Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 15"
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− | We can easily find that <math>\ | + | We can easily find that <math>\tfrac{f(1)}{25}=19,\tfrac{f(2)}{25}=191,\tfrac{f(3)}{25}=1911,</math> and so on. Thus, we claim<math>\text{}^*</math> that <cmath>\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{n-1 \text{ ones}}.</cmath> Now, we find we can easily find that <cmath>\frac{f(1)+f(2)+ \cdots + f(100)}{25}\equiv19+191+911+111 \cdot 97 \equiv 11888 \pmod{1000} \rightarrow \boxed{888}.</cmath> |
− | <math>\text{}^*</math> | + | <math>\text{}^*</math> We proceed by induction. Our base case is <math>\tfrac{f(1)}{25}=\tfrac{475}{25}=19.</math> Our inductive assumption is <cmath>\frac{f(n)}{25}=19\underbrace{111 \cdots 1}_{n-1 \text{ ones}},</cmath> and we wish to prove that this pattern holds for <math>f(n+1).</math> |
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− | ~pinkpig | + | We can easily find that <math>f(n+1)=10f(n)+25.</math> Using our inductive assumption, we obtain <cmath>\frac{f(n+1)}{25}=10 \cdot (19\underbrace{111 \cdots 1}_{n-1 \text{ ones}})+1=19 \cdot \underbrace{111 \cdots 1}_{n-1 \text{ ones}},</cmath> as desired. <math>\mathbb{Q.E.D.}</math> |
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+ | ~Solution by pinkpig, <math>\LaTeX</math>/wording fixes by samrocksnature | ||
==Solution 2 (More Algebraic)== | ==Solution 2 (More Algebraic)== | ||
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<math>\linebreak</math> | <math>\linebreak</math> | ||
~Geometry285 | ~Geometry285 | ||
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+ | ==See also== | ||
+ | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] | ||
+ | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 00:19, 12 July 2021
Problem
For all positive integers define the function to output For example, , , and Find the last three digits of
Solution
We can easily find that and so on. Thus, we claim that Now, we find we can easily find that
We proceed by induction. Our base case is Our inductive assumption is and we wish to prove that this pattern holds for
We can easily find that Using our inductive assumption, we obtain as desired.
~Solution by pinkpig, /wording fixes by samrocksnature
Solution 2 (More Algebraic)
We only care about the last digits, so we evaluate . Note the expression is simply , so factoring a we have . Now, we can divide by to get Evaluate the last digits to get ~Geometry285
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.