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place value

by Abdulla » Sat Nov 01, 2008 5:27 pm
What is the sum of all 3-digit numbers that can be constructed using the digits 3,4 and 5, if each digit can used only once in each number?

Answer :

345-354-435-453-534-543 six ways to arrange them which equals 2664.

This is another approach using the knowledge of place value.
100(24)+10(24)+(24)= 2400+240+24=2664.

My question is from where we got 24 ? What I knew is each number appears twice in hundreds, tens, and ones columns.
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by Neo2000 » Sat Nov 01, 2008 6:24 pm
24 = 12x2
You know that each number appears twice in each of the place that's why the 2
Now, when each number appears it contributes a value so when 3 appears twice it contributes a total value of 6, when 4 appears total value of 8 and when 5, a total value of 10
Total value contributed = 6+8+10 = 24

Simply put (Sum of Numbers)x(No:of times they appear)
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by Abdulla » Sat Nov 01, 2008 9:34 pm
Thanks Neo2000
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