Difference between revisions of "1992 AIME Problems/Problem 8"

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Since all terms of <math>\Delta(\Delta A)</math> are 1, we know that <math>\Delta A</math> looks like <math>(k,k+1,k+2,...)</math> for some <math>k</math>. This means <math>A</math> looks like <math>(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)</math>. More specifically, <math>A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}</math>. Plugging in <math>a_{19}=a_{92}=0</math>, we have the following linear system: <cmath>a_1+91k=-4095</cmath> <cmath>a_1+18k=-153</cmath> From this, we can easily find that <math>k=-54</math> and <math>a_1=\boxed{819}</math>.
 
Since all terms of <math>\Delta(\Delta A)</math> are 1, we know that <math>\Delta A</math> looks like <math>(k,k+1,k+2,...)</math> for some <math>k</math>. This means <math>A</math> looks like <math>(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)</math>. More specifically, <math>A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}</math>. Plugging in <math>a_{19}=a_{92}=0</math>, we have the following linear system: <cmath>a_1+91k=-4095</cmath> <cmath>a_1+18k=-153</cmath> From this, we can easily find that <math>k=-54</math> and <math>a_1=\boxed{819}</math>.
 
Solution by Zeroman
 
Solution by Zeroman
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== Solution 5 ==
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Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that <math>f(19) = f(92) = 0</math> so the quadratic is <math>f(x) = a(x-19)(x-92)</math> for some constant <math>a.</math> Now we use the conditions that the finite difference is <math>1</math> to find <math>a.</math> We know <math>f(19) = 0</math> and <math>f(20) = -72a</math> and <math>f(18) = 74a.</math> Therefore applying finite differences once yields the sequence <math>-74a,-72a</math> and then applying finite differences one more time yields <math>2a</math> so <math>a = \displaystyle\frac{1}{2}.</math> Therefore <math>f() = 9 \cdot 91 = \boxed{819}.</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=7|num-a=9}}
 
{{AIME box|year=1992|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:29, 8 September 2020

Problem

For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$, define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$, whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$. Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$, and that $a_{19}=a_{92}^{}=0$. Find $a_1^{}$.

Solution 1 (uses calculus)

Note that the $\Delta$s are reminiscent of differentiation; from the condition $\Delta(\Delta{A}) = 1$, we are led to consider the differential equation \[\frac{d^2 A}{dn^2} = 1\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \[a_{n} = \frac{1}{2}(n-19)(n-92)\] as we must have roots at $n = 19$ and $n = 92$.

Thus, $a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}$.

Solution 2

Let $\Delta^1 A=\Delta A$, and $\Delta^n A=\Delta(\Delta^{(n-1)}A)$.

Note that in every sequence of $a_i$, $a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...$

Then $a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...$

Since $\Delta a_1 =a_2 -a_1$, $a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}$

$a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153$

$a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095$

Solving, $a_1=\boxed{819}$.

Solution 3

The sequence $\Delta(\Delta A)$ is the second finite difference sequence, and the first $k-1$ terms of this sequence can be computed in terms of the original sequence as shown below.

$\begin{array}{rcl} a_3+a_1-2a_2&=&1\\ a_4+a_2-2a_3&=&1\\ &\vdots\\ a_k + a_{k-2} - 2a_{k-1} &= &1\\ a_{k+1} + a_{k-1} - 2a_k &=& 1.\\ \end{array}$

Adding the above $k-1$ equations we find that

\[(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}\]

We can sum equation $(1)$ from $k=1$ to $18$, finding \[18(a_1-a_2) - a_1 = 153.\tag{2}\]

We can also sum equation $(1)$ from $k=1$ to $91$, finding \[91(a_1-a_2) - a_1 = 4095.\tag{3}\] Finally, $18\cdot (3) - 91\cdot(2)$ gives $a_1=\boxed{819}$.

Kris17

Solution 4

Since all terms of $\Delta(\Delta A)$ are 1, we know that $\Delta A$ looks like $(k,k+1,k+2,...)$ for some $k$. This means $A$ looks like $(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)$. More specifically, $A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}$. Plugging in $a_{19}=a_{92}=0$, we have the following linear system: \[a_1+91k=-4095\] \[a_1+18k=-153\] From this, we can easily find that $k=-54$ and $a_1=\boxed{819}$. Solution by Zeroman


Solution 5

Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that $f(19) = f(92) = 0$ so the quadratic is $f(x) = a(x-19)(x-92)$ for some constant $a.$ Now we use the conditions that the finite difference is $1$ to find $a.$ We know $f(19) = 0$ and $f(20) = -72a$ and $f(18) = 74a.$ Therefore applying finite differences once yields the sequence $-74a,-72a$ and then applying finite differences one more time yields $2a$ so $a = \displaystyle\frac{1}{2}.$ Therefore $f() = 9 \cdot 91 = \boxed{819}.$

See also

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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