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For any natural number n>2 prove that the equation
a^n+b^n=c^n
has no natural solutions for a,b,c
:)
Yes, some general approach is needed, from which any possible combination is naturally derived.
Another number problem (weight 5):
A line contains 32 natural numbers (not necessarily distinct). Prove that between them one can place brackets, signs of addition and multiplication so that the value of obtained expression is evenly divisible by 11000.
Note from me: 11000 = 11 * 2^3 * 5^3.
32 = 11 + 2 + 2 + 2 + 5 + 5 + 5.
It remains to prove the auxiliary statement: between any n numbers we can put brackets and signs (*, +) so that the expression is divisible by n.
The numbers cannot be concatenated (7 and 9 cannot make 79).
But some denominations believe that 0 is a natural number.
A farm ?
Yeah, the great one.))
Avals:
Farm ?
Alcohol?
Alcohol?
))
I found an algebra book from '52.)
Guess what:)
№1234
Pythagoras, when asked about the number of students attending his school,
replied, according to legend, thus: "Half of the pupils study mathematics,
a quarter is music, a seventh is silent and there are also three women"
Attention question: How many disciples did Pythagoras have?
I found an algebra book from '52.)
Guess what:)
№1234
Pythagoras, when asked about the number of students attending his school,
replied, according to legend, thus: "Half of the pupils study mathematics,
a quarter is music, a seventh is silent and there are also three women"
Attention question: How many disciples did Pythagoras have?
28 or 25 pupils and 3 female pupils.
28 or 25 pupils and 3 female pupils.