Difference between revisions of "2023 IOQM/Problem 2"

Revision as of 20:34, 5 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, $b\:=\: a^{2}$ , 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 solution by Unacademy Olympiad Corner: https://www.youtube.com/watch?v=Mm6mXjwU9bY

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

@@ Line 15: / Line 15: @@
 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
-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