Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1959 IMO Problems/Problem 1"
(→‎Solution 1)
 
(29 intermediate revisions by 21 users not shown)
Line 3: Line 3:
 
Prove that the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>.
 
Prove that the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>.
  
 +
== Solution 1 ==
  
−
== Solutions ==
+
Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]]:
  
−
=== First Solution ===
+
<cmath>(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1</cmath>
  
−
We observe that
+
Their greatest common divisor is 1, so <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
  
−
<center>
+
== Solution 2 ==
−
<math>3(14n+3) = 2(21n+4) + 1.</math>
 
−
</center>
 
  
 +
[[Proof by contradiction]]:
  
−
Since a multiple of <math>14n+3</math> differs from a multiple of <math>21n+4</math> by 1, we cannot have any postive integer greater than 1 simultaneously divide <math>14n+3</math> and <math>21n+4</math>.  Hence the [[greatest common divisor]] of the fraction's numerator and denominator is 1, so the fraction is irreducible.  Q.E.D.
+
Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is the [[greatest common factor]] of <math>14n+3</math> and <math>21n + 4</math>. Thus,
−
 
 
−
=== Second Solution ===
 
−
 
 
−
Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]] as follows:
 
−
 
 
 
<center>
 
<center>
−
<math>( 21n+4, 14n+3 ) = ( 7n+1, 14n+3 ) = ( 7n+1, 1 ) = 1 </math>
+
<math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math>
 
</center>
 
</center>
  
−
As in the first solution, it follows that <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
+
Subtracting the second [[equation]] from the first equation we get <math>1\equiv 0\pmod{p}</math> which is clearly absurd.  
  
−
=== Third Solution ===
+
Hence <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
 +
<!--Solution by tonypr-->
 +
 
 +
== Solution 3 ==
  
 
[[Proof by contradiction]]:
 
[[Proof by contradiction]]:
  
−
Let's assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is a [[divisor]] of both the [[numerator]] and the [[denominator]]:
+
Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]].
−
 
 
−
<center>
 
−
<math>14n+3\equiv 0\pmod{p} \implies 42n+9\equiv 0\pmod{p}</math>
 
  
−
<math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math>
+
If a certain fraction <math>\dfrac{a}{b}</math> is reducible, then the fraction <math>\dfrac{2a}{3b}</math> is reducible, too.
−
</center>
+
In this case, <math>\dfrac{2a}{3b} = \dfrac{42n+8}{42n+9}</math>.
  
−
Subtracting the second [[equation]] from the first [[equation]] we get <math>1\equiv 0\pmod{p}</math> which is clearly absurd.  
+
This fraction consists of two consecutives numbers, which never share any factor. So in this case, <math>\dfrac{2a}{3b}</math> is irreducible, which is absurd.  
  
 
Hence <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
 
Hence <math>\frac{21n+4}{14n+3}</math> is irreducible.  Q.E.D.
−
<!--Solution by tonypr-->
+
<!--Solution by juanfkt93 - Juan Friss de Kereki-->
  
  
−
=== Fourth Solution ===
+
== Solution 4 ==
−
Let <math>g = {\rm gcd}(21n + 4, 14n + 3)</math>.  Then <math>g|h</math> where <math>h = {\rm gcd}(42n + 8, 14n + 3) = {\rm gcd}(1, 14n + 3) = 1</math>.  Thus, <math>g = h = 1</math>.  ''Note: This solution, in hindsight, is just the first solution above in a slightly different notation.''
 
−
 
 
−
=== Fifth Solution ===
 
  
 
We notice that:
 
We notice that:
Line 62: Line 54:
 
</center>
 
</center>
  
−
Since the denominator <math> 2(7n+1) + 1 </math> differs from a multiple of the numerator <math>7n+1</math> by 1, then the numerator and the denominator must be relatively prime natural numbers. Hence it follows that <math> \frac{21n+4}{14n+3}</math> is irreducible.
+
Since the denominator <math> 2(7n+1) + 1 </math> differs from a multiple of the numerator <math>7n+1</math> by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that <math> \frac{21n+4}{14n+3}</math> is irreducible.
  
 
Q.E.D
 
Q.E.D
  
−
~ Solution by '''''z6z61im4s99'''''
 
  
−
<center>
+
== Solution 5 ==
−
'''SIXTH SOLUTION'''
+
By [[Bezout's Lemma]], <math>3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1</math>, so the GCD of the numerator and denominator is <math>1</math> and the fraction is irreducible.
−
</center>
 
  
−
we can use PRINCIPLE OF MATHEMATICAL INDUCTION TO EASILY PROVE THE ABOVE STATEMENT.  
+
== Solution 6 ==
 +
To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with <math> \frac{21n}{14n} </math> + <math> \frac{4}{3} </math>.
 +
Simplifying the fraction, we end up with.    <math> \frac{3}{2} </math>. Now combining these fractions with addition as shown before in the problem, we end up with <math> \frac{17}{6} </math>. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us
 +
<math> \frac{25}{17} </math>. Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16).
 +
Trying 2, we end up with <math> \frac{46}{31} </math>. Again, both results are 1 off from being multiples of 15.
 +
Trying 3, we end up with <math> \frac{67}{45} </math>. Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:
  
−
First, we show that the statement is true for n=1. For n=1, 25/17  is irreducible because gcd(25,17) =1.
+
1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but
  
 +
2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us <math> \frac{319}{199} </math>, we keep ending up with results that at the end will only have a common divisor of 1.
  
−
Now, we show that '''If the statement is true for the natural number 'n', then it is also true for natural number 'n+1'.'''
 
