2014 AMC 12B Problems/Problem 8
Problem
In the addition shown below , , , and are distinct digits. How many different values are possible for ?
Solution
From the first column, we see because it yields a single digit answer. From the fourth column, we see that equals and therefore . We know that . Therefore, the number of values can take is equal to the number of possible sums less than that can be formed by adding two distinct natural numbers. Letting , and letting , we have
Solution (Equation Algorithm)
It is intuitively obvious, even to the most casual observer that the problem statement can be rewritten as:
. This equation can be simplified into:
.
Now from here, it should hopefully make sense that by looking at the one's digit of both equations. Factoring out gives:
.
Which equals: .
This simplifies into: .
Therefore .
This means that and or else there would be parts carried over in the equation. The positive integers that satisfy this equation are a minimum and a maximum of . This means that . Giving ~PeterDoesPhysics
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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