Question

The sum of all $4$-digit even numbers formed by the digits $0,1,2,3,5,7$ (without repetition) is

• A. $239760$
• B. $338326$
• C. $447696$
• D. $546086$

Answer 1

The correct option is C $447696$

Different numbers possible with $0$ in the unit's place is
$=5\times 4\times 3=60$

For any digit fixed at ten's place, with $0$ fixed at unit's place, total numbers will be $4\times 3=12$

Same thing will be repeated for hundred's place and thousand's place.

$\therefore$ Sum of numbers with $0$ at unit's place is
$=(1+2+3+5+7)(12\times 1000+12\times 100+12\times 10)+60\times 0=239760$


Now,
Different numbers possible with $2$ in the unit's place is
$=4\times 4\times 3=48$

For any digit (excluding $0$) fixed at ten's place, with $2$ fixed at unit's place, total numbers will be $3\times 3=9$

Same thing for hundred's place.

For thousand's place, total numbers will be $4\times 3=12$

For $0$ fixed at ten's place, hundred's place or thousand place with $2$ fixed at unit's place, total numbers will be $4\times 3=12$

$\therefore$ Sum of numbers with $2$ at unit's place is
$=(1+3+5+7)(12\times 1000+9\times 100+9\times 10)+0(12\times 1000+12\times 100+12\times 10)+48\times 2$

$=207936$

Total sum $=239760+207936=447696$