Difference between revisions of "2023 IOQM/Problem 2"

Revision as of 21:04, 2 October 2023

Find the number of elements in the set

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

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 ~ANDY666

@@ Line 5: / Line 5: @@
 ==Solution1(Quick)==
-Finding the no. of [[elements]] in the [[set]] means finding no. of ordered pairs of(<math>a</math>, <math>b</math>)
+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>.