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> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>.) | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ~Technodoggo |
Revision as of 14:22, 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