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by sanju09 » Tue Feb 24, 2009 4:04 am
What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?

A. 126

B. 1386

C. 3108

D. 308

E. 13986
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Source: — Problem Solving |

by sureshbala » Tue Feb 24, 2009 4:38 am
The formula for this is

r^(n-1) x (Sum of the digits) x (1111...n times) (where n is the number of the digits in the number that is being formed and r is the number of digits being used. In this problem n=r)

3^2 x (1+5+8) x (111) = 13986.

Folks, soon I will come up with a lesson on handling these questions and the logic behind these forumulae so that you can handle the exceptional cases like 0 being used in the digits.
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