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

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

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New post 28 Jun 2009, 19:01
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A
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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: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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New post 28 Nov 2015, 20:45
5
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  [#permalink]

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New post 29 Jun 2009, 04:49
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5
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 ;)
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Re: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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New post 29 Jun 2009, 04:40
1
Is it C ..
This is how I did it ..
Keeping 1 as Hundred digit ..
158+185=343 -----1
Keeping 5 as Hundred digit ..
518+581=1099-------2
Keeping 8 as Hundred digit ..
815+851=1666-------------3
Adding 1+2+3 = 3108 ..
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Re: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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New post 29 Jun 2009, 07:01
2
Since problem permits repetition. There are 27 numbers that satisfies.
e.g.:
Lets say: first digit is 1, then numbers can be:
111
115
118
151
155
158
181
185
188
Same is true when first digits are 5 and 8. As you can see, there are nine 1 in first digit. Nine 5 in first digit. And nine 8 in first digit. Same is true for the other digits.
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Re: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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New post 29 Jun 2009, 14:08
2
maliyeci wrote:
Since problem permits repetition. There are 27 numbers that satisfies.
e.g.:
Lets say: first digit is 1, then numbers can be:
111
115
118
151
155
158
181
185
188
Same is true when first digits are 5 and 8. As you can see, there are nine 1 in first digit. Nine 5 in first digit. And nine 8 in first digit. Same is true for the other digits.


One more tip: when we add up all the numbers, we can start with the hundreds. We know each number (1, 5, & 8) will appear in the hundreds place a total of 9 times. So let's see how many hundreds we have.
1X9=9
5X9=45
8X9=72
Add this up we have a total of 126 hundreds, or also expressed as 12,600. We see that only one answer could possibly match the size of this sum, which is (E), so without calculating the exact sum, we already know (E) is the only possible choice. On a real test however, the writers could make life difficult by adding a few answer choices that are close to this sum (i.e. 11,950, or 14,088, etc).
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Re: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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New post 29 Jun 2009, 22:30
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  [#permalink]

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New post 29 Jun 2009, 23:29
2
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 ;)


Great ! Kudos to you.

Another approach is intelligent guess, based on which I would have opted E. Explanation:
Total possibilities = 3*3*3 =27
Now, taking examples of numbers starting with 8.
Sum of any four 3-digit numbers starting with 8 > 3200,

We know that there are 9 possible nos starting with 8 (apart form other 18 numbers), so sum would certainly be much much greater then 3200.

All other options, except E is less then 3200. (Btw, one can eliminate A, B and D on the 1st glance itself)
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Re: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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New post 02 Jul 2009, 12:44
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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
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Re: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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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  [#permalink]

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New post 14 Jul 2009, 13:27
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  [#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

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

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New post 14 Jan 2016, 10:40
Here is my approach. I did not use any advanced concepts/tricks. Just did logical reasoning and POE.

Since any digit can repeat more than once, start by building as high a number as possible.

Thus 888.
888 itself eliminates options A and D. Now we have 3 options remaining.

Now which other (large value) numbers can I form that adhere to all the rules given in the question? ( The idea is to see if i can come up with numbers that go higher than any of the answer choices and thus eliminate some answer choices).

885
I can also do 858
and on the same lines: 855

Now I don't need to compute the exact sum. the sum is definitely greater than 3200 ( because there are 4 numbers starting with 8. 8*4 =32. Since all numbers are 3 digit, I get 3200).

Luckily for us, 3200 is higher than 4 of the answer choices. In other words, this eliminates 4 answer choices. Thus the remaining option is the correct answer: Opt E.
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Re: What is the sum of all 3 digit positive integers that can be formed  [#permalink]

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New post 24 Aug 2016, 19:51
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  [#permalink]

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New post 31 Mar 2018, 14:54
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


111+555+888=1554
1554/3=518
518*27=13986
E
Re: What is the sum of all 3 digit positive integers that can be formed &nbs [#permalink] 31 Mar 2018, 14:54
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