Difference between revisions of "2023 IOQM/Problem 2"
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<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>. | ||
| − | <math>\implies</math> <math>x | + | <math>\implies</math> <math>x+\frac{6}{x}=5</math>. Upon simplifying, we get <math>x^{2}-5x+6=0</math> |
<math>\implies</math> <math>(x-2)(x-3)=0</math> | <math>\implies</math> <math>(x-2)(x-3)=0</math> | ||
| Line 15: | Line 15: | ||
So, <math>x</math> equals to 2 or 3 | 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> | + | 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 | + | 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 | 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> | + | 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 | Total no. of these pairs = 11 | ||
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==See Also== | ==See Also== | ||
| + | [[IOQM]] | ||
| + | |||
| + | [[Mathematics competitions]] | ||
| + | |||
| + | '''Please note that all problems on this page are copyrighted by THE [https://www.mtai.org.in | MTA(I)]''' | ||
Latest revision as of 01:48, 4 May 2024
Problem
Find the number of elements in the set
![\[\lbrace(a.b)\in N: 2 \leq a,b \leq2023,\:\: \log_{a}{b}+6\log_{b}{a}=5\rbrace\]](http://latex.artofproblemsolving.com/7/5/f/75fe09afd4da734f976243ddeda521b906b7a04b.png)
Solution1(Quick)
Finding the no. of elements in the set means finding no. of ordered pairs of (
, 
)
Then, 
.
. 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 (2, 4);(3, 9);(4, 16);........(44, 1936)
Total no. of these pairs = 43
For = 3, Following a 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 ~Andy666
Video Solutions
Video solution by cheetna: https://www.youtube.com/watch?v=z0NPa0tUzZk
Video solution by Unacademy Olympiad Corner: https://www.youtube.com/watch?v=Mm6mXjwU9bY
Video solution by Vedantu Olympiad School: https://www.youtube.com/watch?v=4DJXtR4VHEA
Video solution by Olympiad Wallah: https://www.youtube.com/watch?v=4HSjmY7d3nA
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
See Also
Please note that all problems on this page are copyrighted by THE | MTA(I)
