Difference between revisions of "1989 AIME Problems/Problem 3"
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<center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center> | <center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center> | ||
− | == Solution == | + | == Solution 1 == |
− | Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], <math>0.d25d25d25\ldots = \sum_{ | + | We can express <math>0.\overline{d25}</math> as <math>\frac{100d+25}{999}</math>. We set up the given equation and isolate <math>n:</math> \begin{align*} \frac{100d+25}{999} &= \frac{n}{810}, \\ \frac{100d+25}{111} &= \frac{n}{90}, \\ 9000d + 2250 &= 111n. \end{align*} We then set up the following modular congruence to solve for <math>d:</math> \begin{align*}9000d + 2250 &\equiv 0 \pmod {111}, \\ 9d + 30 &\equiv 0 \pmod {111}, \\ 3d + 10 &\equiv 0 \pmod {37}, \\ 36d + 120 &\equiv 0 \pmod {37}, \\ -d &\equiv -9 \pmod {37}, \\ d &\equiv 9 \pmod {37}.\end{align*} Since <math>d</math> is a digit, it must be equal to <math>9</math> based on our above constraint. When <math>d=9, n = \boxed{750}.</math> |
+ | |||
+ | ~vaisri | ||
+ | |||
+ | == Solution 2 == | ||
+ | Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], <math>0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}</math>. Thus <math>\frac{n}{810} = \frac{100d + 25}{999}</math> so <math>n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}</math>. Since 750 and 37 are [[relatively prime]], <math>4d + 1</math> must be [[divisible]] by 37, and the only digit for which this is possible is <math>d = 9</math>. Thus <math>4d + 1 = 37</math> and <math>n = \boxed{750}</math>. | ||
(Note: Any repeating sequence of <math>n</math> digits that looks like <math>0.a_1a_2a_3...a_{n-1}a_na_1a_2...</math> can be written as <math>\frac{a_1a_2...a_n}{10^n-1}</math>, where <math>a_1a_2...a_n</math> represents an <math>n</math> digit number.) | (Note: Any repeating sequence of <math>n</math> digits that looks like <math>0.a_1a_2a_3...a_{n-1}a_na_1a_2...</math> can be written as <math>\frac{a_1a_2...a_n}{10^n-1}</math>, where <math>a_1a_2...a_n</math> represents an <math>n</math> digit number.) | ||
− | == Solution | + | == Solution 3 == |
To get rid of repeating decimals, we multiply the equation by 1000. We get <math>\frac{1000n}{810} = d25.d25d25...</math> We subtract the original equation from the second to get <math>\frac{999n}{810}=d25</math> We simplify to <math>\frac{37n}{30} = d25</math> Since <math>\frac{37n}{30}</math> is an integer, <math>n=(30)(5)(2k+1)</math> because <math>37</math> is relatively prime to <math>30</math>, and d25 is divisible by <math>5</math> but not <math>10</math>. The only odd number that yields a single digit <math>d</math> and 25 at the end of the three digit number is <math>k=2</math>, so the answer is <math>\boxed{750}</math>. | To get rid of repeating decimals, we multiply the equation by 1000. We get <math>\frac{1000n}{810} = d25.d25d25...</math> We subtract the original equation from the second to get <math>\frac{999n}{810}=d25</math> We simplify to <math>\frac{37n}{30} = d25</math> Since <math>\frac{37n}{30}</math> is an integer, <math>n=(30)(5)(2k+1)</math> because <math>37</math> is relatively prime to <math>30</math>, and d25 is divisible by <math>5</math> but not <math>10</math>. The only odd number that yields a single digit <math>d</math> and 25 at the end of the three digit number is <math>k=2</math>, so the answer is <math>\boxed{750}</math>. | ||
− | ==Solution | + | ==Solution 4== |
