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

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

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New post 04 Jan 2013, 07:43
8
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.

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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 formed us  [#permalink]

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New post 05 Jan 2013, 09:58
1
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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What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 05 Jan 2013, 10:40
5
3
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 = 13986. E
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#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: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 02 Oct 2013, 03:24
1
4
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.

Similar questions to practice:
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find-the-sum-of-all-the-four-digit-numbers-which-are-formed-88357.html
find-the-sum-of-all-3-digit-nos-that-can-be-formed-by-88864.html
if-the-three-unique-positive-digits-a-b-and-c-are-arranged-143836.html
what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html
what-is-the-sum-of-all-4-digit-numbers-that-can-be-formed-94836.html
the-sum-of-the-digits-of-64-279-what-is-the-141460.html
there-are-24-different-four-digit-integers-than-can-be-141891.html
the-addition-problem-above-shows-four-of-the-24-different-in-104166.html

Hope this helps.
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 31 Oct 2015, 08:31
1
8
The Ans is E.

This can also be solved by using a formula

Sum of N numbers

With repetition = N ^ ( N - 1 ) * Sum of the Numbers * 111..... N no. of times

With out repetition = ( N - 1 ) ! * Sum of the Numbers * 111.... N no. of times.

Here repetition is allowed, therefore N is 3 ( No. of digits given ) and sum of No is 14.

3^2 * 14 * 111 = 13986.
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 27 Nov 2015, 00:06
3
Used POE, No need to solve the whole question to get an exact value
Explanation:

since three digit numbers formed by 1, 5, 8 would be :
Lets start with numbers starting with 8 : 888, 885, 881, 855, 851, 815, 811
sum of these numbers is greater than Options A , B , C , D
Hence Ans: E.

Keep it simple ppl

Good luck!!
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 28 Nov 2015, 20:45
6
Hi All,

There's a great 'pattern-matching' shortcut built into this question that can help you to avoid much of the 'math work' involved.

We're told to use the digits 1, 5 and 8 to form every possible 3-digit number (including those with duplicate digits) and then take the sum of those numbers.

Since the digits can be repeated, we're dealing with the numbers that fall into the range of 111 to 888, inclusive. There are (3)(3)(3) = 27 total numbers and 1/3 of those numbers will begin with an 8. From THAT deduction, we know that the sum of those 9 numbers will be greater than (9)(800) = 7200. There's only one answer that fits that description...

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

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New post 28 Dec 2015, 14:26
1
With 1 in the hundreds place, there are 9 numbers that could be created, since repetition is allowed. So 1 occurs 9 times in hundreds place = 900
Similarly it occurs 9 times in tens place = 90
and 9 times in ones place = 9

999 * 1 = 999

Similarly for 5 and 8

999 ( 1 + 5 + 8) = 999 * 14 = (1000 - 1) * 14 = 14,000 - 14 = 13,986
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 04 Jan 2016, 15:17
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(1+5+8) * 9 = 126
(1+5+8) * 9 * 10 = 1260
(1+5+8) * 9 * 10 * 10 =12600

add 126+1260+12600= 13,986
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 23 Aug 2016, 20:52
2
R2I4D 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



I took a crude approach, the smallest number is 111 and largest is 888 and the number of integers are 3 * 3 * 3 = 27.
the smallest is 111 and largest is 888, so the mean is around 500, so 500*27 = 13,500.
E is the answer.


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

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New post 24 Aug 2016, 19:51
1
sdrandom1 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



There are in total 3 * 3 * 3 = 27 numbers in these combinations. The smallest is 111 and largest is 888, so the avg is approx, 500.
So the sum is approx 27 * 500 = 13,500 , close to E.
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 31 Aug 2016, 22:16
1
2
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

Before solving, I knew that the number will be big but did not know how big.
As I did not know a fancy way of solving this, I just tried to find the number if the digits are NOT allowed to repeat.
Then it would be,

158
185
518
581
815
851

And just by rough estimation, it looked like it will be in 3000 range. As the question stem says digits ARE allowed to repeat, meaning,
there should be more than 3000. The only answer more than 3000 was E.
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What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 13 May 2019, 18:29
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


All these big formulas that people have used to solve this question are super impressive. However, it is much easier to answer this question using common sense and leveraging the answer choices.

If you simply write out the possibilities of three digit integers that would begin with an eight, such as 888, 818, 851, 815, 885, 858 (not sure if I'm missing some here), you'll see that the sum of just the 8s is greater than ~4800. Logically, therefore, once you take into account all the combinations, the number will be much bigger than ~4,800. All but one answer choice remains; therefore, E is the correct answer.
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 19 Aug 2019, 22:59
1
Narenn wrote:
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



The solution given has a typo error. It’s written as 13896, whereas it should be 13986
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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New post 19 Aug 2019, 23:03
drroy wrote:
Narenn wrote:
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



The solution given has a typo error. It’s written as 13896, whereas it should be 13986

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Fixed the typo. Thank you.
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Re: What is the sum of all 3 digit positive integers that can be formed us  [#permalink]

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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]

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New post 16 Nov 2019, 14:27
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

111+555+888=1554 sum
1554/3=518 mean
3^3=27 numbers
27*518=13986 sum
E
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What is the sum of all 3 digit positive integers that can be formed us   [#permalink] 16 Nov 2019, 14:27

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