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Difference between revisions of "2023 AMC 8 Problems/Problem 25"
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==Solution 1==
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<math>1\leq a_1\leq10</math>,
 
<math>1\leq a_1\leq10</math>,
  
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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{\text{(D)}11}</math>
 
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{\text{(D)}11}</math>
 
-apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 
-apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 +
 +
==Solution 2==
 +
Let the common difference between consecutive <math>a_i</math> be <math>d</math>. Then, since <math>a_15 - a_1 = 14d</math>, we find from the first and last inequalities that <math>231 \le d \le 249</math>. As <math>d</math> must be an integer, this means <math>d = 17</math>. Plugging this into all of the given inequalities so we may extract information about <math>a_1</math> gives
 +
<cmath>1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.</cmath>
 +
The second inequality tells us that <math>a_1 \le 3</math>, while the last inequality tells us <math>3 \le a_1</math>, so we must have <math>a_1 = 3</math>. Finally, to solve for <math>a_14</math>, we simply have <math>a_14 = a_1 + 13d = 3 + 221 = 224</math>, so our answer is <math>\boxed{\textbf{(A)}\ 8}</math>.

Revision as of 18:11, 24 January 2023

Solution 1

$1\leq a_1\leq10$,

$13\leq a_2\leq20$,

$241\leq a_15\leq250$,

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 $\boxed{\text{(D)}11}$ -apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Solution 2

Let the common difference between consecutive $a_i$ be $d$. Then, since $a_15 - a_1 = 14d$, we find from the first and last inequalities that $231 \le d \le 249$. As $d$ must be an integer, this means $d = 17$. Plugging this into all of the given inequalities so we may extract information about $a_1$ gives \[1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.\] The second inequality tells us that $a_1 \le 3$, while the last inequality tells us $3 \le a_1$, so we must have $a_1 = 3$. Finally, to solve for $a_14$, we simply have $a_14 = a_1 + 13d = 3 + 221 = 224$, so our answer is $\boxed{\textbf{(A)}\ 8}$.

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