Question

If 4 digit numbers are formed by using the digits 1, 2, 3, 6, 8, 9, any digit being repeated any number of times, then the sum of the numbers so formed is

• A. (1+2+3+6+8+9).$\angle 3$.1111
• B. (1+2+3+6+8+9).6³. 1111
• C. (1+2+3+6+8+9).3⁶. 1111
• D. None of these

Answer 1

Answer: (B)

(1+2+3+6+8+9).6³. 1111

Answer 2

Sum of digits in the unit’s place of all the number.

= (1 + 2 + 3 + 6 + 8 + 9) 6³

($\because$When one particular digit is fixed in unit’s place, each of the remaining three places can be filled in 6 ways).

Similarly, the sum of digits in any other place is also the same. Hence, the sum of all the numbers.

= (1+2+3+6+8+9) x 6³ x (1+10+100+1000).

Alternatively, using the formula no. 26 given in the competition window, the required sum.

= (1+2+3+6+8+9) x 6³ x 111.