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Re: Sum of all 3-digit nos with 1, 2 & 3 [#permalink]

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07 Jan 2010, 02:56

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As nothing has been mentioned in the question, we'll assume that repetition of numbers are allowed. Hence the total number of 3 digit numbers that can be formed from 1,2,3 = 3*3*3 = 27.

Now out of these 27 numbers each of the digits 1,2,3 will occur at each of the hundred's, ten's and unit's position 9 times. e.g. starting from all the numbers having 1 in hundred position we have the following numbers - 111 112 113 121 122 123 131 132 133 Similar would be the sequence for numbers starting with 2 and 3 and if we count we'll find that 1,2,3 occurs at hundreds position 9 times each; at tens position 9 times each and at units position 9 times each. Hence the sum of all these 27 numbers = [(1+2+3) * 100 + (1+2+3) * 10 + (1+2+3) * 1 ] * 9 = 6 * 111 * 9 = 54 * 111 = 5994. Hope this clarifies or else I will elaborate it more. Thanks!

Re: Sum of all 3-digit nos with 1, 2 & 3 [#permalink]

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07 Jan 2010, 04:29

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Hey thnx. It helps.

But thr's another explanation:

No of such 3-digit nos = 27 (which is Ok) 1st no = 111, Last no = 333 (these r also Ok) Hence their average = (111 + 333)/2 = 222 (couldn't understand how this formula is applied. I thought this holds true for an AP series only) So, Sum = Number of nos X Average of the nos = 27 X 222 = 5994

hi sudip what i can of think as a way is... total nos=3*3*3=27..... so sum will have 27 nos .... so each no 1,2,3 will be used (27/3)9 times in each digits place (hundreds,tens and ones) ...units digit=9*(1+2+3)=54, so 4.. tens digit=9*(1+2+3)=54(+5)=9, so 9.."+5" is the carried tens digit from 54 of step 1 hundreds digit=9*(1+2+3)=54(+5)=59, so the no is 5994..
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Re: Sum of all 3-digit nos with 1, 2 & 3 (no repeat) [#permalink]

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08 Jan 2010, 01:53

Thnx.

By the way, what happens to the same problem if we are not allowed to repeat any of the digits in any particular no formed from by the digits (i.e. 111 or 221 or 133 etc are not to be considered)?

if the digits are not to be repeated.. total nos=3*2*1=6.. so each no 2 times.. no is =(3+2+1)*2*100+(3+2+1)*2*10+ (3+2+1)*2*1=1200+120+12=1332
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Q. Find the sum of all 3-digit nos that can be formed by 1, 2 and 3 (Ref: Kaplan. I didn't understand the explanation there. Can u pl help?)

GMAT will tell you in advance whether repetition is allowed or not. Or the wording will make it obvious.

Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:

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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Re: Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink]

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17 Jul 2014, 00:25

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Re: Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink]

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14 Aug 2015, 07:10

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Re: Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink]

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15 Aug 2015, 00:36

jusjmkol740 wrote:

Find the sum of all 3-digit nos that can be formed by 1, 2 and 3

(Ref: Kaplan. I didn't understand the explanation there. Can u pl help?)

with digits 1, 2, 3 you can form 3*3*3=27 distinct three digit numbers. If you observe the unit digits, it is a repetition of 1,2, and 3, each digit repeating 9 times The sum of 1,2, and 3 is 6. Therefore, when you add all 27 unit digits you get 9*6=54 If you add the ten's digits, again you get 54 If you add the hundred's digits, again you get 54 Now, you can mentally add the answer is 5994.

Re: Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink]

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23 Oct 2016, 21:36

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