| >> 3. An old but nice problem. Find all real solutions to the following:
>> x x x
>> (x+1) + (x+2) = (x+3) .
What follows is not a proof, merely what I found by a quick examination
of a few trial cases.
It seems like the only solution over the reals is x = 2; substituting
2 for x gives 3^2 + 4^2 = 5^2 -- gee, that looks familiar!
If x = -1, -2, or -3, then zero is being raised to a negative power,
and so I am ruling those x's out. If x < -1 and is not an integer,
then (x + 1)^x is a negative number raised to a fractional power,
so out they go, too. Then I tried -4, -5, -6, ... and the values
of
(x+1)^x + (x+2)^x - (x+3)^x
did not become zero. For x > -1 there were no domain problems,
and again by examining values I found the root x = 2 and it appeared
that there were no other roots.
I suppose a real proof would involve checking the derivative to
verify that the growth trends guarantee no other roots.
Dan
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| (4) Find the limit points of the following sequences:
(a) {n*sin(2*pi*n!*e)}
(b) {n*sin(2*pi*n!/e)}
Spoilers follow.
What the fractional part of n!*e? Since e = 1 + 1/1! + 1/2! + 1/3! + ...,
fract(n!*e) = 1/(n+1) + 1/((n+1)*(n+2)) + .... Since this is small, we have
n*sin(2*pi*fract(n!*e)) = n*2*pi*( 1/(n+1) + 1/((n+1)*(n+2)) + ...). Thus,
the limit as n -> inf is just 2*pi.
What the fractional part of n!/e? Since 1/e = 1 - 1/1! + 1/2! - 1/3! + ...,
fract(n!/e) = -1/(n+1) + 1/((n+1)*(n+2)) + ... for n even, and
fract(n!/e) = +1/(n+1) - 1/((n+1)*(n+2)) + ... for n odd.
Since this is small, ..., there are two limit points: � 2*pi.
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| (5) Find the largest even number that can't be written as the sum
of two odd composite numbers.
Solution follows.
Let the even number be 2*m. Let x = (19*m + 15) mod 30. Then
m+x mod 5 = (20*m + 15) mod 5 = 0, and
m-x mod 3 = (15 - 18*m) mod 3 = 0, and
m+x mod 2 = (20*m + 15) mod 2 = 1, and
m-x mod 2 = (15 - 18*m) mod 2 = 1.
Thus, m-x is an odd multiple of 3, and m+x is an odd multiple of 5,
and (m-x) + (m-x) = 2*m. If m-x is greater than 3, then m-x is an
odd composite number (it has another factor besides the 3), and if
m+x is greater than 5, then it is also an odd composite number.
This construction suffices for m > max(3+x, 5-x) >= max(31, 5) = 31.
For smaller m, we find the decompositions: 62=27+35, 60=9+51, 58=9+49,
56=21+25, 54=9+45, 52=25+27, 50=15+35, 48=9+39, 46=21+25, 44=9+35, 42=9+33,
and 40=15+25. But there is no such decomposition for 38 -- the answer --
each of the possible decompositions into two odd summands include a prime, viz:
38 = 19+19 = 17+21 = 15+23 = 13+25 = 11+27 = 9+29 = 7+31 = 5+33 = 3+35 = 1+37.
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