Difference between revisions of "2023 IOQM/Problem 2"

(Solution1(Quick))
(Solution1(Quick))
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==Solution1(Quick)==
 
==Solution1(Quick)==
Finding the no. of [[elements]] in the [[set]] means finding no. of ordered pairs of (<math>a</math>, <math>b</math>)
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Finding the no. of [[elements]] in the [[set]] means finding no. of [[ordered pairs]] of (<math>a</math>, <math>b</math>)
  
 
<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>.
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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>a\:=\: b^{2}</math>, Hence all such pairs are of the form (<math>b^{2}</math>,<math>b</math>)
  
Where each number lies between 2 and 2023(inclusive). All such pairs are (4, 2);(9, 3);(16, 4);........(1936, 44)
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Where each number lies between 2 and 2023 (inclusive). All such pairs are (4, 2);(9, 3);(16, 4);........(1936, 44)
  
 
Total no. of these pairs = 43
 
Total no. of these pairs = 43
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Thus, there are 43+11=<math>\boxed{54}</math> elements in the set
 
Thus, there are 43+11=<math>\boxed{54}</math> elements in the set
  
~ SANSGANKRSNGUPTA AND ~ANDY666
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~ SANSGANKRSNGUPTA AND ~ANDY666z

Revision as of 21:05, 2 October 2023

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, $a\:=\: b^{2}$, Hence all such pairs are of the form ($b^{2}$,$b$)

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

Total no. of these pairs = 43

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

Total no. of these pairs = 11

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

~ SANSGANKRSNGUPTA AND ~ANDY666z