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

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umeshpatil

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04 Jan 2013, 07:43

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asimov

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?

A. 126
B. 1386
C. 3108
D. 308
E. 13986

One quickest way to answer this question !!

As we are using digits 1,5, 8 and digits are allowed to repeat. Each of the unit, tenth and hundredth digit can be used by each of three digits.
So, Total possible numbers with these digits=3 X 3 X 3 =27.

First, As we have 27 three digit number, Sum will be for sure more than 2700.. Eliminate options A,B,D :lol:

Second, If you imagine numbers with the given digits 1,5,8. We have numbers like 888,885,855,858,851. Sum is for sure more than 4000. Eliminate option C. :lol:

You are left with answer E.

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consider giving a +kudo if this helps :-D

Narenn

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05 Jan 2013, 10:40

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asimov

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?

A. 126
B. 1386
C. 3108
D. 308
E. 13986

Answer to this question is easier to guess than to calculate. e.g. if we take 8 at hundreds place we would get at least 9 nos. So 800 * 9 = 7200 which surpasses every option but E.

Thru conventional method
(1+5+8)9 = 126
(1+5+8)9*10=1260
(1+5+8)9*100=12600

126 + 1260 + 12600 = 13986. E

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01 Oct 2013, 18:44

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aismirnov

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

How did you know how to do this? I mean, how did you learn? I have 4 weeks to go until my GMAT and I haven't gotten any better at these. I have no idea how to even begin approaching these problems, and none of these formulas make any sense to me.

Bunuel

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02 Oct 2013, 03:24

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aismirnov

Direct formulas are here: what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html#p862674 Please ask if anything there is unclear.

Similar questions to practice:
find-the-sum-of-all-the-four-digit-numbers-formed-using-the-103523.html
find-the-sum-of-all-the-four-digit-numbers-which-are-formed-88357.html
find-the-sum-of-all-3-digit-nos-that-can-be-formed-by-88864.html
if-the-three-unique-positive-digits-a-b-and-c-are-arranged-143836.html
what-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html
what-is-the-sum-of-all-4-digit-numbers-that-can-be-formed-94836.html
the-sum-of-the-digits-of-64-279-what-is-the-141460.html
there-are-24-different-four-digit-integers-than-can-be-141891.html
the-addition-problem-above-shows-four-of-the-24-different-in-104166.html

Hope this helps.

goldfinchmonster

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31 Oct 2015, 08:31

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The Ans is E.

This can also be solved by using a formula

Sum of N numbers

With repetition = N ^ ( N - 1 ) * Sum of the Numbers * 111..... N no. of times

With out repetition = ( N - 1 ) ! * Sum of the Numbers * 111.... N no. of times.

Here repetition is allowed, therefore N is 3 ( No. of digits given ) and sum of No is 14.

3^2 * 14 * 111 = 13986.

mohitmohan23

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27 Nov 2015, 00:06

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Used POE, No need to solve the whole question to get an exact value
Explanation:

since three digit numbers formed by 1, 5, 8 would be :
Lets start with numbers starting with 8 : 888, 885, 881, 855, 851, 815, 811
sum of these numbers is greater than Options A , B , C , D
Hence Ans: E.

Keep it simple ppl

Good luck!!
Cheers

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28 Nov 2015, 20:45

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Hi All,

There's a great 'pattern-matching' shortcut built into this question that can help you to avoid much of the 'math work' involved.

We're told to use the digits 1, 5 and 8 to form every possible 3-digit number (including those with duplicate digits) and then take the sum of those numbers.

Since the digits can be repeated, we're dealing with the numbers that fall into the range of 111 to 888, inclusive. There are (3)(3)(3) = 27 total numbers and 1/3 of those numbers will begin with an 8. From THAT deduction, we know that the sum of those 9 numbers will be greater than (9)(800) = 7200. There's only one answer that fits that description...

Final Answer:

Show Spoiler

GMAT assassins aren't born, they're made,
Rich

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15 Nov 2019, 19:05

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asimov

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?

A. 126
B. 1386
C. 3108
D. 308
E. 13986

Method 1: Sum Each Column of Digits

We need positive integers having 3 digits.

S = __ __ __

We can make 3*3*3 = 27 such positive integers since we can fill in each of the 3 spaces in 3 ways. Now imagine writing these 27 numbers one below the other to add.

158
185
...
...
x 27 combinations

When we add them, noticing the symmetry we know that there will be 9 1's in units digits, 9 5's and 9 8's. So units digits will add up to (1+5+8)*9.

Similarly, tens digits will add up (1+5+8)*9*10.
Similarly, hundreds digits will add up (1+5+8)*9*100

Adding all of them:
(1+5+8)*9 + (1+5+8)*9*10 + (1+5+8)*9*100 = (1+5+8) * 9 * (1+10+100) = 14 * 9 * 111 = 13,986

Method 2: Direct Formula

Sum of all n digit numbers formed by n non-zero digits with repetition being allowed is:

n^(n−1)∗(sum of the digits)∗(111... n times)

9 * 14 * 111 = 13,986

Answer: E

BrentGMATPrepNow

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asimov

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?

A. 126
B. 1386
C. 3108
D. 308
E. 13986

Let's first calculate how many 3-digit numbers can be created using only the digits 1, 5 and 8.
There are 3 ways to choose a hundreds digit.
There are 3 ways to choose a tens digit.
There are 3 ways to choose a ones digit.
So, the total number of ways to create a three digit number = (3)(3)(3) = 27

Now let's focus on the hundreds digits of our 27 numbers.
9 (aka 1/3) of the 27 numbers will have hundreds digit 1. 9 x 100 = 900
9 (aka 1/3) of the 27 numbers will have hundreds digit 5. 9 x 500 = 4500
9 (aka 1/3) of the 27 numbers will have hundreds digit 8. 9 x 800 = 7200
900 + 4500 + 7200 = 12,600

So, if we ignore the tens and ones digits, the sum of our 27 numbers is already 12,600, which means the TOTAL sum must be greater than 12,600

Answer: E

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