There are 24 different four-digit integers than can be formed using the digits 2, 3, 4 and 5 exactly once in each integer. The addition problem above shows 4 such integers. What is the sum of all 24 such integers?
Choose the correct choice from the following choices.
A) 24,444
B) 28,000
C) 84,844
D) 90,024
E) 93,324
Fast Solution:   First stage: we will look at the numbers that contain only 1 digit and the rest is zeros, e.g.: 5000,0200 (=200) ,0040 (=40)   - Let's look at the thousand digit:   * We have: 4 options 5000,4000,3000,2000 = (5+4+3+2)*1000= 14*1000   * We will use each option 3! times (because _X_,__,__,__ - you have 3! options to place the rest 3 numbers)   * that means that the total is :14*1000*3!   - We can do the same for the hundreds digit, tens digits and ones.   Second stage: We will sum every number   14*3!*1000 +   14*3!*100 +   14*3!*10 +   14*3!*1   -------------------   14*3!*1111 -&gt; 14*6*1111=93,324   Backsolving and Picking Numbers aren’t feasible here, so we’ll have to use some (not-so) Straightforward Math. Consider the column on the right-hand side, representing the ones digit of each integer. There are 24 integers and 4 different digits, so each of the 4 digits must appear 6 times in the ones column. Thus, the sum of the ones place for the 24 integers is (6 × 2) + (6 × 3) + (6 × 4) + (6 × 5), or 6 × 14 = 84. Now we know that the correct choice must end with a 4, so eliminate choice (B).   The same pattern holds for the tens column EXCEPT that we have to add8to represent the amount carried over from the 84 in the ones column. So, the tens column must add up to   84 + 8= 
<gadget id="calculator">  84 + 8</gadget>
<output>92</output>
92. Now we see that we must have a 2 in the tens place, so eliminate choices (A) and (C).   For the hundreds place, the sum will be 84 + the 9 carried over from the 92 in the tens column; 84 + 9= 
<gadget id="calculator">84 + 9</gadget>
<output>93</output>
93, so we must have a 3 in the hundreds place. Answer Choice (E) is the only possibility.   Alternatively, if the ones column adds up to 84, the tens column will be the same thing multiplied by 10, or 840. Similarly, the hundreds column will be 8,400 and the thousands column will be 84,000. Adding up these four numbers, we get 84 + 840 + 8,400 + 84,000= 93,324.answer is 93,324
<result>E</result>
There are 24 different four-digit integers than can be formed using the digits 2, 3, 4 and 5 exactly once in each integer. The addition problem above shows 4 such integers. What is the sum of all 24 such integers?