## Wednesday, October 10, 2012

### Cool math Olympics - II

Cool math Olympics - 1960

1. Determine all three-digit numbers N having the property that N is divisible by 11. and N / 11 is equal to the sum of the squares of the digits of N.
2. For what values of the variable x does the following inequality hold: (4x^{2})/(1 -\sqrt{1 + 2x})^{2} < 2x + 9.
3. In a given right triangle ABC the hypotenuse BC of length a is divided into n equal parts (n an odd integer). Let \alpha be the acute angle subtending, from A that segment which contains the midpoint of the hypotenuse. Let h be the length of the altitude to the hypotenuse of the triangle.
Prove: tan(\alpha) = (4nh) / ((n^{2} - 1) * a).
4. Construct triangle ABC given h_{a}, h_{b} (the altitudes from A and B) and m_{a}, the median from vertex A.
5. Consider the cube ABCDA^{1}B^{1}C^{1}D^{1} (with face ABCD directly above face A^{1}B^{1}C^{1}D^{1}).
(a) Find the locus of the midpoints of segments XY where X is any point of AC and Y is any point of B^{1}D^{1}.
(b) Find the locus of points Z which lie on the segments XY of part (a) with ZY = 2XZ.
6. Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let V1 be the volume of the cone and V2 the volume of the cylinder.
(a) Prove that V1 != V2.
(b) Find the smallest number k for which V1 = kV2, for this case, construct the angle subtended by a diameter of the base of the cone at the vertex of the cone.
7. An isosceles trapezoid with bases a and c and altitude h is given.
(a) On the axis of symmetry of this trapezoid, find all points P such that both legs of the trapezoid subtend right angles at P:
(b) Calculate the distance of P from either base.
(c) Determine under what conditions such points P actually exist. (Discuss various cases that might arise.)

### Cool Math Olympics - 1959 Key

1. Solution: GCD of a and b is the greatest common divisor of a and b. It is known that, what divides a and b also divides a - b. So, GCD of a and b should be same as that of a and a - b (when a > b) or that of b and a - b. Considering that, and also from the observation that, some rational number is irreducible, iff, the numerator and denominator have a GCD of 1, it follows that:

GCD(21n + 4, 14n + 3) = 1                                    <=>
GCD(21n + 4, (21n + 4) - (14n + 3)) = 1               <=>
GCD(21n + 4, 7n + 1) = 1

Now, calculate GCD normally, like we do, divide take remainder and repeat the procedure with the remainder and the divider. The remainder in 21n + 4 when divided by 7n + 1 is 1. which means the GCD is 1. It immediately follows that, the GCD of original numbers is also 1, hence, those are irreducible.

2. Square the expression on both sides, noting that A >= 0 (since square root is a function, and results of that function are always non-negative).

\sqrt{(x + \sqrt{2x - 1})} + \sqrt{(x - \sqrt{2x - 1})} = A                                       <=>
(\sqrt{(x + \sqrt{2x - 1})} + \sqrt{(x - \sqrt{2x - 1})})^{2} = A^{2}, (A >= 0)     <=>
2x +  2 * \sqrt{(x + \sqrt{2x - 1})(x - \sqrt{2x - 1})} = A^{2}, (A >= 0)               <=>

But here, notice that the expression inside square roots is nothing but (x^2 - 2x + 1, which is a square expression). Now, again;

2x + 2|x - 1| = A^{2}, (A >= 0, x >= 1/2)               <=>

(|x| is the absolute value of x).
The condition x >= 1/2 is because of \sqrt{2x - 1}, where the one inside square-root cant be negative.

Now substitute values for A,

(i) A = 1, x + |x - 1| = 1/2.

Notice that, |x - 1| cant be 1 - x. If it is so, A^{2}/2 = 1 always, regardless of the actual value of A. So, this yields another condition:

If A^{2} / 2 != 1, then x >= 1.

Consider these conditions into the expression (i), and you get x = 3/4, which is not according to the conditions. So, (i) does not have a solution.

(ii) A = 1/2, x + |x - 1| = 1/8.

