student26 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
PS78602.01
Method 1: Sum Each Column of Digits
We need positive integers having 4 digits.
S = __ __ __ __
Since no repetition allowed, we can make 4*3*2*1 = 24 such positive integers.
Now imagine writing these 24 numbers one below the other to add.
1234
1243
...
...
x 24 combinations
When we add them, noticing the symmetry we know that there will be 6 1's in units digits, 6 2's, 6 3's and 6 4's. So units digits will add up to (1+2+3+4)*6.
Similarly, tens digits will add up (1+2+3+4)*6*10
Similarly, hundreds digits will add up (1+2+3+4)*6*100
Similarly, thousands digits will add up (1+2+3+4)*6*100)
Adding all of them:
(1+2+3+4)*6 + (1+2+3+4)*6*10 + (1+2+3+4)*6*100 + (1+2+3+4)*6*1000 = (1+2+3+4)*6 * (1 + 10 + 100 + 1000) = 10 * 6 * 1111 = 66,660
ANSWER: E
Method 2: Direct Formula
Sum of all n digit numbers formed by n non-zero digits without repetition is:
(n−1)!∗(sum of the digits)∗(111... n times)
= (4-3)! * (1 + 2 + 3 + 4) * (1111)
= 6 * 10 * 1111
= 66,660
Answer: E