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sum of these 24 integers?

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sum of these 24 integers?

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1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660

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Junior | Next Rank: 30 Posts
Joined
11 Jun 2019
Posted:
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Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.

The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660

Answer: e

OR

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the nn digits without repetition is: (n-1)!-(sum of the digits)-(111... n times)(n-1)!-(sum of the digits)-(111... n times)

2. Sum of all the numbers which can be formed by using the nn digits (repetition being allowed) is: nn-1-(sum of the digits)-(111... n times)nn-1-(sum of the digits)-(111... n times).

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rajeet123 wrote:
1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660

_________________
Brent Hanneson – Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for free Question of the Day emails
And check out all of these free resources

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rajeet123 wrote:
1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
Since we're adding 24 numbers, we know that:
Six numbers will be in the form 1---
Six numbers will be in the form 2---
Six numbers will be in the form 3---
Six numbers will be in the form 4---

Let's first see what the sum is when we say all 24 numbers are 1000, 2000, 3000 or 4000
The sum = (6)(1000) + (6)(2000) + (6)(3000) + (6)(4000)
= 6(1000 + 2000 + 3000 + 4000)
= 6(10,000)
= 60,000

Since the 24 numbers are actually greater than 1000, 2000, etc, we know that the actual sum must be greater than 60,000

Answer: E

Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for free Question of the Day emails
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For any set that is SYMMETRICAL ABOUT THE MEDIAN:
sum = (count)(median)

rajeet123 wrote:
1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exactly once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
The set is composed of integers in the following ranges:
1234...1432
2134...2431
3124...3421
4123...4321.

Each range contains the same number of integers.
Thus, the median of the set is equal to the average of the two integers in red:
(2431 + 3124)/2 = 5555/2.

Notice that the set is SYMMETRICAL ABOUT THE MEDIAN:
...2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241...
Thus:
sum = (count)(median) = 24 * 5555/2 = 12 * 5555 = 66,660.

The correct answer is E.

For a similar problem that can be solved with the same line of reasoning, check my post here:
http://www.beatthegmat.com/on-consecutive-non-consecutive-series-t85395.html

Alternate solution:

Each digit will appear in each position 24/4 = 6 times.
Thus, in each position, there will be six 1's, six 2's, six 3's, and six 4's.
Sum of the digits in each position = 6(1+2+3+4) = 60.

Sum of the thousands place = 60*1000 = 60,000.
Sum of the hundreds place =60*100 =6,000.
Sum of the tens place = 60*10 = 600.
Sum of the units place = 60*1 = 60.

Sum of all the integers = 60000 + 6000 + 600 + 60 = 66,660.

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Post
rajeet123 wrote:
1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
Since there are 24 different integers, each of the digits (1, 2, 3 and 4) will appear in each of the place values (thousands, hundreds, tens and ones) exactly 6 times. Therefore, the sum of the 24 integers will be as follows:

6(1000 + 2000 + 3000 + 4000) + 6(100 + 200 + 300 + 400) + 6(10 + 20 + 30 + 40) + 6(1 + 2 + 3 + 4)

6(10,000) + 6(1,000) + 6(100) + 6(10)

60,000 + 6,000 + 600 + 60

66,660

Answer: E

_________________

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

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