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Difference between revisions of "2023 IOQM/Problem 2"
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<math>\log_{a}{b}=x</math> Then, <math>\log_{b}{a}=\frac{1}{x}</math>.
 
<math>\log_{a}{b}=x</math> Then, <math>\log_{b}{a}=\frac{1}{x}</math>.
  
−
<math>\implies</math>  <math>x</math>+<math>\frac{6}{x}</math> =5. Upon simplifying, we get <math>x^{2}-5x+6=0</math>  
+
<math>\implies</math>  <math>x+\frac{6}{x}=5</math>. Upon simplifying, we get <math>x^{2}-5x+6=0</math>  
  
 
<math>\implies</math> <math>(x-2)(x-3)=0</math>
 
<math>\implies</math> <math>(x-2)(x-3)=0</math>
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So, <math>x</math> equals to 2 or 3
 
So, <math>x</math> equals to 2 or 3
  
−
For <math>x</math> = 2, it implies that <math>\log_{a}{b}=2</math>. So, <math>a\:=\: b^{2}</math>, Hence all such pairs are of the form (<math>b^{2}</math>,<math>b</math>)
+
For <math>x</math> = 2, it implies that <math>\log_{a}{b}=2</math>. So, <math>b\:=\: a^{2}</math>, Hence all such pairs are of the form (<math>a</math>,<math>a^{2}</math>)
  
−
Where each number lies between 2 and 2023 (inclusive). All such pairs are (4, 2);(9, 3);(16, 4);........(1936, 44)
+
Where each number lies between 2 and 2023 (inclusive). All such pairs are (2, 4);(3, 9);(4, 16);........(44, 1936)
  
 
Total no. of these pairs = 43
 
Total no. of these pairs = 43
  
−
For <math>x</math> = 3, Following a similar pattern, we get the pairs as {2,8}...{12,1728} (<math>b^{3}</math>,<math>b</math>)
+
For <math>x</math> = 3, Following a similar pattern, we get the pairs as {2,8}...{12,1728} (<math>a</math>,<math>a^{3}</math>)
  
 
Total no. of these pairs = 11
 
Total no. of these pairs = 11
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==Video Solutions==
 
==Video Solutions==
 +
Video solution by cheetna: https://www.youtube.com/watch?v=z0NPa0tUzZk
 +
 +
Video solution by Unacademy Olympiad Corner: https://www.youtube.com/watch?v=Mm6mXjwU9bY
 +
 +
Video solution by Vedantu Olympiad School: https://www.youtube.com/watch?v=4DJXtR4VHEA
 +
 +
Video solution by Olympiad Wallah: https://www.youtube.com/watch?v=4HSjmY7d3nA
 +
 +
Video solution by : Motion Olympiad Foundation Class 5th - 10th: https://www.youtube.com/watch?v=oVaeHceHXsQ
 +
 +
'''Please note that above videos solutions are in Hindi, some in English and some in mixed(Hindi + English).'''
 +
 +
~SANSGANKRSNGUPTA
 +
 +
==See Also==
 +
[[IOQM]]
 +
 +
[[Mathematics competitions]]
 +
 +
'''Please note that all problems on this page are copyrighted by THE [https://www.mtai.org.in | MTA(I)]'''

Latest revision as of 01:48, 4 May 2024

Problem

Find the number of elements in the set

\[\lbrace(a.b)\in N: 2 \leq a,b \leq2023,\:\: \log_{a}{b}+6\log_{b}{a}=5\rbrace\]

Solution1(Quick)

Finding the no. of elements in the set means finding no. of ordered pairs of ($a$, $b$)

$\log_{a}{b}=x$ Then, $\log_{b}{a}=\frac{1}{x}$.

$\implies$ $x+\frac{6}{x}=5$. Upon simplifying, we get $x^{2}-5x+6=0$

$\implies$ $(x-2)(x-3)=0$

So, $x$ equals to 2 or 3

For $x$ = 2, it implies that $\log_{a}{b}=2$. So, $b\:=\: a^{2}$, Hence all such pairs are of the form ($a$,$a^{2}$)

Where each number lies between 2 and 2023 (inclusive). All such pairs are (2, 4);(3, 9);(4, 16);........(44, 1936)

Total no. of these pairs = 43

For $x$ = 3, Following a similar pattern, we get the pairs as {2,8}...{12,1728} ($a$,$a^{3}$)

Total no. of these pairs = 11

Thus, there are 43+11=$\boxed{54}$ elements in the set

~ SANSGANKRSNGUPTA AND ~Andy666

Video Solutions

Video solution by cheetna: https://www.youtube.com/watch?v=z0NPa0tUzZk

Video solution by Unacademy Olympiad Corner: https://www.youtube.com/watch?v=Mm6mXjwU9bY

Video solution by Vedantu Olympiad School: https://www.youtube.com/watch?v=4DJXtR4VHEA

Video solution by Olympiad Wallah: https://www.youtube.com/watch?v=4HSjmY7d3nA

Video solution by : Motion Olympiad Foundation Class 5th - 10th: https://www.youtube.com/watch?v=oVaeHceHXsQ

Please note that above videos solutions are in Hindi, some in English and some in mixed(Hindi + English).

~SANSGANKRSNGUPTA

See Also

IOQM

Mathematics competitions

Please note that all problems on this page are copyrighted by THE | MTA(I)

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