Here, we get x = 9/16, since this is also less than 1, (ii) also has no solution.

(iii) A = 2, x + |x - 1| = 2. Here you get, x = 3/2, which satisfies all conditions. So, only this combination has solutions for x, which is given by 3/2.

3. Remember cos(2x) = 2cos^{2}(x) - 1. And, in a quadratic equation ax^{2} + bx + c = 0, the sum of roots is given by -b/a and product is given by c/a. First, derive formulae for cos(2x_{1}) + cos(2x_{2})  and cos(2x_{1})cos(2x_{2}) in terms of cos(x_{1}) + cos(x_{2}) and cos(x_{1})cos(x_{2}), where x_{1} and x_{2} are roots of original equation.

cos(2x_{1}) + cos(2x_{2}) = 2cos^{2}(x_{1}) - 1 + 2cos^{2}(x_{2}) - 1         <=>
= 2(cos(x_{1}) + cos(x_{2}))^{2} - 4cos(x_{1})cos(x_{2}) - 2.

(Since x^2 + y^2 = (x + y)^2 - 2xy).

cos(2x_{1})cos(2x_{2}) = (2cos^{2}(x_{1}) - 1) * (2cos^{2}(x_{2}) - 1)                 <=>
= 4cos^{2}(x_{1}) * cos^{2}(x_{2}) - 2cos^{2}(x_{1}) - 2cos^{2}(x_{2}) + 1  <=>
= 4(cos(x_{1})cos(x_{2}))^{2} - 2(cos(x_{1}) + cos(x_{2}))^{2} + 4cos(x_{1})cos(x_{2})  + 1.

Substituting

cos(x_{1}) + cos(x_{2}) = -b/a and cos(x_{1})cos(x_{2}) = c/a, you get the following:

-b^{1}/a^{1} = 2 * (-b/a)^{2} - 4 * (c/a) - 2 <=>
= (2b^{2} - 4ca -2a^{2})/a^{2}

c^{1}/a^{1} = 4 * (c/a)^{2} - 2 * (-b/a)^{2} + 4 * (c/a) + 1 <=>
= (4c^{2} - 2b^{2} + 4ca + a^{2}) / a^{2}

So, one can use: a^{1}  = a^{2}, b^{1} = (-2b^{2} + 4ca + 2a^{2}) and c^{1} = (4c^{2} - 2b^{2} + 4ca + a^{2}).

The resulting expression is, a^{2} cos^{2}(2x) + (-2b^{2} + 4ca + 2a^{2}) cos(2x) + (4c^{2} - 2b^{2} + 4ca + a^{2}) = 0.

When you substitute, a = 4, b = 2, c = -1, you get. the roots for original equation:

\alpha = (-1 + \sqrt{5}) / 4, and \beta = (-1 - \sqrt{5}) / 4.

Surprisingly, the equation in cos(2x) also yields the same roots. and in fact, the sane equation when simplified by eliminating common factors between co-efficients.

However, there is no need to be surprised. Because, calculate cos(2x) for \alpha, it will be \beta and cos(2x) for \beta will be \alpha. So, both the equations are right, and compatible. (Its a neat trick substitution).

4. Consider an angle in the right triangle to be \theta, where \theta != 90^{o}. The median divides the hypotenuse into 2 equal lengthed parts. Since the lengh of hypotenuse is c, one side will be c * cos(\theta) and another will be c * sin(\theta). The median divides the right triangle into two other triangles. Consider the triangle inside which we have angle \theta. Using the cosine rule of lengths of sides, gives the following:

(To brush up: Cosine rule says that, if angle between sides length a and b is \theta, the third side in the triangle is given by sqrt{a^{2} + b^{2} - 2ab * cos(\theta)}).

