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

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What is the sum of all 3 digit positive integers that can be [#permalink]

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29 Apr 2009, 00:06
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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
[Reveal] Spoiler: OA
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Re: sum of 3 digit #s [#permalink]

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29 Apr 2009, 05:24
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E

summing
units (1+5+8)*9 +
tens (1+5+8)*9*10 +
hundreds (1+5+8)*9*100 =

= 126+1,260+12,600 = 13,986
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Re: sum of 3 digit #s [#permalink]

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29 Apr 2009, 12:26
hi aismirnov, can you elaborated on your explanation a bit. I'm a bit weak with these types of problems. for example, what did you use tens (1+5+8)*9*10, etc.

tia
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Re: sum of 3 digit #s [#permalink]

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29 Apr 2009, 23:24
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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
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Re: sum of 3 digit #s [#permalink]

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03 May 2009, 13:12
The OA is E as well.

Thanks I understand it now.
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Re: sum of 3 digit #s [#permalink]

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14 Oct 2009, 17:47
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I don't have your specific method but by POE I still can have E.
for 8xy alone we have 888,881,885,818,855,851,858,815,811. The total of them is larger than 7200 so there is only option E left
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Re: sum of 3 digit #s [#permalink]

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09 Aug 2010, 13:21
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but what if the digits repeat within a number??
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Re: sum of 3 digit #s [#permalink]

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09 Aug 2010, 13:44
I also reached by POE. Can someone please explain an easier way to get to the correct answer.
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Re: sum of 3 digit #s [#permalink]

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09 Aug 2010, 20:33
this is how i approached

888
588
188
158
518
118
558
These are the number that can have 8 as the units digit. here, 8(7) = 56 so units dig is 6
similarly, for 5 as the units dig - 5(7) = 35 so units dig is 5
for 1 as the units dig - 1(7) = 7 as the unit dig.
therfore, in the sum of these dig, the units dig will be 6+5+7 = 8

Please tell me where i am wrong
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Re: sum of 3 digit #s [#permalink]

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02 Feb 2011, 02:15
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I sowe really good formula for solving this problem in some notes downloaded from this forum. I just cannot find it, so I appologize to the author.

The formula says:

Repetition allowed:
SUM of digits * (n^n-1)*(11111 ...number composed of n 1digits)

Repetition NOT allowed:
SUM of digits * (n-1)!*(11111 ...number composed of n 1digits)

Here we have 3 digits. n is 3.
Sum of digits 1+5+8=14

Repetition allowed:
14*(3^2)*111=13986

Repetition not allowed:
14*2*111=3108
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Re: sum of 3 digit #s [#permalink]

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02 Feb 2011, 02:37
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craky wrote:
I sowe really good formula for solving this problem in some notes downloaded from this forum. I just cannot find it, so I appologize to the author.

The formula says:

Repetition allowed:
SUM of digits * (n^n-1)*(11111 ...number composed of n 1digits)

Repetition NOT allowed:
SUM of digits * (n-1)!*(11111 ...number composed of n 1digits)

Here we have 3 digits. n is 3.
Sum of digits 1+5+8=14

Repetition allowed:
14*(3^2)*111=13986

Repetition not allowed:
14*2*111=3108

It should be:

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

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What is the sum of all 3 digit positive integers that can be [#permalink]

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20 Dec 2012, 03:15
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.

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Re: What is the sum of all 3 digit positive integers that can be [#permalink]

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04 Jan 2013, 05:50
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 [#permalink]

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04 Jan 2013, 06:43
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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

One quickest way to answer this question !!

As we are using digits 1,5, 8 and digits are allowed to repeat. Each of the unit, tenth and hundredth digit can be used by each of three digits.
So, Total possible numbers with these digits=3 X 3 X 3 =27.

First, As we have 27 three digit number, Sum will be for sure more than 2700.. Eliminate options A,B,D

Second, If you imagine numbers with the given digits 1,5,8. We have numbers like 888,885,855,858,851. Sum is for sure more than 4000. Eliminate option C.

You are left with answer E.

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Re: What is the sum of all 3 digit positive integers that can be [#permalink]

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05 Jan 2013, 08:58
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1, 5 and 8 are allowed to be used 9 times as hundreds, tenths and units digit.

So you can line up:

9x100
9x 10
9x 1
9x500
9x 50
9x 5
9x800
9x 80
9x 8

When lining these up, you should quickly realize that it's bigger than 10.000 and pick your answer without going further.
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Re: What is the sum of all 3 digit positive integers that can be [#permalink]

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05 Jan 2013, 09:40
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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

Answer to this question is easier to guess than to calculate. e.g. if we take 8 at hundreds place we would get at least 9 nos. So 800 * 9 = 7200 which surpasses every option but E.

Thru conventional method
(1+5+8)9 = 126
(1+5+8)9*10=1260
(1+5+8)9*100=12600

126 + 1260 + 12600 = 13896. E
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Re: sum of 3 digit #s [#permalink]

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01 Oct 2013, 17:44
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: sum of 3 digit #s [#permalink]

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02 Oct 2013, 02:24
Expert's post
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AccipiterQ wrote:
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.

Direct formulas are here: what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html#p862674 Please ask if anything there is unclear.

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Hope this helps.
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