Difference between revisions of "2023 IOQM/Problem 2"
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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>) | + | 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) | + | 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 ~ | + | ~ SANSGANKRSNGUPTA AND ~ANDY666z |
Revision as of 21:05, 2 October 2023
Problem
Find the number of elements in the set
Solution1(Quick)
Finding the no. of elements in the set means finding no. of ordered pairs of (, )
Then, .
+ =5. Upon simplifying, we get
So, equals to 2 or 3
For = 2, it implies that . So, , Hence all such pairs are of the form (,)
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 = 3, Following the similar pattern, we get the pairs as {2,8}...{12,1728} (,)
Total no. of these pairs = 11
Thus, there are 43+11= elements in the set
~ SANSGANKRSNGUPTA AND ~ANDY666z