The geometric mean of sides is, sqrt{c * cos(\theta) * c * sin(\theta)}
= c * sqrt{sin(2 * \theta) / 2}

Using cosine rule,

c^{2} * sin(2 * \theta) / 2 = c^{2} / 4 + c^{2} * cos^{2} (\theta) - 2 * c/2 * c * cos(\theta) * cos(\theta) <=>
c^{2} * sin(2 * \theta) / 2 = c^{2} / 4 + c^{2} * cos^{2} (\theta) - c^{2} * cos^{2} (\theta)     <=>
c^{2} * sin(2 * \theta) / 2 = c^{2} / 4,

So, the side length is c / 2 as well (note that the length of parts of hypotenuse after being divided by the median, are also same as c/2). Now, to the angle \theta.

(sin(2 * \theta)) / 2 = 1/ 4 <=>
sin(2 * \theta) = 1/2       <=>
2 * \theta = 30^{o} or 2 * \theta = 150^{o} (since sin(30^{o}) = sin(150^{o}) = 1/2).
=>
\theta = 15^{o} or \theta = 75^{o} (In fact, both angles are complementary in a right angled triangle).

Basically, to construct such a triangle, with any c, choose an angle to be 15^{o} in the right triangle.

5. Now to the last problem, this is a geom problem. I will present anal. geom. solution except for the last part. (Which is too easy in co-ordinate geometry).

Consider this theorem, well known in circles of (pun is intended), ananlytical geom. If a Jyaa, AB in a circle makes an angle \theta at the center C of the circle, (namely, <ACB is \theta), then on any point on the circle, to the side of C, it makes an angle of < \theta / 2). With the help of this theorem, you can easily solve this problem.

First of all, notice that <ANM and <MNB are both 45^{o}, since <APM and <MQB are 90^{o} (P and Q are also centers of circum-circle for a square).

Consider <NAM = \theta. <NMA = 135^{o} - \theta. <NMB = 45^{o} + \theta and <NBM = 90^{o} - \theta. Apply sine rule now, using NM which is same in both triangles, \delta NMA and \delta NMB.

(NM)/sin(\theta) = (AM)/sin(45^{o}) and (NM)/sin(90^{o} - \theta) = (MB)/sin(45^{o})

This gives, sin(\theta) = (NM)/(sqrt{2} * AM) and cos(\theta) = (NM)/(sqrt{2} * MB), which both can be combined to give,

tan(\theta) = (MB)/(AM).

But, notice that, (MB)/(AM) = (MB)/(MC) = (FM)/(AM). so, this, along with the facts that <NAM = \theta and <NBM = 90^{o} - \theta, gives that, A, N and F are co-linear, as well as, B, C and N are co-linear.

(ii) Consider for now, A is the origin, and AB is the x axis. Consider AM = l_{1} and AB = l. Consider the point R at (l/2, -l/2). The angle <BMR = 135^{o} - \theta since the opposite angle <AMN is of same measure.

tan(<BMP) = tan(135^{o} - \theta) = -tan(45^{o} + \theta) = (1 + tan(\theta))/(tan(\theta) - 1).

We also know that.

tan(\theta) = (l - l_{1}) / l_{1}, substituting it.

tan(<BMP) = l / (l - 2 * l_{1}) and notice that with mid-point S of AB, MS length = l/2 - l_{1}.

This gives length of SR = (MS) * tan(<BMP) = (l/2 - l_{1}) * l / (l - 2 * l_{1})  <=>
SR = (l - 2 * l_{1}) / 2 * l / (l - 2 * l_{1}) = l / 2

which is irrespective of any \theta or equivalently l_{1}.

(iii) Using co-ord geom, P is (l_{1} / 2 , l_{1} / 2) and Q is (l_{1} + (l - l_{1})/2, (l - l_{1})/2) which is same as, ((l + l_{1})/2, (l - l_{1})/2). So, their mid-point is given by ((l + (2 *  l_{1}))/4, l/2). So, that means locus is, y = l/2.

## Monday, October 1, 2012

### Cool Math Olympics

International Math Olympiad problems will be presented along with solutions here in "Cool Math Olympics". Olympiad problems from 1959 (first time Math Olympiad started) to current year will be presented and solved here. First set of Olympiad problems and solutions (one day gap) for 1959 Olympiad is here at "1959 Cool Math Olympics". Olympians, or wannabe Olympians, participate in our "Cool Math Olympics". You performance is much appreciated.

