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

Question

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?)

Bunuel · Accepted Answer

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) .

ashueureka · Answer

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 = 
  * 9
= 6 * 111 * 9 = 54 * 111 = 5994.
Hope this clarifies or else I will elaborate it more.
Thanks!

jusjmkol740 · Answer

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

chetan2u · Answer

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..

jusjmkol740 · Answer

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)?

chetan2u · Answer

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

mathivanan · Answer

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.

Gknight5603 · Answer

total numbers: 3*3*3=27
total no. of digits in all numbers = 27*3= 81
each 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 10 100
2 20 200
3 30 300

total sum= 9(1+10+100)
9(2+20+200)
9(3+30+300)
= 5994

Posted from my mobile device

Khan98 · Answer

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

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