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find the sum of aall possible four digit numbers that can be formed usind the digits 2,4,5,6,7 and 8 with no repeated digits.

endurance

by endurance at September 18, 2011

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Choosing a number (I'll choose 2), you have 5*4*3=60 numbers which have that number as the first digit. Likewise, there are also 60 numbers with that number as the second, third, and fourth digit. This is true for all six numbers. Every number that starts with is 2000+some amount, second digit is 200+ some amount, etc., so the total of all of the digits represented by a 2 is 60*2*1,000+60*2*100+60*2+10+60*2*1=60*2*(1111). Using all of the digits Yields 60*2*1111+60*4*1111+... =60*1111*(2+4+5+6+7+8) =66,660*32 =2,133,120. So your final answer is 2,133,120. There's a good chance that there's an easier way to do this, but I'm confident that this works.

MathiusTheGreat MathiusTheGreat September 18, 2011

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