1. Prove that the fraction (21n+4) / (14n+3) is irreducible for every natural number n. (Hint: What is the definition of GCD? And if a and b have GCD of 1, what about a and a - b or, b and a - b).
2. For what real values of x is

\sqrt{(x + \sqrt{2x - 1})} + \sqrt{(x - \sqrt{2x - 1})} = A,
given (a) A =
1 (b) A = 1/2, (c) A = 2
where only non-negative real numbers are admitted for square roots? (Hint: Too many square roots, looks like it requires raising to power two multiple times, or is it?)
3. Let a, b, c be real numbers. Consider the quadratic equation in cos(x):
a * cos^{2}(x) + b * cos(x) + c = 0
Using the numbers a, b, c form a quadratic equation in cos(2x), whose roots are the same as those of the original equation. Compare the equations in cos(x) and cos(2x) for a = 4, b = 2, c = -1. (Hint: Use formulas for sum and product of roots, also apply same for cos(2x)).
3a. Bouns question: What happened when you used a = 4, b = 2, c = -1 and why?
4. Construct a right triangle with given hypotenuse c such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle. (Hint: Parameterize by \theta).
4a. What is the angle \theta got? Is it dependent on c?
4b. What is the geometric mean? What is it same as?
5. An arbitrary point M is selected in the interior of the segment AB. The squares AMCD and MBEF are constructed on the same side of AB; with the segments AM and MB as their respective bases. The circles circumscribed about these squares, with centers P and Q; intersect at M and also at another point N. Let N^{1} denote the point of intersection of the straight lines AF and BC.
(a) Prove that the points N and N^{1} coincide.
(b) Prove that the straight lines MN pass through a fixed point S independent of the choice of M.
(c) Find the locus of the midpoints of the segments PQ as M varies between A and B.
(Hint: Use Cartesian Co-ordinates. Like many such problems, choose origin and axes to minimize the paraphernalia).
5d. Bonus problem. There is a way to solve 5a, using analytical geometry.
5e. Prove that AF and BC intersect making an angle of 90^{0}.
5f. Prove also using analytical geometry, (use \theta where tan(\theta) = (AM)/(MB) as well) that there is such a fixed point like S.
5g. Argue that such a fixed point, would be on the perpendicular bisector of AB (use symmetry arguments, possibly?)
6. Two planes, P and Q intersect along the line p. The point A is given in the plane P and the point C in the plane Q; neither of these points lies on the straight line p. Construct an isosceles trapezoid ABCD (with AB parallel to CD) in which a circle can be inscribed, and with vertices B and D lying in the planes P and Q respectively. (Don't know what this problem is about. Whats the idea of this problem, anyway? What is "in which a circle can be inscribed", totally vague).

1959 Cool Math Olympics complete. Boy, these things take time, Phew. I personally prefer analytical geometry approach to problems than Co-ordinate geom approach, that's just dry, what do you people think? Though these problems are way back in 1959, Olympics solving took time for me, partly because I dealt with these math subjects, long long long time ago (15 years). However, these problems rank as "brilliant" for their formulation, where the results are pretty surprising (Especially 4 and 5). Hats-off to those people who had such great geometric insights. (Spoiler: Pretty much everything in geometry has an interesting locus, or passes through a fixed point I think. What do you people think?)

People who could crack these (with or without hints) can start telling their friends that they are good enough to do it. Start the celebrations, and the geekery of you telling others about these problems. Next time, with 1960 Cool Math Olympics.

## Saturday, September 29, 2012

### Cool Math - I

Current set of cool math problems (solutions later). See if you can see through them. Best of luck.

Note: I have solutions to a lot of these questions, almost all. But, it could be that some solution slipped by time, since I used these long time ago. In that case, I might not have some solutions.

