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.
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