Difference between revisions of "2014 AMC 12B Problems/Problem 8"
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<cmath>D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}</cmath> | <cmath>D = 3,4,5,6,7,8,9 \implies \boxed{\textbf{(C)}\ 7}</cmath> | ||
− | ( | + | ==Solution (Equation Algorithm)== |
+ | |||
+ | It is intuitively obvious, even to the most casual observer that the problem statement can be rewritten as: | ||
+ | |||
+ | <math>10^4A + 10^4B + 10^3B + 10^3C + 10^2B + 10^2A + 10C + 10D + B + A = 10^4D + 10^3B + 10^2D + 10D + D</math>. This equation can be simplified into: | ||
+ | |||
+ | <math>10^4A + 10^4B + 10^3C + 10^2B + 10^2A + 10C + B + A = 10^4D + 10^2D + D</math>. | ||
+ | |||
+ | Now from here, it should hopefully make sense that <math>A + B = D</math> by looking at the one's digit of both equations. Factoring out <math>A + B</math> gives: | ||
+ | |||
+ | <math>10^4(A+B) + 10^3C + 10^2(A+B) + 10C + (B + A) = 10^4D + 10^2D + D</math>. | ||
+ | |||
+ | Which equals: | ||
+ | <math>10^4(D) + 10^3C + 10^2(D) + 10C + D = 10^4D + 10^2D + D</math>. | ||
+ | |||
+ | This simplifies into: | ||
+ | <math>10^3C + 10C = 0</math>. | ||
+ | |||
+ | Therefore <math>C = 0</math>. | ||
+ | |||
+ | This means that <math>A + B = D</math> and <math>D < 10</math> or else there would be parts carried over in the equation. The positive integers that satisfy this equation are a minimum <math>(2, 1)</math> and a maximum of <math>(4, 5)</math>. This means that <math>D = 3</math> <math>, 4</math> <math>, 5</math> <math>, 6</math> <math>, 7</math> <math>, 8</math> <math>, 9</math>. Giving | ||
+ | <math>\boxed{\textbf{(C)}\ 7}</math> | ||
+ | |||
+ | ~PeterDoesPhysics | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2014|ab=B|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:45, 11 August 2024
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.