*1. Consider a 4n * 4n square. Now, consider rectangles of integer dimensions which would fit inside this square (max dimension of them can be 4n). What is the probability (considering very large n), that these rectangles have an area lesser than or equal to 4n^{2} chosen over all possible dimensions (that would fit inside the square).
*2.  Prove that, if a number n has the prime number expansion form of p_{1}^{k_{1}}*p_{2}^{k_{2}}*p_{3}^{k_{3}}..., the sum of its factors is (where 1 and n are also considered factors) is given by (p_{1}^{k_{1}+1} - 1) / (p_{1} - 1) * (p_{2}^{k_{2}+1} - 1) /(p_{2} - 1) * ...
*3. From the above definition of factors, prove that a perfect number has sum of its factors, equal to twice the number.
**4. Prove fermat's little theorem: If p is a prime, a^{p} \equiv a (mod p). (Hint: Use binomial expansion to power p, and mathematical induction on a).
5. Prove that if p is not a prime, then 2^{p} - 1 is not a prime.
**6. Prove that all even perfect numbers are of the form, 2^{p−1} * (2^{p}−1) where (2^{p}−1) is a prime.
**7. Prove that if both p and (2p + 1) are primes, (2p + 1) will be a factor of (2^{p}−1).
8. A triangular number (n th) is a number, which is got by summing the first n, positive integers. Prove that  every number of the form 2^{p−1} * (2^{p}−1), is a triangular number.
9. Derive a formula for 'sum of first n odd cubes', and prove that every number of the form 2^{p−1} * (2^{p}−1), is a sum of first n odd cubes.
10. Even perfect numbers (except 6) give a remainder of 1, when divided with 9.
11. Subtracting 1 from a perfect number, and dividing it by 9, always gives a perfect number.
12. Summing all digits in a number, and summing all the digits in the resulting number, and repeating the process till a single digit is obtained is called, getting the "digital root" of a number. Prove that a positive number is divisible by 9, iff, its digital root is 9.
13. Any number of the form 2^{m−1} * (2^{m}−1), where m is any odd integer, leaves a remainder of 1 when divided with 9.
*14. An Ore's harmonic number, is a number whose factors have a harmonic mean of an integer. Prove that every perfect number is an Ore's harmonic number.
*15. Prove that, for any perfect number, sum of reciprocals of its factors is equals to two.
*16. Prove that for any integer M, the product of arithmetic mean of its factors and harmonic mean of the factors, is the number M itself.
*17. Any odd perfect number N is of the form, N = q^{\alpha} * p_{1}^{2e_{1}} *  p_{2}^{2e_{2}} *  p_{3}^{2e_{3}} ...p_{k}^{2e_{k}}. Where q, p_{1}, p_{2} are all distinct primes, with q \equiv \alpha \equiv 1 (mod 4).
18. Continuing on 17, the smallest prime factor of N is less than (2k + 8) / 3.
19. Continuing on 18, prove also that N < 2^{4^{k+1}}.
20. Continuing on 19, prove also that the largest prime factor of N is greater than 10^{8}.
21. Continuing on 20, prove also that q^{\alpha} > 10^{62} or p_{j}^{2e_{j}} > 10^{62} for some j.
22. Continuing on 21, prove also that the largest prime factor of N is greater than 10^{8}.
23. Continuing on 22, prove also that the second largest prime factor is greater than 10^{4}, and the third largest prime factor is greater than 100.
24. Continuing on 23, prove also that if N has at least 101 prime factors and at least 9 distinct prime factors. If 3 is not one of the factors of N, then N has at least 12 distinct prime factors.
25. Prove that an odd perfect number is not divisible by 105.
26. Every odd perfect number is of the form N ≡ 1 (mod 12), N ≡ 117 (mod 468), or N ≡ 81 (mod 324).
27. The only even perfect number of the form x3 + 1 is 28.
28. 28 is also the only even perfect number that is a sum of two positive integral cubes
29. Prove that no perfect number can be a square number.
30. The number of perfect numbers less than n is less than c\sqrt{n}, where c > 0 is a constant.