  
 +
== Solution 7 ==
 +
Let <math>\gcd (21n+4,14n+3)=a.</math> So for some co-prime positive integers <math>x,y</math> we have <cmath>21n+4 = ax \qquad (1)</cmath> <cmath>14n+3 = ay \qquad (2)</cmath> Multiplying <math>(1)</math> by <math>2</math> and <math>(2)</math> by <math>3</math> and then subtracting <math>(1)</math> from <math>(2)</math> we get <cmath>42n+9-(42n+8)=3ay-2ax</cmath> <cmath>\implies a(3y-2x)=1.</cmath> We must have <math>a=1</math>, since <math>a</math> is an positive integer. Thus, <math>\gcd(21n+4,14n+3)=1</math> which means the fraction is irreducible, as needed.
  
−
PROOF:
+
== Video Solution  by OmegaLearn==
 +
https://youtu.be/zfChnbMGLVQ?t=1266
  
−
Let the given statement be true for the natural number 'n', which means gcd (21n+4, 14n+3) =1
+
~ pi_is_3.14
  
−
Now we need to show that the statement is true for natural number 'n+1', which means we need to show gcd (21n+25, 14n+17) =1
+
https://www.youtube.com/watch?v=IpUh5cLP454&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=1
 +
- AMBRIGGS
  
−
let us take a look at that, 
+
https://youtu.be/Pfg4MVUML64
  
−
gcd (21n+25, 14n+17) =1
+
- little fermat
  
−
-> gcd [21n+4+(7*3), 14n+3+(7*2)] = 1, which is true since gcd (21n+25, 14n+17) =1 and adding some multiple of some number 'k=7' to both the numbers doesn't change the gcd.
+
{{alternate solutions}}
  
−
Hence, the statement is true for natural number 'n+1'.
 
−
 
−
Hence, the statement is true for each natural number.
 
−
 
−
 
−
 
−
For this fraction to be irreducible,
 
−
{{alternate solutions}}
 
 
== See Also ==
 
== See Also ==
 
{{IMO box|year=1959|before=First question|num-a=2}}
 
{{IMO box|year=1959|before=First question|num-a=2}}

Latest revision as of 09:59, 23 July 2023

Problem

Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$.

Solution 1

Denoting the greatest common divisor of $a, b$ as $(a,b)$, we use the Euclidean algorithm:

\[(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1\]

Their greatest common divisor is 1, so $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.

Solution 2

Proof by contradiction:

Assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction where $p$ is the greatest common factor of $14n+3$ and $21n + 4$. Thus,

$21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}$

Subtracting the second equation from the first equation we get $1\equiv 0\pmod{p}$ which is clearly absurd.

Hence $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.

Solution 3

Proof by contradiction:

Assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction.

If a certain fraction $\dfrac{a}{b}$ is reducible, then the fraction $\dfrac{2a}{3b}$ is reducible, too. In this case, $\dfrac{2a}{3b} = \dfrac{42n+8}{42n+9}$.

This fraction consists of two consecutives numbers, which never share any factor. So in this case, $\dfrac{2a}{3b}$ is irreducible, which is absurd.

Hence $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.


Solution 4

We notice that:

$\frac{21n+4}{14n+3} = \frac{(14n+3)+(7n+1)}{14n+3} = 1+\frac{7n+1}{14n+3}$

So it follows that $7n+1$ and $14n+3$ must be coprime for every natural number $n$ for the fraction to be irreducible. Now the problem simplifies to proving $\frac{7n+1}{14n+3}$ irreducible. We re-write this fraction as:

$\frac{7n+1}{(7n+1)+(7n+1) + 1} = \frac{7n+1}{2(7n+1)+1}$

Since the denominator $2(7n+1) + 1$ differs from a multiple of the numerator $7n+1$ by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that $\frac{21n+4}{14n+3}$ is irreducible.

Q.E.D


Solution 5

By Bezout's Lemma, $3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1$, so the GCD of the numerator and denominator is $1$ and the fraction is irreducible.

Solution 6

To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with $\frac{21n}{14n}$ + $\frac{4}{3}$. Simplifying the fraction, we end up with. $\frac{3}{2}$. Now combining these fractions with addition as shown before in the problem, we end up with $\frac{17}{6}$. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us $\frac{25}{17}$. Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16). Trying 2, we end up with $\frac{46}{31}$. Again, both results are 1 off from being multiples of 15. Trying 3, we end up with $\frac{67}{45}$. Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:

1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but

2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us $\frac{319}{199}$, we keep ending up with results that at the end will only have a common divisor of 1.


Solution 7

Let $\gcd (21n+4,14n+3)=a.$ So for some co-prime positive integers $x,y$ we have \[21n+4 = ax \qquad (1)\] \[14n+3 = ay \qquad (2)\] Multiplying $(1)$ by $2$ and $(2)$ by $3$ and then subtracting $(1)$ from $(2)$ we get \[42n+9-(42n+8)=3ay-2ax\] \[\implies a(3y-2x)=1.\] We must have $a=1$, since $a$ is an positive integer. Thus, $\gcd(21n+4,14n+3)=1$ which means the fraction is irreducible, as needed.

Video Solution by OmegaLearn

https://youtu.be/zfChnbMGLVQ?t=1266

~ pi_is_3.14

https://www.youtube.com/watch?v=IpUh5cLP454&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=1 - AMBRIGGS

https://youtu.be/Pfg4MVUML64

- little fermat

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1959 IMO (Problems) • Resources
Preceded by
First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/Problem_1&oldid=195995"