Difference between revisions of "2023 AMC 10B Problems/Problem 23"
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| + | ==Solution 1== | ||
| + | Since one of the terms was either <math>1</math> more or <math>1</math> less than it should have been, the sum should have been <math>222-1=221</math> or <math>222+1=223.</math> | ||
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| + | The formula for an arithmetic series is <math>an+d\left(\dfrac{(n-1)n}2\right)=\dfrac n2\left(a+d(n-1)\right).</math> This can quickly be rederived by noticing that the sequence goes <math>a,a+d,a+2d,a+3d,\dots,a+(n-1)d</math>, and grouping terms. | ||
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| + | We know that <math>\dfrac n2(2a+d(n-1))=221</math> or <math>223</math>. Let us now show that <math>223</math> is not possible. | ||
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| + | If <math>\dfrac n2(2a+d(n-1))=223</math>, we can simplify this to be <math>n(a+d(n-1))=223\cdot2.</math> Since every expression in here should be an integer, we know that either <math>n=2</math> and <math>a+d(n-1)=223</math> or <math>n=223</math> and <math>a+d(n-1)=2.</math> The latter is not possible, since <math>n\ge3,d>1,</math> and <math>a>0.</math> The former is also impossible, as <math>n\ge3.</math> Thus, <math>\dfrac n2(2a+d(n-1))\neq223\implies\dfrac n2(2a+d(n-1))=221</math>. | ||
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| + | We can factor <math>221</math> as <math>13\cdot17</math>. Using similar reasoning, we see that <math>221\cdot2</math> can not be paired as <math>2</math> and <math>221</math>, but rather must be paired as <math>13</math> and <math>17</math> with a factor of <math>2</math> somewhere. | ||
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| + | Let us first try <math>n=13.</math> Our equation simplifies to <math>2a+12d=34\implies a+6d=17.</math> We know that <math>d>1,</math> so we try the smallest possible value: <math>d=2.</math> This would give us <math>a=17-2\cdot6=17-12=5.</math> (Indeed, this is the only possible <math>d</math>.) | ||
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| + | There is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer <math>a+d+n=5+2+13=\boxed{\textbf{(B) }20.}</math> | ||
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| + | For the sake of completeness, we can explore <math>n=17.</math> It turns out that we reach a contradiction in this case, so we are done. | ||
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| + | ~Technodoggo | ||
Revision as of 13:22, 15 November 2023
Solution 1
Since one of the terms was either 
more or less than it should have been, the sum should have been 
or 
The formula for an arithmetic series is 
This can quickly be rederived by noticing that the sequence goes 
, and grouping terms.
We know that 
or 
. Let us now show that is not possible.
If 
, we can simplify this to be 
Since every expression in here should be an integer, we know that either 
and 
or 
and 
The latter is not possible, since 
and 
The former is also impossible, as 
Thus, 
.
We can factor 
as 
. Using similar reasoning, we see that 
can not be paired as 
and , but rather must be paired as 
and 
with a factor of somewhere.
Let us first try 
Our equation simplifies to 
We know that 
so we try the smallest possible value: 
This would give us 
(Indeed, this is the only possible 
.)
There is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer 
For the sake of completeness, we can explore 
It turns out that we reach a contradiction in this case, so we are done.
~Technodoggo
