2025 AMC 10A Problems/Problem 19: Difference between revisions
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Knowing this, we can use this pattern to quickly find the solution. We know that the first number will always be -1, so we can ignore that. Knowing this, all you have to do now is to add up the second and third numbers 13 times. This is already done for us three times, so we just have to do it ten more. Note that you do not have to do this process for all the numbers, only the second and third. | Knowing this, we can use this pattern to quickly find the solution. We know that the first number will always be -1, so we can ignore that. Knowing this, all you have to do now is to add up the second and third numbers 13 times. This is already done for us three times, so we just have to do it ten more. Note that you do not have to do this process for all the numbers, only the second and third. | ||
Doing this 13 times, you get the following string of numbers (starting from the first one): 1, 4, 6, 7, 6, 4, 1, -3, -8, -14, -21, and finally, <imath>\boxed{\text{(A) }-29}.</imath> | |||
Doing this 13 times(since <imath>2^12=\frac{12288}{3}=4096</imath>), you get the following string of numbers (starting from the first one): 1, 4, 6, 7, 6, 4, 1, -3, -8, -14, -21, and finally, <imath>\boxed{\text{(A) }-29}.</imath> | |||
~i_am_not_suk_at_math | ~i_am_not_suk_at_math | ||
Revision as of 16:46, 6 November 2025
(Problem goes here)
Solution 1
Consider the polynomial 
When we multiply this polynomial by 
we are essentially doing the operation given in the problem (When we multiply 
by a term of degree 
in the yielded expression is the sum of 
and 
in This effect is visible in Pascal's Triangle).
So, if we let the coefficients of 
be the zero row of the array, then the 
row is just the coefficients of 
The next thing to note is that the sum of the coefficients in any polynomial is just 
Therefore, the sum of the entries in the row of the array is 
Letting this equal 
we get 
We are looking for the 
term in the 
row.
The row is given by the coefficients of 
Since the degree of the resulting expression is 
the third term in the row is just the coefficient of 
in the expression, which is 
~Tacos_are_yummy_1
Solution 2
If we take a look at the first few rows, we notice that the sum of the terms in each row 
is equal to the twice the sum of row 
. We note the first row is 
so recognize 
must be equal to times a power of 
. 
. Therefore, we are looking for the rd term from the left in the 
th row. From here, we
~Squidget(note: this solution is incomplete, will complete soon)
Solution 3 - ⚡ Fast
Add all the numbers up on the first row. You get 3. Add all the numbers up on the second row. You get 6. Notice that as the rows keep going, that number keeps doubling. When you repeat this process, you realize that you reach 12288 on the 
one.
Knowing this, we can use this pattern to quickly find the solution. We know that the first number will always be -1, so we can ignore that. Knowing this, all you have to do now is to add up the second and third numbers 13 times. This is already done for us three times, so we just have to do it ten more. Note that you do not have to do this process for all the numbers, only the second and third.
Doing this 13 times(since 
), you get the following string of numbers (starting from the first one): 1, 4, 6, 7, 6, 4, 1, -3, -8, -14, -21, and finally, 
~i_am_not_suk_at_math
