Maxswe wrote:
2345
2354
2435
+5432
There are 24 different four-digit integers that 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?
A. 24,444
B. 28,000
C. 84,844
D. 90,024
E. 93,324
Notice that each of the 24 integers can be
broken down onto a nice sum.
For example, 2345 = 2000 + 300 + 40 + 5, and 4253 = 4000 + 200 + 50 + 3
Also recognize that, of the 24 different four-digit integers than can be formed, SIX will start with 2 (e.g., 2354, 2543, etc), SIX will start with 3 (e.g., 3245, 3524, etc), SIX will start with 4, and SIX will start with 5.
So, if we BREAK DOWN each integer, we'll have SIX 2000's, SIX 3000's, SIX 4000's, and SIX 5000's
Let's see what we get if we add SIX 2000's, SIX 3000's, SIX 4000's, and SIX 5000's
2000 + 3000 + 4000 + 5000 = 14,000
6(14,000) = 84,000
In the same manner, we should recognize that, of the 24 different four-digit integers than can be formed, SIX will have a 2 in the hundreds position (e.g., 3254, 5243, etc), SIX will have a 3 in the hundreds position (e.g., 2345, 5324, etc), SIX will have a 4 in the hundreds position, and SIX will have a 5 in the hundreds position
Let's see what we get if we add SIX 200's, SIX 300's, SIX 400's, and SIX 500's
200 + 300 + 400 + 500 = 1400
6(1400) = 8,400
IMPORTANT: So far, we've just added PART OF the total sum.
So far, our sum = 84,000 + 8,400 = 92,400
Since we haven't finished adding ALL PARTS of the total sum, we can conclude that
the correct answer is GREATER THAN 92,400Answer: E
Cheers,
Brent
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