31. Numbers where the sum is less than the number itself are called deficient, and where it is greater than the number, abundant. Give an example of each.
32. In number theory, a practical number or panarithmic number is a positive integer n such that all smaller positive integers can be represented as sums of distinct divisors of n. Prove that any even perfect number and any power of two is a practical number.
33. Prove that prime numbers are infinite. (Use proof by contradiction).
34. Prove that 2n_{C_{n}} is divisible by a prime number n, iff, n's p-ary repesentation contains all digits less than p / 2.
35. Prove that if N is a perfect number, none of its multiples or factors is a perfect number.
36. when written in decimal notation, how many trailing zeros are present in 32!. Can you generalize it for n.
37. Prove that only square numbers have odd number of factors.
38. Prove that the number of factors of a number n, is always not greater than 2 * \lfloor\sqrt{n}\rfloor.
39. Continuing on 38, is there any number for which the number of factors is equal to the upper bound? Why?
40. when written in decimal notation, what digit do powers of 5 end with? What about powers of 6?
41. Prove that in x^{4} + y^{4} = z^{4} (x, y and z are positive integers), at least one of x or y is divisible by 5.
42. Consider a number with maximum number of factors between 1 and n^{2}. Prove that such a number is not divisible by any prime number p >= n, for n > 4.
43. Prove that n! cant be a square number for any n > 1. (My proof involves the proven conjecture that there exists a prime number between 2n and 3n, for n > 1).
44. Every prime number p > 3 satisfies the following: p^{2} = 24*k + 1 for some positive number k.
45. Also, every prime number p > 30 satisfies the following: p^{4} = 240*k + 1 for some positive number k.
46. Prove that a composite number (N) definitely has a prime factor (<= \lfloor\sqrt{n}\rfloor).
47. Prove that the fraction of "the number of divisors of first n prime numbers", in any N numbers is same as (1 - 1/2) * (1 - 1/3) * (1 - 1/5) * ... (Denominators are all first n prime numbers. Hint: Generalize Eratosthenes Sieve).
48. Using +, -, *, /, \sqrt, . (decimal), ! (factorial) and ^ (to the power of), find a way to represent 73 using four fours.
49. Do same as 48, for 77.
50. Do same as 48, for 87.
51. Do same as 48, for 93.
52. Do same as 48, for 99.
53. There is a way to represent all positive numbers using repeated trigonometric functional application on a single 4. Could you get it?
54. Prove that successive terms of a fibonacci sequence are relatively prime.
55. Prove that if the n th fibonacci number is denoted as F_{n}, F_{n} | F_{mn} for any m, n.
56. Consider the fibonacci sequence modulo some positive number k. Prove that modulo k, the fibonacci sequence has zeros equally spaced (periodically arranged).
57. Prove that the n th prime number, is less than n^{2}, for n > 1.
58. Prove that, in every 6 numbers (numbers themselves greater than 6), there can only be 2 prime numbers.
59. Continuing 58, also prove that every 15 numbers, there can only be 4 primes.
60. Continuing 59, also prove that every 35 numbers, only 8 can be primes.
61. Continuing 60, can you generalize it.
62. Continuing 61, can you generalize it as a formula representing number of prime numbers from 1 to N.
63. Continuing 62, we seem to sort of know when the number of primes exceeds the known prime numbers so far. So, does it give a way of finding where next prime would lie? (Well, it did not give a precise enough formula for me. Looked very promising though.)
64. Prove that the fraction of numbers relatively prime to 3 are 2/3. Prove that the fraction of numbers relatively prime to 5 are 4/5. Also, prove that the fraction of numbers which are relatively prime to 15 are 8/15.
65. Continuing on 64, what do these results indicate?
