Fractions, Decimals & Percents: Chapter 1

[mazhar.hussain]

Fractions, Decimals & Percents: Chapter 1 Problem Set: Q2

What is the sum of all the possible 3-digit numbers taht can be constructed using the digits 3,4, & 5, if each digit can be used only once in each number??
Hi guys,

I actually solved this problem manually by summing up the six possibilities. In the solution, however, a shorcut is given in which the integer 24 is used. I couldn't understand the logic behind this one.

Can anyone kindly explain the detailed reasoning???

[ayashlaha]

Hi Mazhar,

The sum of the 3 digits is 3+4+5 = 12
Since each digit occupies a position 2 times, we double that number to get 24.

then we use the shortcut
100(24) + 10(24) + 1(24)

Alternately you could write it out as:
100(3+3+4+4+5+5) + 10(3+3+4+4+5+5) + 1(3+3+4+4+5+5)

Hope this helps.

Amar.

[ibrahim.alfawzan]

even though the sum is easy in this one .. there is another way ..

you have 6 posibileties for in each digit (3,3,4,4,5,5) and there sum is 24

so

24 in the units ,, we add the 2 to the tens and leave the 4

24 in the tens (+ the 2 we added from the units = 26) ,, leave the 6 and move the 2 to the hundreds

24 in the hundreds (+ the 2 we added from the tens = 26)

the final no. 2664

[Ben Ku]

The two responses are correct. I think when you listed the possibilities, you can see that each digit shows up twice in each place value. The sum of each digit showing up twice (3+3+4+4+5+5) is 24.

Hope that helps.

[jerly_vivek]

For such questions there is a generic formula also:
(Sum of n numbers)*(n-1)! * [10^n-1]/[10-1]

i applied on above question
(24)*(2)*111 = 2664

[esledge]

Vivek, I've never seen that formula--it's interesting, and I want to spend some more time thinking about why it works. Does it require that the shuffled terms (integers) be consecutive?