divide 1740.
8. Thus, the largest GCD possible from the options given is 60.

Therefore, the largest possible value for the greatest common divisor of the seven distinct positive integers is 60. Thus, the answer is \(60\).

Answer 2

To solve this problem, we need to find the greatest common divisor (GCD) of the 7 distinct positive integers whose sum is 1740. Our objective is to maximize this GCD. 

1. Let these integers be \(a_1, a_2, a_3, a_4, a_5, a_6, a_7\), and let their GCD be \(g\). Thus, each \(a_i\) can be expressed as \(a_i = g \times b_i\), where \(b_i\) are distinct positive integers.

2. Substituting the expressions into the sum yields: \(g \times b_1 + g \times b_2 + g \times b_3 + g \times b_4 + g \times b_5 + g \times b_6 + g \times b_7 = 1740\).

3. Factoring out \(g\) gives: \(g \times (b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7) = 1740\), which implies \(g \times S = 1740\) where \(S = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7\).

4. To maximize \(g\), \(S\) should be minimized. Since \(b_i\) are distinct positive integers, the minimum value of \(S\) with 7 terms is achieved with \(b_i = 1, 2, 3, 4, 5, 6, 7\), giving \(S = 28\).

5. If \(S = 28\), then \(g = \frac{1740}{28} = 60\).

6. Therefore, with this setup, the largest possible GCD of these integers is 60.

The correct answer is 60.