66. Continuing on 65, can you generalize it (the fraction of relatively prime numbers to a number n) for an arbitrary number n (Consider the prime power expansion of n).
67. Show that if a number is deficient, all of its factors are deficient too.
68. Show that if a number is abundant, all of its multiples are abundant too.
69. Based on 67 and 68, devise a method to stop searching for perfect numbers, when starting search from 1 and going upwards.
70. Prove that no prime number is a perfect number.
71. Prove that no number of the form p_{1}^{k_{1}}, where p_{1} is a prime can be a perfect number.
72. Prove that no number of the form p_{1}^{k_{1}} * p_{2}^{k_{2}}, where p_{1} > 2 and p_{2} > p_{1} are primes can be a perfect number.
73. Prove that, if all prime factors of a perfect number (N) are more than p, prove that N must have at least log_{p/(p-1)}^{2} number of prime factors.
74. Prove that, for any prime number p, a^{(p-1)*n} + b^{(p-1)*n} = c^{(p-1)*n} for any positive n, and positive numbers a, b and c, implies that at least one of a, b and c are divisible by p.
75. Prove that any prime factor of (a^{5} - b^{5}) / (a - b) is of the form of 10*k + 11, for some non-negative k, or it is 5.
76. Generalize 75, generalize about any prime factor of (a^{p} - b^{p}) / (a - b) where p is a prime number.
77. Continuing on 75, also prove that 5 is a factor of that expression, iff (a - b) is divisible by 5.
78. Generalize similarly for a generic prime exponent p.
79. Depending on 75-78, consider a equality of the form a^{5} - b^{5} = c^{5} - d^{5}. Prove that if (a - b) is divisible by a prime number p not of the form 10*k + 11, then (c - d) also is divisible by p. (Taxicab problem related).
80. Can we say something about a (or b) in the equation, a^{p} + b^{p} = c^{p}. What observation can be made about it?
81. How many 2 digit numbers are divisible by the digits in them. (If same digit occurs multiple times, divide multiple times with the digit. Exclude zeroes from divisors)
82. How many three digit numbers are divisible by all digits in them
83. If n! + 1 = m^{2} has solutions, then n! (consider n beyond trivial ones for which we know some solution, like, 4, 5 and 7) can be factorized into two numbers a and b such that a - b = 2, and ab = n!.
84. Continuing on 83, a and b are such that, one of them is divisible by 2 but not by 4.
85. Continuing on 84, a and b are such that, every prime number n > p > 2, is a factor of exactly one of them, but not both.
86. Continuing on 85, a is of the form 2 * op_{1}^{k_{1}} * op_{2}^{k_{2}} * op_{3}^{k_{3}} * ... and b is of the form 2^{k} * op_{a}^{k_{a}} * op_{b}^{k_{b}} * op_{c}^{k_{c}} * ... where op_{i} are all distinct odd primes.
87. Continuing on 86, you have to prove that for large k, op_{1}^{k_{1}} * op_{2}^{k_{2}} * op_{3}^{k_{3}} ... + 1 (or -1), where op_{i} are odd-primes cannot have high powers of 2 as a factor (for solving the Brocard's problem). Proof, anyone?
88. Continuing on 87, prove that 3^{n} + 1 is divisible by either 2 or 4, but not by higher powers of 2.
89. Continuing on 88, prove that 5^{n} + 1 is always divisible by 2, but not by higher powers of 2.
90. Continuing on 89, prove that p^{2n} + 1, where p is a prime number, is always divisible by 2, but not by 4.
91. Continuing on 90, prove that p^{2n+1} + 1 is divisible by the same power of 2, that divides p + 1.
92. Continuing on 91, prove that 3^{2n+1} - 1 is always divisible by 2, but not by higher powers of 2.
93. Continuing on 92, prove that 5^{2n+1} - 1 is always divisible by 4, but not by higher powers of 2.
94. Continuing on 93, prove that p^{2n + 1} - 1, where p is a prime number, is always divisible by the same power of 2, that divides p - 1.
95. Continuing on 94, prove that p^{2n} - 1, where p is a prime number, is always divisible by same power of 2 (r), unless n is a power of 2 itself.
96. Continuing on 95, prove that p^{2^{n}} - 1, where p is a prime number, is always divisible by 2^{n+k}, where 2^{k} divides p^{2i} - 1 maximally, where i is not of the form 2^{o}.
