Alex55
For each set of three distinct nonzero digits, consider the sum of all positive three-digit integers that can be formed by the digits. For example, for the three digits 1, 2, and 3, the sum of all positive three-digit integers that can be formed by the digits is 123 + 132 + 213 + 231 + 312 + 321 = 1,332. How many different integers are equal to such a sum?
A. 9
B. 11
C. 14
D. 19
E. 24
The sum of the 6 different permutations of abc =
100a+10b+c
+100a+10c+b
+100b+10a+c
+100b+10c+a
+100c+10a+b
+100c+10b+c
=222(a+b+c)
Therefore, the sum of a+b+c determines the sum of the 6 different permutations of abc
How many different sums of a+b+c are there, and how many sums of 6 different permutations of abc are there?
The minimum case of the sum of a+b+c, 1+2+3=6, the sum of six different permutations=6*222
The maximum case of the sum of a+b+c, 7+8+9=24, the sum of six different permutations=24*222
1, 2, 3, 4, 5, 6, 7, 8, 9 numbers are used arbitrarily, then the sum of a+b+c can be obtained continuously from 6 to 24, and there are 24-6+1=19 possibilities
There are 19 different sums of a+b+c, so there are 19 sums of 6 different permutations of abc