Xavier Aptitude Test 2023 Paper

Let ' $\mathit{m}$ ' be the Greatest common divisor and be the 7 unique numbers.
When 'm' divides , we get seven unique quotients. Let the quotients be $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f}$, and g . It is given,
$\mathit{A}+\mathit{B}+\mathit{C}+\mathit{D}+\mathit{E}+\mathit{F}+\mathit{G}=1740$
$\mathit{m}\left(\mathit{a}+\mathit{b}+\mathit{c}+\mathit{d}+\mathit{e}+\mathit{f}+\mathit{g}\right)=1740$
$1740=2^2\times 3\times 5\times 29$

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 $60\left(1+2+3....+6+8\right)=1740$
Largest possible G.C.D is 60 .
The answer is option B.