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OG no.176

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Master | Next Rank: 500 Posts
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by magical cook » Sun Jul 30, 2006 7:08 pm
Could anyone please explain why the answer for below is 66,660?
Thanks in advance!
Jane


1234
1342
1324
・
・
・
the addition problem above shows four of the 24 different integers that can be formed by each of the 1,2,3 and 4 exactly once in each integer. what is the sum of 24 integers?
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Source: — Problem Solving |
Junior | Next Rank: 30 Posts
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The Answer

by achal.kumble » Sun Jul 30, 2006 9:50 pm
Using the digits 1,2,3 & 4, 24 four digit numbers can be formed.
Out of these 24 numbers in 6(24/4) numbers each digit occurs once in the thousands place, once in the hundreds place, once in the tens place and once in the units place.
Hence, the digit 1 occurs 6 times in the 1000's place, 6 times in the 100's place, 6 times in the 10's place and 6 times in the units place.
So will the other digits.
Let's start with the digit 1:
The sum of oll the numbers with 1 occuring in ANY place will be 1111*6
Similarly for 2 it will be 2222*6
And so on for 3 & 4.
Hence the sum of all the numbers will be (1111+2222+3333+4444)*6=66660.
Hope this helps!
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Master | Next Rank: 500 Posts
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Thanks!!!

by magical cook » Sun Jul 30, 2006 11:35 pm
Hi achal,

Thanks so much for your help! (I wish I was smart like u!!)

Again, thank you for your great help!

Jane
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by gmatjedi » Tue May 25, 2010 4:09 pm
another way to solve:

primary principle:

sum=avg*numbers

lowest number =1234
maximal number=4321
number of integers using 4, 3, 2, 1= 4*3*2*1=24

sum= [(1234+4321)/2]*24
=12*5555
=66660
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