Russia Math Olympiad Question You should know this trick!!
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- Опубликовано: 26 авг 2024
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If you are watching this without searching like this.
I used only logarithms and fractions comparsion to solve it in a faster and clever way:
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1) Transform the two numbers in natural logarithms:
50^50 = 50 ln(50)
49^51 = 51 ln(49)
2) Create a fraction with previous logarithms and separate it in two fractions:
50 ln(50) / [ 51 ln(49) ] -->
--> ( 50 / 51 ) * [ ln(50) / ln(49) ]
3) Confront the first rational fraction and the second logarithmic fraction with the value 1:
50 / 51 < 1
ln(50) / ln(49) > 1
4) The absolute difference of numbers are always 1 (in this example):
| 50 - 51 | = 1
| 49 - 50 | = 1
5) Because the natural logarithm function grows slower for every positive real number:
ln(x) < x for any x € R+
ln(x+1) < (x +1) for any x € R+
6) Now we can say that every fraction with numerator bigger that its denominator is always
bigger that the same fraction with logarithmed numerator and denominator for any number
greater that the constant e:
(x +1) / x > ln(x + 1) / ln(x) for x > e
x / (x +1) < ln(x) / [ ln(x +1) ] for x > e
7) That makes the confront at step 3 more accurate thanks to step 4:
51 / 50 > ln(51) / ln(50) --> 50 / 51 < ln(50) / ln(51)
ln(50) / ln(49) is closer to 1 than 50 / 51 due to step 5
50 / 51 is more under the value 1
ln(50) / ln(49) is less up the number 1
8) Finally we can confront the initial fraction at step 2:
( 50 / 51 ) * [ ln(50) / ln(49) ] -->
--> (a value more under 1) * (a value less up 1) = (a value under 1) -->
--> ( 50 / 51 ) * [ ln(50) / ln(49) ] < 1
9) The fraction at step 2 has the denominator bigger that numerator due to being < 1:
The numerator is 50 * ln(50) and the denominator is 51 * ln(49)
Result: 49^51 > 50^50
nice approach
@AshaTomar-v6r Your logic is the same I wrote in my comment with the difference that I proved it mathematically. Also I mentioned the fact for numbers greater than e ( 2,718 ) that proof is safe to use. Generic proof are from step 3 to 6.
Thanks, this was really helpful! You should definitely go into the field if you're this smart!
Glad you think so!
solved it very similarly
great
Amazing. Cheers from Portugal
thank you
Wow 😲😲 . I dont have that much patience finding em.
Thats why I am here for you. :)
Are these videos for your class or just for utube?
Ii love to do maths in class too.
Can be solved using log also right?
may be yes...i havent tried log method.....will try this too soon.
@@BZKnowHow will be waiting! 💚
are ise asaan maine kar diya dimaa me hi
i have solvd it in a verry very very short and easy method
I study in class 12th but could have definately solved it because i made the trick to solve questions related to multiplication and powers in class 8th
today evening I will post the video on it
please see it mam
by the wqay sorry if I sound rude but I believe that everybody should learn something new
yes you are right everybody has own logic to solve problems.