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Difference between revisions of "2004 AIME I Problems/Problem 2"
m (→‎Solution 2)
Line 18: Line 18:
 
First, we note that for set <math>A</math>
 
First, we note that for set <math>A</math>
  
−
<cmath>\frac{m(f + l)}{2} = 2m</cmath>,
+
<cmath>\frac{m(f + l)}{2} = 2m</cmath>
  
 
Where <math>f</math> and <math>l</math> represent the first and last terms of <math>A</math>.  This comes from the sum of an arithmetic sequence.
 
Where <math>f</math> and <math>l</math> represent the first and last terms of <math>A</math>.  This comes from the sum of an arithmetic sequence.
Line 26: Line 26:
 
Doing the same for set B, and setting up the equation with <math>b</math> and <math>e</math> being the first and last terms of set <math>B</math>,  
 
Doing the same for set B, and setting up the equation with <math>b</math> and <math>e</math> being the first and last terms of set <math>B</math>,  
  
−
<cmath>m(b+e) = m</cmath>.
+
<cmath>m(b+e) = m</cmath>
  
 
and so <math>b+e = 1</math>
 
and so <math>b+e = 1</math>

Revision as of 18:56, 6 February 2015

Problem

Set $A$ consists of $m$ consecutive integers whose sum is $2m,$and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$. Find $m.$

Solution

Let us give the elements of our sets names: $A = \{x, x + 1, x + 2, \ldots, x + m - 1\}$ and $B = \{y, y + 1, \ldots, y + 2m - 1\}$. So we are given that \[2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,\] so $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$. Also, \[m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,\] so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$.

Then by the given, $99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|$. $m$ is a positive integer so we must have $99 = \frac{m - 3}2$ and so $m = \boxed{201}$.

Solution 2

The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.

First, we note that for set $A$

\[\frac{m(f + l)}{2} = 2m\]

Where $f$ and $l$ represent the first and last terms of $A$. This comes from the sum of an arithmetic sequence.

Solving for $f+l$, we find the sum of the two terms is $4$

Doing the same for set B, and setting up the equation with $b$ and $e$ being the first and last terms of set $B$,

\[m(b+e) = m\]

and so $b+e = 1$

Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set $A$ has half the number of elements as set $B$, and the difference between the greatest terms of the two two sequences is $99$ (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where $x$ is the last term of set A:

\[2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)\]

Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to $4$ and $1$ respectively (add $x$ and $(-x+4)$ to see what i mean).

Solving this equation we find $x = 102$. We know the first and last terms have to sum to $4$ so we find the first term of the sequence is $-98$. Now, the solution is in clear sight, we just find the number of integers between $-98$ and $102$, inclusive, and it is $201$.

Note how this method is not very algebra heavy. It seems like a lot by the amount of text but really the first two steps are quite simple.

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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