Difference between revisions of "2023 IOQM/Problem 2"

Revision as of 21:02, 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 4: / Line 4: @@
 <cmath>\lbrace(a.b)\in N: 2 \leq a,b \leq2023,\:\: \log_{a}{b}+6\log_{b}{a}=5\rbrace</cmath>
-==Solution quick==
+==Solution1(Quick)==
+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>x</math>+<math>\frac{5}{x}</math> =6. Upon solving the equation
+<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-2)(x-3)=0</math>
-So, x equals to 2 or 3
+So, <math>x</math> equals to 2 or 3
-For 2, it's obvious that there are {2,4}...{44,1936}
+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 3, it's obvious that there are {2,8}...{12,1728}
+Where each number lies between 2 and 2023(inclusive). All such pairs are (4, 2);(9, 3);(16, 4);........(1936, 44)
-Thus, there are 43+11=54 pairs
+Total no. of these pairs = 43
+For <math>x</math> = 3, Following the similar pattern We get the pairs as {2,8}...{12,1728} (<math>b^{3}</math>,<math>b</math>)
+Total no. of these pairs = 11
+Thus, there are 43+11=<math>\boxed{54}</math> elements in the set
+~ SANSGANKRSNGUPTA AND ~ANDY666