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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 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
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Re: sum of 3 digit #s [#permalink]

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New post 29 Apr 2009, 06:24
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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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New post 29 Apr 2009, 13:26
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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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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New post 03 May 2009, 14:12
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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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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New post 09 Aug 2010, 14:21
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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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New post 09 Aug 2010, 14:44
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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New post 09 Aug 2010, 21:33
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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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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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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can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
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What is the sum of all 3 digit positive integers that can be [#permalink]

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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 [#permalink]

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New post 04 Jan 2013, 06:50
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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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 :lol:

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. :lol:

You are left with answer E.

----------------
consider giving a +kudo if this helps :-D

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

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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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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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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: sum of 3 digit #s [#permalink]

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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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