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Sum of three-digit integers

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Sum of three-digit integers

by pkw209 » Thu Apr 15, 2010 1:43 pm
What's the fastest way of completing this problem?

There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?

a) 2,704

b) 2,990

c) 5,404

d) 5,444

e) 5,994
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by kevincanspain » Thu Apr 15, 2010 2:57 pm
Think of the units column: there will be 9 numbers ending in each of 1, 2 and 3

Thus the sum of the units column is 9(1 + 2 + 3) =54


In fact, the sums of the tens column and the hundreds column will also be 54

Thus the sum will be 54(100+10+1)= 54(111), more than 10% more than 5400

Choose E

You could also think that the average of these 27 integers should be 222 and 222 x 27 is aprox 20/9 x 100 x 27 i.e. 6000

Note 2.22222= 2 + 2/9 = 20/9
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by kstv » Fri Apr 16, 2010 3:11 am
The answer choices are a) 2,704 b) 2,990 c) 5,404 d) 5,444 e) 5,994
the tens and the unit digit are not repeating in the values of c d and e which are in close range
trying to find the sum of the tens and unit digit
i.e. all the two digits consisting of 1, 2 & 3 should be enough
there are 9 diff. two digit nos.
Rank the Min no alongside Max no
11 ---- 33 = 44
12 ---- 32 = 44
13 --- 31 = 44
21 --- 23 = 22
---- 22 ---= 22
ie 44*3 + 22*3 = 3(66)
but these nine numbers will have 3 diff digits (1, 2 or 3) in the hundredth place
so 3(66) x 3 = ?94 so opton E
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by Fiver » Fri Apr 16, 2010 4:49 am
pkw209 wrote:What's the fastest way of completing this problem?

There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?

a) 2,704

b) 2,990

c) 5,404

d) 5,444

e) 5,994
My solution is in the same lines as that of Kevin, albeit a little more detailled.

The sum of the units place is (1+2+3)*9 = 54. Keep 4 carry over 5 to the tens place.
The sum of the tens place is again (1+2+3)*9+5(carried over) = 59. Keep 9 and carryover 5.
The sum of the hundreds place is again (1+2+3)*9+5(carried over) = 59.

Hence the final result is 5994
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by sandy_online » Fri Apr 16, 2010 5:49 am
there is formula for this particular problem:

sum of integers that are formed by the permutations of n digits
sum is given by equation:

= (sum of digits)*(n-1)!*(111... n times) if repetition is not allowed.

= (sum of digits)*(n)^(n-1)*(111... n times) if repetition is allowed.


for given problem: sum = (1+2+3)*3^(3-1)*(111) = 6*9*111 = 5994
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by pops » Fri Apr 16, 2010 9:22 pm
brilliant.. thanks... that was what I was trying to recall mate !
sandy_online wrote:there is formula for this particular problem:

sum of integers that are formed by the permutations of n digits
sum is given by equation:

= (sum of digits)*(n-1)!*(111... n times) if repetition is not allowed.

= (sum of digits)*(n)^(n-1)*(111... n times) if repetition is allowed.


for given problem: sum = (1+2+3)*3^(3-1)*(111) = 6*9*111 = 5994
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by narisipalli » Sun Apr 18, 2010 10:03 pm
pkw209 wrote:What's the fastest way of completing this problem?

There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?

a) 2,704

b) 2,990

c) 5,404

d) 5,444

e) 5,994
I came up with a different approach and somehow it worked.
I considered (1 + 2 + 3) = 6 which is divisible by 3 ( Sum of the digits in the three-digits integers). After this I scanned through the answer choices and checked which option agrees with divisibility rule of 3 and E is the only option.

I am not sure if this approach is applicable to similar problems or is there an inherent flaw with this.
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by akhpad » Mon Apr 19, 2010 12:21 am
sandy_online wrote:there is formula for this particular problem:

sum of integers that are formed by the permutations of n digits
sum is given by equation:

= (sum of digits)*(n-1)!*(111... n times) if repetition is not allowed.

= (sum of digits)*(n)^(n-1)*(111... n times) if repetition is allowed.


for given problem: sum = (1+2+3)*3^(3-1)*(111) = 6*9*111 = 5994
I would like to add few things. Above formulas are correct but careful while using.
Ex: Different five digit numbers are formed with help of the digits 0, 2, 3, 4, and 6. Find the sum of all possible five digit numbers.
Here, zero cannot come at extreme left. We cannot apply above formula directly.

OR suppose, there are given some restriction with few integers.
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by jonny5k » Mon Apr 19, 2010 1:36 am
My intuitive answer was:

222*27 = 5,994. I used 222 as average of all the 27 numbers. In this case I was lucky, I suppose.
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by narisipalli » Mon Apr 19, 2010 10:18 am
jonny5k wrote:My intuitive answer was:

222*27 = 5,994. I used 222 as average of all the 27 numbers. In this case I was lucky, I suppose.
jonny5k, its one of the correct approaches.

As Sum = Average / Number of elements. Calculating the average is pretty straightforward, with 111 being the lowest and 333 being the highest numbers possible and these numbers would be distributed on either sides of the median which makes it the average.

Average = (111 + 333)/2 = 222
Sum = 222*27 = 5994[/spoiler]
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