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What is the sum of all 3 digit positive integers that can be formed us

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mbaiseasy
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   20 Dec 2012, 04:15
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iwillwin wrote:
What is the sum of all 3 digit positive numbers 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


Here is a formula to know the sum of possible arrangements when a digit is not allowed to repeat:

\((n-1)!*sumofdigits*111 = (3-1)!*(1+5+8)*111=28*111=3108\)

But we know that digits are allowed to repeat. Thus, sum is much greater than 3108.

Answer: E
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   04 Jan 2013, 06:50
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I approach this particular problem without the formulae. Can somebody please help me if this is correct --> If you know that the numbers are allowed to repeat then the possible numbers are 3*3*3 = 27 (instead of 3*2*1 when repetition is not allowed), then you know that there will be 9 ones, 9 fives, 9 eights. So for the first position you can have the 9+45+72 = 12600, then all the answer choices will fall except for E. If you calculate further you get 12600 + 01260 + 00126 = 13,986. Bunuel, Karishma or someone else can you please confirm if this is correct?
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   29 Jun 2009, 22:30
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maliyeci wrote:
There can be 27 numbers. Every digit has three possibilities because question permits repetition. So it becomes a very good summation problem.
If there are 27 digits. There are 27 hundred digits, 27 tens digit and 27 ones digit to be summed. Of 27 hundred digits, 9 of them are 1, 9 of them are 5 and the last 9 are 8. Same is true for the other two digits.
Thus the sum is:
9x100+9x500+9x800+9x10+9x50+9x80+9x1+9x5+9x8=
999x(1+5+8)=999x14=13986
E ;)


Now these are my early days here. I even have some problems to use this site :) When I saw this squestion, I thought how is it possible to do that, to add twenty seven numbers, but thanks to maliyeci!!

I learned a new approach today to add numbers!! +1 :)
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   14 Jul 2009, 09:15
walker wrote:
maliyeci, good aproach
+1

My take:
As we have equal probability for each digit to be included and total number of integers is 3^3=27, we can write our sum as:

S = 27 * (1+5+8)/3 * 111 = 14*999 = 14000 - 14 = 13986



Could you please explain how did you get 111?

Thanks.
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   14 Jul 2009, 13:27
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skim wrote:
Could you please explain how did you get 111?

Thanks.


Of course,

(1+5+8)/3 - "average" digit.
(1+5+8)/3 * 111 - another way to write 3-digit number formed from "average digit": xyz = (1+5+8)/3 (1+5+8)/3 (1+5+8)/3 or (1+5+8)/3 * 111
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   30 Dec 2009, 04:36
xcusemeplz2009 wrote:
there will 9 times 1 , 9 times 5 and nine times 8 at each place in a 3 digit no. with repetition

sum will be =100*9*(1+58) + 10*9*(1+5+8) + 1*9*(1+5+8)=13986

in order to know how it is 9 times

tot no. of ways =3*3*3=27
so in tot 27 words will be formed where each digit among 1,5,8 wil be repeated equal no. of times i.e 27/3=9
hence we have 9 , 1's ; 9 5's and 9 8's at each level


Wow. I have no idea what is happening here. Would you (or someone else) mind explaining a bit more...perhaps expounding on your method a little bit?

Perhaps, the same deal, with a two digit number using 1 and 2, with digits allowed to repeat?

[Answer: 11+12+21+22 = 66]

How can I use xcusemeplz2009's method to arrive at the same answer?
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   01 Oct 2013, 18:44
aismirnov wrote:
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?

Imagine, we have got all these possible numbers written down - there are in total 3^3 numbers (each digit can be either 1 or 5 or 8)

there are 3*3 options for having a number XY1
there are 3*3 options for having a number XY5
there are 3*3 options for having a number XY8

there are 3*3 options for having a number X1Z
there are 3*3 options for having a number X5Z
there are 3*3 options for having a number X8Z

there are 3*3 options for having a number 1YZ
there are 3*3 options for having a number 5YZ
there are 3*3 options for having a number 8YZ

we can sum units, tens and hundreds independently:
summing units gives (1+5+8)*3*3
summing tens gives (1+5+8)*10*3*3
summing hundreds gives (1+5+8)*100*3*3



How did you know how to do this? I mean, how did you learn? I have 4 weeks to go until my GMAT and I haven't gotten any better at these. I have no idea how to even begin approaching these problems, and none of these formulas make any sense to me.
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Re: What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   15 Nov 2019, 19:05
asimov wrote:
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



Method 1: Sum Each Column of Digits



We need positive integers having 3 digits.

S = __ __ __

We can make 3*3*3 = 27 such positive integers since we can fill in each of the 3 spaces in 3 ways. Now imagine writing these 27 numbers one below the other to add.

158
185
...
...
x 27 combinations

When we add them, noticing the symmetry we know that there will be 9 1's in units digits, 9 5's and 9 8's. So units digits will add up to (1+5+8)*9.

Similarly, tens digits will add up (1+5+8)*9*10.
Similarly, hundreds digits will add up (1+5+8)*9*100

Adding all of them:
(1+5+8)*9 + (1+5+8)*9*10 + (1+5+8)*9*100 = (1+5+8) * 9 * (1+10+100) = 14 * 9 * 111 = 13,986


Method 2: Direct Formula



Sum of all n digit numbers formed by n non-zero digits with repetition being allowed is:

n^(n−1)∗(sum of the digits)∗(111... n times)

9 * 14 * 111 = 13,986

Answer: E
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What is the sum of all 3 digit positive integers that can be formed us [#permalink] New post   04 Sep 2022, 07:57
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asimov wrote:
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


Let's first calculate how many 3-digit numbers can be created using only the digits 1, 5 and 8.
There are 3 ways to choose a hundreds digit.
There are 3 ways to choose a tens digit.
There are 3 ways to choose a ones digit.
So, the total number of ways to create a three digit number = (3)(3)(3) = 27

Now let's focus on the hundreds digits of our 27 numbers.
9 (aka 1/3) of the 27 numbers will have hundreds digit 1. 9 x 100 = 900
9 (aka 1/3) of the 27 numbers will have hundreds digit 5. 9 x 500 = 4500
9 (aka 1/3) of the 27 numbers will have hundreds digit 8. 9 x 800 = 7200
900 + 4500 + 7200 = 12,600

So, if we ignore the tens and ones digits, the sum of our 27 numbers is already 12,600, which means the TOTAL sum must be greater than 12,600

Answer: E
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