Similar to Solution 2, we start off by writing that <math> \frac{1000n}{810} = d25.d25d25 \dots</math> .Then we subtract this from the original equation to get: | Similar to Solution 2, we start off by writing that <math> \frac{1000n}{810} = d25.d25d25 \dots</math> .Then we subtract this from the original equation to get: | ||
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Since n is an integer, we have that <math>37 \mid d25 \cdot 30</math>. | Since n is an integer, we have that <math>37 \mid d25 \cdot 30</math>. | ||
− | Since <math>37</math> is prime, we can apply Euclid's Lemma | + | Since <math>37</math> is prime, we can apply [[Euclid's Lemma]] to realize that <math>37 \mid d25 </math>, since <math>37 \nmid 30</math>. Then we can expand <math>d25</math> as <math>25 \cdot (4d +1)</math>. Since <math>37 \nmid 25 </math>, by Euclid, we can arrive at <math>37 \mid 4d+1 \Longrightarrow d=9</math>. From this we know that <math>n= 25 \cdot 30 = \boxed{750}</math>. (This is true because <math>37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750</math>) |
~qwertysri987 | ~qwertysri987 | ||
− | + | ==Solution 5== | |
− | ==Solution | + | Write out these equations: |
− | Write out | ||
− | <math>\frac{n}{ | + | <math>\frac{n}{810} = \frac{d25}{999}</math> |
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-jackshi2006 | -jackshi2006 | ||
+ | |||
+ | Note: We know <math>n < 810</math> since <math>\frac{n}{810} < 1,</math> so it suffices to check for numbers under <math>810</math> not <math>1000.</math> | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/1-iWPCWPsLw?t=600 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Latest revision as of 14:38, 11 August 2024
Contents
Problem
Suppose is a positive integer and
is a single digit in base 10. Find
if
![$\frac{n}{810}=0.d25d25d25\ldots$](http://latex.artofproblemsolving.com/0/d/2/0d2ffb8013a922ca8e69ae1b3c9575aa9b68e7bf.png)
Solution 1
We can express as
. We set up the given equation and isolate
\begin{align*} \frac{100d+25}{999} &= \frac{n}{810}, \\ \frac{100d+25}{111} &= \frac{n}{90}, \\ 9000d + 2250 &= 111n. \end{align*} We then set up the following modular congruence to solve for
\begin{align*}9000d + 2250 &\equiv 0 \pmod {111}, \\ 9d + 30 &\equiv 0 \pmod {111}, \\ 3d + 10 &\equiv 0 \pmod {37}, \\ 36d + 120 &\equiv 0 \pmod {37}, \\ -d &\equiv -9 \pmod {37}, \\ d &\equiv 9 \pmod {37}.\end{align*} Since
is a digit, it must be equal to
based on our above constraint. When
~vaisri
Solution 2
Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, . Thus
so
. Since 750 and 37 are relatively prime,
must be divisible by 37, and the only digit for which this is possible is
. Thus
and
.
(Note: Any repeating sequence of digits that looks like
can be written as
, where
represents an
digit number.)
Solution 3
To get rid of repeating decimals, we multiply the equation by 1000. We get We subtract the original equation from the second to get
We simplify to
Since
is an integer,
because
is relatively prime to
, and d25 is divisible by
but not
. The only odd number that yields a single digit
and 25 at the end of the three digit number is
, so the answer is
.
Solution 4
Similar to Solution 2, we start off by writing that .Then we subtract this from the original equation to get:
Since n is an integer, we have that .
Since is prime, we can apply Euclid's Lemma to realize that
, since
. Then we can expand
as
. Since
, by Euclid, we can arrive at
. From this we know that
. (This is true because
)
~qwertysri987
Solution 5
Write out these equations:
Thus divides 25 and 30. The only solution for this under 1000 is
.
-jackshi2006
Note: We know since
so it suffices to check for numbers under
not
Video Solution by OmegaLearn
https://youtu.be/1-iWPCWPsLw?t=600
~ pi_is_3.14
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.