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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.

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

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

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

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: \(n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)\).

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?

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

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04 Jan 2013, 06:43

4

This post received KUDOS

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.

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

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.

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.

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.

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

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08 Oct 2014, 15:30

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

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11 Oct 2015, 13:54

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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