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Difference between revisions of "2002 AMC 12P Problems/Problem 14"

(Created page with "== Problem == How many positive integers <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer? <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B...")
 
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== Problem ==
 
== Problem ==
−
How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
+
Find <math>i + 2i^2 +3i^3 + ... + 2002i^{2002}.</math>
  
−
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math>
+
<math>
 +
\text{(A) }-999 + 1002i
 +
\qquad
 +
\text{(B) }-1002 + 999i
 +
\qquad
 +
\text{(C) }-1001 + 1000i
 +
\qquad
 +
\text{(D) }-1002 + 1001i
 +
\qquad
 +
\text{(E) }i
 +
</math>
  
−
== Solution ==
+
== Solution 1 ==
−
If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+
Note that <math>i^4 = 1</math>, so <math>i^n = i^{4m+n}</math> for all integers <math>m</math> and <math>n</math>. In particular, <math>i = 1</math>, <math>i^2 = -1</math>, and <math>i^3 = -i</math>.
 +
We group the positive and negative real terms together and group the positive and negative imaginary parts together.
 +
 
 +
The positive real terms have exponents on <math>i</math> that are multiples of 4. Therefore, the positive real part evaluates to  
 +
<cmath>4 + 8 + ... + 2000</cmath>
 +
The negative real terms have exponents on <math>i</math> that are of the form <math>4k + 2</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>-(2 + 6 + ... + 2002)</cmath>
 +
The positive imaginary terms have exponents on <math>i</math> that are of the form <math>4k + 1</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>(1 + 5 + ... + 2001)i</cmath>
 +
The negative imaginary terms have exponents on <math>i</math> that are of the form <math>4k + 3</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>-(3 + 7 + ... + 1999)i</cmath>
 +
 
 +
Putting everything together, we have <cmath>i + 2i^2 + ... + 2002i^{2002} = (-2 + 4 - 6 + 8 - ... + 2000 - 2002) + (1 - 3 + 5 - 7 + ... - 1999 + 2001)i</cmath>
 +
 
 +
Group every 2 consecutive terms as shown below <cmath>((-2 + 4)+(-6 + 8) + ... + (-1998 + 2000)-2002) + ((1 - 3)+(5 - 7) + ... + (1997 - 1999) + 2001)i</cmath>
 +
 
 +
Now we evaluate each small bracket with 2 terms. We get <math>500(2) = 1000</math> in the real part and <math>500(-2) = -1000</math> in the imaginary part. Therefore, the sum becomes <math>(1000 - 2002) + (-1000 + 2001)i = \boxed {\text{(D) }-1002 + 1001i}</math>.
 +
 
 +
Note: This problem is similar to [[2009 AMC 12A Problems/Problem 15|2009 AMC 12A Problem 15]]
  
 
== See also ==
 
== See also ==
−
{{AMC12 box|year=2000|num-b=6|num-a=8}}
+
{{AMC12 box|year=2002|ab=P|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:10, 15 July 2024

Problem

Find $i + 2i^2 +3i^3 + ... + 2002i^{2002}.$

$\text{(A) }-999 + 1002i \qquad \text{(B) }-1002 + 999i \qquad \text{(C) }-1001 + 1000i \qquad \text{(D) }-1002 + 1001i \qquad \text{(E) }i$

Solution 1

Note that $i^4 = 1$, so $i^n = i^{4m+n}$ for all integers $m$ and $n$. In particular, $i = 1$, $i^2 = -1$, and $i^3 = -i$. We group the positive and negative real terms together and group the positive and negative imaginary parts together.

The positive real terms have exponents on $i$ that are multiples of 4. Therefore, the positive real part evaluates to \[4 + 8 + ... + 2000\] The negative real terms have exponents on $i$ that are of the form $4k + 2$ for integers $k$. Therefore, the negative real part evaluates to \[-(2 + 6 + ... + 2002)\] The positive imaginary terms have exponents on $i$ that are of the form $4k + 1$ for integers $k$. Therefore, the negative real part evaluates to \[(1 + 5 + ... + 2001)i\] The negative imaginary terms have exponents on $i$ that are of the form $4k + 3$ for integers $k$. Therefore, the negative real part evaluates to \[-(3 + 7 + ... + 1999)i\]

Putting everything together, we have \[i + 2i^2 + ... + 2002i^{2002} = (-2 + 4 - 6 + 8 - ... + 2000 - 2002) + (1 - 3 + 5 - 7 + ... - 1999 + 2001)i\]

Group every 2 consecutive terms as shown below \[((-2 + 4)+(-6 + 8) + ... + (-1998 + 2000)-2002) + ((1 - 3)+(5 - 7) + ... + (1997 - 1999) + 2001)i\]

Now we evaluate each small bracket with 2 terms. We get $500(2) = 1000$ in the real part and $500(-2) = -1000$ in the imaginary part. Therefore, the sum becomes $(1000 - 2002) + (-1000 + 2001)i = \boxed {\text{(D) }-1002 + 1001i}$.

Note: This problem is similar to 2009 AMC 12A Problem 15

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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