2023 AMC 8 Problems/Problem 25

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 $\boxed{\textbf{(A)}\ 8}$ . ~eibc

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2023 AMC 8 Problems/Problem 25

Solution 1

Solution 2