What is the sum of all 3 digit positive integers that can be formed us

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?

Imagine, we have got all these possible numbers written down - there are in total 3^3 numbers (each digit can be either 1 or 5 or 8)

there are 3*3 options for having a number XY1
there are 3*3 options for having a number XY5
there are 3*3 options for having a number XY8

there are 3*3 options for having a number X1Z
there are 3*3 options for having a number X5Z
there are 3*3 options for having a number X8Z

there are 3*3 options for having a number 1YZ
there are 3*3 options for having a number 5YZ
there are 3*3 options for having a number 8YZ

we can sum units, tens and hundreds independently:
summing units gives (1+5+8)*3*3
summing tens gives (1+5+8)*10*3*3
summing hundreds gives (1+5+8)*100*3*3

E

summing
units (1+5+8)*9 +
tens (1+5+8)*9*10 +
hundreds (1+5+8)*9*100 =

= 126+1,260+12,600 = 13,986

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).

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

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 3-digit 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)

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.

Of course,

(1+5+8)/3 - "average" digit.
(1+5+8)/3 * 111 - another way to write 3-digit number formed from "average digit": xyz = (1+5+8)/3 (1+5+8)/3 (1+5+8)/3 or (1+5+8)/3 * 111

I don't have your specific method but by POE I still can have E.
for 8xy alone we have 888,881,885,818,855,851,858,815,811. The total of them is larger than 7200 so there is only option E left

there will 9 times 1 , 9 times 5 and nine times 8 at each place in a 3 digit no. with repetition

sum will be =100*9*(1+58) + 10*9*(1+5+8) + 1*9*(1+5+8)=13986

in order to know how it is 9 times

tot no. of ways =3*3*3=27
so in tot 27 words will be formed where each digit among 1,5,8 wil be repeated equal no. of times i.e 27/3=9
hence we have 9 , 1's ; 9 5's and 9 8's at each level

Wow. I have no idea what is happening here. Would you (or someone else) mind explaining a bit more...perhaps expounding on your method a little bit?

Perhaps, the same deal, with a two digit number using 1 and 2, with digits allowed to repeat?

[Answer: 11+12+21+22 = 66]

How can I use xcusemeplz2009's method to arrive at the same answer?

let me try again

let the no. be xyz
Place value of no will be 100x+10y+Z.............eqn 1

TOTAL NO OF WAYS IN WHICH WE CAN MAKE A NO. WITH GIVEN CONDITION IS 3*3*3=27( as digits are getting repeated)

now at unit place i.e Z if we fix 1 , then for y we have 3 options and for x we have 3 options , so total number where unit digit is one can be formed in 3*3=9 ways

similarly for other digits at unit place can be done in 9 ways

hence at unit place a digit is getting repeated 9 times

and in the same manner in tenths place repetion for all digits will be 9 each

now the task is to find the sum
sum of digits at z place is 9*1+9*5+9*8 , WHICH
IS SAME FOR Y AND X
for that we need to put all the options in eqn 1 format
i.e 100*9[1+5+8] + 10*9[1+5+8] + 9*[1+5+8]=13986

simple way is to remeber the quick formula to find out the repetition

find out tot no. ways in which the no. can be formed and divide it by the no. of digit , then mutiply this factor with the summation of all the dig and the place value

for second ex : 1,2

two dig no with repetn can be formewd in 2*2=4 ways , repetion factor=4/2 =2 ( tot no. of ways / tot no of dig)

sum=10(place value) * 2 (rep factor) *(1+2 summation of dig)+2*3=60+6=66

Pretty impressive formula used by xcusemeplz2009. +1 for that!

Nevertheless the question can be solved with POE as it is quite logical....

digits in scope = 1,5,8..
with this we can get highest 3 or 4 numbers - 888, 885, 881, 858.
Adding just these 4 numbers gives us = 3512
and we have 23 more numbers to be added... Hence the correct answer would be E of course!

Cheers!
JT

I sowe really good formula for solving this problem in some notes downloaded from this forum. I just cannot find it, so I appologize to the author.

The formula says:



Here we have 3 digits. n is 3.
Sum of digits 1+5+8=14

Repetition allowed:
14*(3^2)*111=13986

Repetition not allowed:
14*2*111=3108

Here is a formula to know the sum of possible arrangements when a digit is not allowed to repeat:

\((n-1)!*sumofdigits*111 = (3-1)!*(1+5+8)*111=28*111=3108\)

But we know that digits are allowed to repeat. Thus, sum is much greater than 3108.

Answer: E

I approach this particular problem without the formulae. Can somebody please help me if this is correct --> If you know that the numbers are allowed to repeat then the possible numbers are 3*3*3 = 27 (instead of 3*2*1 when repetition is not allowed), then you know that there will be 9 ones, 9 fives, 9 eights. So for the first position you can have the 9+45+72 = 12600, then all the answer choices will fall except for E. If you calculate further you get 12600 + 01260 + 00126 = 13,986. Bunuel, Karishma or someone else can you please confirm if this is correct?