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What is the sum of all 3 digit positive integers that can be formed [#permalink]
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28 Jun 2009, 19:01
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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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28 Jun 2009, 22:10
subscribing to the thread to know the answer.I had no idea of these kind of questions.



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Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]
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28 Jun 2009, 22:12
Me neither. Awaiting response from someone who knows.



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Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]
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29 Jun 2009, 04:40
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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=10992 Keeping 8 as Hundred digit .. 815+851=16663 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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29 Jun 2009, 04:49
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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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29 Jun 2009, 06:53
OA is E.
But I am still not clear with the explanation. Can you please explain in more detail...thanks in advance.



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Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]
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29 Jun 2009, 07:01
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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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29 Jun 2009, 14:08
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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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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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29 Jun 2009, 23:29
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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 3digit 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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02 Jul 2009, 12:44
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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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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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 3digit 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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18 Nov 2011, 14:57
Great explanations guys
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Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]
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24 Nov 2011, 11:36
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 explanation .. +1 kudos.
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Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]
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27 Nov 2015, 00:06
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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.
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Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]
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28 Nov 2015, 20:45
Hi All, There's a great 'patternmatching' 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 3digit 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... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: What is the sum of all 3 digit positive integers that can be formed [#permalink]
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28 Dec 2015, 14:26
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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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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