Puggy....Could you answer these plz?

Question

1) Prove that
(1 / 15) < (1 / 2) (3 / 4) (5 / 6) ... (99 / 100) < (1 / 10)

2) Consider a 25 x 25 square grid in each of whose 625 squares is placed either 'a+1' or 'a-1'. Suppose a(i) denotes the product of all the numbers in the (i)th row and b(j) denotes the product of all the numbers in the (j)th column. Prove that a(1) + b(1) + a(2) + b(2) + ... + a(25) + b(25) is not equal to 0.

3) If p and p^2 + 2 are primes, prove that p^3 + 2 is also prime.

Thanks a lot!!!

Answer 1

I respect Puggy's prowess, but there are other
maths answerers here as well.
Here's the answer to no. 3).
The only prime that satisfies the hypothesis
of the problem is p = 3.
Why? Well if p = 1(mod 3), p²+2 = 1+2 = 0(mod 3).
If p = 2(mod 3), p²+2 = 1+2 = 0(mod 3).
That leaves p = 0(mod 3), and if p is such a
prime, then p = 3.
If p = 3, 3²+2 = 11 is prime and so is 3³+2 = 29.
So the result is true for all primes satisfying
the conditions of the problem!
Here's a start to no. 1. I couldn't quite
get the result wanted, but I came close.
Unfortunately, I have no time left to do
the extra calculations to get the result wanted.
Anyway, let me illustrate the method I used
and indicate how it can be refined.
Problem:
Prove that
(1 / 15) < (1 / 2) (3 / 4) (5 / 6) ... (99 / 100) < (1 / 10).

1). Take reciprocals.
So the problem is to prove that
10 < (2/1)(4/3)(6/5)...(100/99) <15.

2). Take logs to get
log(10) < (log 2-log 1) + (log 4-log 3) + ...
+(log 100-log 99) < log(15)

3. Call the sum we wish to estimate S.
It is enough to show that
2.3026 < S <2.7080
because log 10 <2.3026
and log 15 >2.7080

4. We can express each term of S as a definite integral.
We have
log 2 -log 1 = ∫(1..2) dx/x
log 4 -log 3 = ∫(3..4) dx/x
.........................................
log 100 - log 99 = ∫(99..100) dx/x

5. My failure to get the desired result now stems
from the fact that I used very crude approximations
to the integrals and added them.
By using more refined upper and lower Riemann
sums, we should get the required inequalities.
Any way, here's what I did.
I used the fact that for any definite integral
m(b-a) <= ∫(a..b) f(x) dx <= M(b-a),
where m is the minimum of f on [a,b] and M is
the maximum of f on [a,b].
In our case, each b-a is 1 and since 1/x
is a decreasing function of x, M is at
the left end point of each interval, m at the right
end point.
So, estimating S gives
S > 1/2 + 1/4 + ... + 1/100
= (1/2)*(1 + 1/2 + ... + 1/50).
Using the harmonic sum calculator from the
University of Utah,
we get
S > (1/2)*(4.99) >2.449
and exp(2.449) is about 9.249.
Also S < 1 + (1/3) + ... +(1/99)
= 1 + 1/2 + ... + 1/100 - 2.449
<5.1874 - 2.449 < 2..7384
and exp(2.7384) is about 15.4623.
So crude approximations have brought
us close to the answer and perhaps finer
approximations to the integrals will do the trick!

Sorry, I have no time left to work on no. 2.

Answer 2

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Answer 3

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Answer 4

Do your own homework!