 Intern Joined: 25 Nov 2009 Posts: 47 Location: India
Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink]
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Question Stats: Hide Show timer StatisticsFind 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?)
Math Expert Joined: 02 Sep 2009 Posts: 87960
Re: Sum of all 3-digit nos with 1, 2 & 3 [#permalink]
sudip135 wrote: 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). _________________
Intern Joined: 20 Dec 2009 Posts: 9
Re: Sum of all 3-digit nos with 1, 2 & 3 [#permalink] 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 -111112113121122123131132133Similar 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!
Intern Joined: 25 Nov 2009 Posts: 47 Location: India
Re: Sum of all 3-digit nos with 1, 2 & 3 [#permalink] 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
Math Expert Joined: 02 Aug 2009 Posts: 10680
Re: Sum of all 3-digit nos with 1, 2 & 3 [#permalink]
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.. _________________
Intern Joined: 25 Nov 2009 Posts: 47 Location: India
Re: Sum of all 3-digit nos with 1, 2 & 3 (no repeat) [#permalink] 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)?
Math Expert Joined: 02 Aug 2009 Posts: 10680
Re: Sum of all 3-digit nos with 1, 2 & 3 [#permalink] 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 _________________
Manager Joined: 10 Jun 2015 Posts: 94
Re: Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink]
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 timesThe sum of 1,2, and 3 is 6. Therefore, when you add all 27 unit digits you get 9*6=54If you add the ten's digits, again you get 54If you add the hundred's digits, again you get 54Now, you can mentally addthe answer is 5994.
Manager Joined: 25 Oct 2019 Posts: 152 Location: India GPA: 4
Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink] total numbers: 3*3*3=27total no. of digits in all numbers = 27*3= 81each no. will appear = 81/3 = 27 times ...9 times at unit place, 9 times at 10th, 9 times at 100th place...the place value of the digits will vary each time.such as 1 10 1002 20 2003 30 300total sum= 9(1+10+100)9(2+20+200)9(3+30+300)= 5994 Posted from my mobile device
Intern Joined: 20 Jan 2022 Posts: 9 Location: India
Re: Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink] By using three digit 1,2,3 we would have maximum possible arrangements are 3! i.e 6 possible arrangements In these 6 arrangements each digit will repeat similar time in each place.So sum at unit place of all digit 3x2+2x2+1x2=12 Hence sum of all digit =100x12+10x12+12x1=1332 Posted from my mobile device
Re: Find the sum of all 3-digit nos that can be formed by 1, 2 [#permalink] 19 Feb 2022, 17:50 |
What is the sum of all 3 digit numbers formed by using the digits 1 2 3 without repetition?
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