I happened to find my quite old calculator the other day. So I put 17^20 into it, and it said that it is 3. At least if one turns it upside down, otherwise it is E, which would be 14 in hexadecimal. And hex is beyond its specifications. So we do need another method. (It kinda looks like a little pony if one turns it 90 degrees. Like a dead one the other way.)
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17^20>16^20, 64^13>63^13
17^20>4^40, > 4^39>63^13
Of course! One should perhaps not go down to base 2 immediately, but just halve it and look for an obvious answer.
17"20 / 63^13
= 289^10 / 63^13
= (289/63)^10 / 63^3
> 4^10 / 63^3
= 4x64^3 / 63^3
> 4 > 1
tu as commis une erreur à5min pour la forme binomiale,on devrait avoir >1.Merci
I happened to find my quite old calculator the other day. So I put 17^20 into it, and it said that it is 3. At least if one turns it upside down, otherwise it is E, which would be 14 in hexadecimal. And hex is beyond its specifications. So we do need another method. (It kinda looks like a little pony if one turns it 90 degrees. Like a dead one the other way.)
Predicate: 17^20 > *16^20 > 64^13* > 63^13 ---> 16^20 > (4*16)^13 ---> 16^7 > 4^13 -->
4^14 > 4^13 qed.
*@ SyberMath* 17^20 > 16^20 = (4^2)^20 = 4^40 > 4^39 = (4^3)^13 = 64^13 > 63^13
Therefore, 17^20 > 63^13.
Nice
17^20 > 16^20 = 2^80
63^13 < 64^13 = 2^78
63^13 < 2^78 < 2^80 < 17*20
17^2^10or6^1^1^1 17^12^10or6 1^12^2^5or2^3 (x ➖ 3x+2).1^2^1or 2^1 (x ➖ 2x+1). 2^1>2^3 17^20>63^13
I used the powers of 2 to solve this - couldn't believe how easy it was!
idk hahhaa my head dizzy
17^13*17^7 or 63^13..17^7 or (63/17)^13...ma si ha che (63/17)^13(63/17)^13..quindi risulta 17^20>63^13....mah,non sono sicurissimo del ragionamento