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

Let 'm' be the Greatest common divisor and A, B, C, D, E, F & G be the 7 unique numbers.

When 'm' divides A, B, ......& G, we get seven unique quotients. Let the quotients be a, b, c, d, e, f, and g.

It is given,

A + B + C + D + E + F + G = 1740

m(a + b + c + d + e + f + g) = 1740

$$\ \ 1740=2^2\times3\times5\times29$$

m(a + b + c + d + e + f + g) = $$\ \ 1740=2^2\times3\times5\times29$$

29 is a prime number and cannot be factorised further.

Therefore, largest possible value m can take is 60.

Least possible sum of 7 unique numbers is 1 + 2 + 3 +...+ 7 = 28

Replacing 7 with 8, we will get 29.

This implies 60(1 + 2 + 3 .....+ 6 + 8) = 1740

Largest possible G.C.D is 60.

The answer is option B.