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Difference between revisions of "2023 AMC 8 Problems/Problem 25"

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<math>1\leq a_1\leq10</math>
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<math>13\leq a_2\leq20</math>
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<math>241\leq a_15\leq250</math>
  
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We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two–<math>241-20=221</math>, and the maximum–<math>250-13=237</math>. There are <math>13</math> 7 differences between them, so only <math>17</math> and <math>18</math> work, as <math>17*13=221</math>, so <math>17</math> satisfied <math>221\leq 13x\leq237</math>. The number <math>18</math> is similarly found. <math>19</math>, however, is too much.
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Now, we check with the first and last equations using the same method. We know <math>241-10\leq 14x\leq250-1</math>. Therefore, <math>231\leq 14x\leq249</math>. We test both values we just got, and we can realize that <math>18</math> is too large to satisfy this inequality. On the other hand, we can now find that the difference will be <math>17</math>, which satisfies this inequality.
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The last step is to find the first term. We know that the first term can only be from <math>1</math> to <math>3</math>, since any larger value would render the second inequality invalid. Testing these three, we find that only <math>a_1=3</math> will satisfy all the inequalities. Therefore, <math>a_14=13\cdot17+3=224</math>. The sum of the digits is therefore <math>\boxed(D){8}</math>.
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-apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Revision as of 18:06, 24 January 2023

$1\leq a_1\leq10$ $13\leq a_2\leq20$ $241\leq a_15\leq250$

We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two–$241-20=221$, and the maximum–$250-13=237$. There are $13$ 7 differences between them, so only $17$ and $18$ work, as $17*13=221$, so $17$ satisfied $221\leq 13x\leq237$. The number $18$ is similarly found. $19$, however, is too much.

Now, we check with the first and last equations using the same method. We know $241-10\leq 14x\leq250-1$. Therefore, $231\leq 14x\leq249$. We test both values we just got, and we can realize that $18$ is too large to satisfy this inequality. On the other hand, we can now find that the difference will be $17$, which satisfies this inequality.

The last step is to find the first term. We know that the first term can only be from $1$ to $3$, since any larger value would render the second inequality invalid. Testing these three, we find that only $a_1=3$ will satisfy all the inequalities. Therefore, $a_14=13\cdot17+3=224$. The sum of the digits is therefore $\boxed(D){8}$. -apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

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