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Difference between revisions of "2020 CIME I Problems/Problem 12"
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==Solution==
 
==Solution==
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{{solution}}
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We simply test out some values first. Denote <math>P(n)=\prod_{i=0}^n a_i</math>; we must find <math>P(10)</math>. Additionally, notice that <math>P(n)=a_nP(n-1)</math>.
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We clearly see that <math>P(0)=a_0=11_2</math>. The sum of the digits is <math>2</math>.
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Next, we can quickly calculate that <math>P(1)=a_1P(0)=100111_2</math>, whose sum of digits is <math>4</math>.
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Then, we can find (after more effort) that <math>P(2)=a_2P(1)=10010010110111_2</math>, whose sum of digits is <math>8</math>.
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The above hints towards a pattern; denote the sum of digits of <math>P(n)</math> as <math>S(n)</math>; we quickly realize due to the <math>2,4,8</math> pattern that we can guess <math>S(n)=2^{n+1}</math>. Calculating <math>P(3)</math> confirms this theory, so we assume that it is true and plug in:
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<cmath>S(10)=2^{11}=2048\rightarrow \boxed{048}</cmath>
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
==See also==
 
==See also==
 
{{CIME box|year=2020|n=I|num-b=11|num-a=13}}
 
{{CIME box|year=2020|n=I|num-b=11|num-a=13}}
 
{{MAC Notice}}
 
{{MAC Notice}}

Latest revision as of 02:08, 16 April 2025

Problem 12

Define a sequence $a_0, a_1, a_2, ...$ by \[a_i=\underbrace{1\ldots1}_{2^{i}\text{ 1's}}\underbrace{0\ldots0}_{(2^i-1)\text{ 0's}}1_2,\] where $a_i$ is expressed in binary. Let $S$ be the sum of the digits when $a_0 a_1 a_2 \cdots a_{10}$ is expressed in binary. Find the remainder when $S$ is divided by $1000$.

Solution

We simply test out some values first. Denote $P(n)=\prod_{i=0}^n a_i$; we must find $P(10)$. Additionally, notice that $P(n)=a_nP(n-1)$.

We clearly see that $P(0)=a_0=11_2$. The sum of the digits is $2$.

Next, we can quickly calculate that $P(1)=a_1P(0)=100111_2$, whose sum of digits is $4$.

Then, we can find (after more effort) that $P(2)=a_2P(1)=10010010110111_2$, whose sum of digits is $8$.

The above hints towards a pattern; denote the sum of digits of $P(n)$ as $S(n)$; we quickly realize due to the $2,4,8$ pattern that we can guess $S(n)=2^{n+1}$. Calculating $P(3)$ confirms this theory, so we assume that it is true and plug in: \[S(10)=2^{11}=2048\rightarrow \boxed{048}\]

~ eevee9406

See also

2020 CIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions

Template:MAC Notice

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