97. Continuing on 96, one can give simpler formulation of problem in 83. For this, prove that 2^{n} is never a factor of n!.
98. Continuing on 97, prove that n! always has a factor of the form 2^{(1 - \delta) * n}, where \delta decreases with increasing n.
99. Continuing on 98 and 83, for such an n to exist, you have to prove that for large n, op_{1}^{k_{1}} * op_{2}^{k_{2}} * op_{3}^{k_{3}} ... + 1 (or -1), where op_{i} are odd-primes should divide 2^{(7 * n)/8} ((7 * n) / 8 is for instance).
100. Prove that 2^{p−1} * (2^{p}−1) is an even perfect number whenever 2^{p}−1 is a prime (Euclid).
101. Prove that, if p is an odd number such that p + 1 = 2^{k_{1}}*o_{1}, where o_{1} is an odd number and k_{1} > 1, and q is another similar number (like p), then pq + 1 is divisible by 2 and not by 4.
102. Prove that, if p is an odd number such that p + 1 = 2*o_{1}, where o_{1} is an odd number, and q is another similar number (like p), then pq + 1 is divisible by 2 and not by 4.
103. Prove that, if p is an odd number such that p - 1 = 2^{k_{1}}*o_{1}, where o_{1} is an odd number and k_{1} >= 1, and q is another similar number (like p), then pq - 1 is divisible by min(k_{1}, k_{2}) unless k_{1} = k_{2}.
104. Prove that, if p is an odd number such that p + 1 = 2^{k_{1}}*o_{1}, where o_{1} is an odd number, and q is another similar number (like p), then pq + 1 is divisible by highers powers of 2 than (2^{1}), iff, exactly one of k_{1} and k_{2} is equal to 1.
105. Prove that, op_{1}^{k_{1}} * op_{2}^{k_{2}} * op_{3}^{k_{3}} * ... + 1 where op_{i} are odd primes, can be reduced to something that divides 2 but not 4, or, something of the form (2*o_{1} - 1) * (2^{k}*o_{2} - 1) + 1 for some odd numbers o_{1} and o_{2}.
106. State and prove the conditions where the expression in 105 divides only 2 but not 4.
107. Consider the case in 105, where o_{1} = 1. In this case, prove that it does not result in an n such that n! + 1 = m^{2}. (k is not as big as (7*n)/8).
108. Continuing on 103, prove that all prime numbers between n/i and n/(i+1) (where i >= 1) have the same power in n!.
109. Continuing on 103, prove that only the prime numbers between n/2 and n have an exponent of 1, in the prime number expansion of n.
110. Continuing on 10, consider (2^{k}*o_{2} - 1) = 3^{k_{1}} (Since any such prime number would do). Prove that if 3^{k_{1}} divides 2^{l} - 1, then l is such that l = (4 * k_{1} - 2).
111. Prove that, \pi . p_{i}^(1 / (p_{i} - 1)) (Here, \pi means product of), where p_{i} are prime numbers (starting from 2), does not converge as i increases.
112. Continuing on 111, prove that the lower-bound of the expression in 111, is, (2\pi n)^{1/(2n)} . n/e where e is the natural logarithm base (Hint: Remember something, yeah!).
113. Prove that, if 2n numbers are present, where the minimum difference between then is \delta and maximum difference is a polynomial function of \delta with constant term zero, then any difference between two terms formed by multiplying n each numbers is always divisible by \delta.
114. Prove that the exponent of prime number p in n! is always not less than (n + 1)/(p -1) * (1 - 1/(p^{\lfloor log_{p}^{n} \rfloor})) - \lfloor log_{p}^{n} \rfloor.
115. Prove that the exponent of prime number p in n! is always not greater than n/(p -1) * (1 - 1/(p^{\lfloor log_{p}^{n} \rfloor}))
116. Prove that, all numbers between n/i and n/(i + 1) have same power in prime number expansion of n!, whenever i < \sqrt (n).
117. Continuing on 116, prove that, for two prime numbers p_{1} and p_{2}, belonging to same interval (p_{1} > p_{2}), and having exponent k in prime number expansion of n!, the ratio of p_{1}^{k} to p_{2}^{k} is always less than e (the natural log base).
118. Also prove that, when a solution to Brocard's problem exists, the prime terms in expansion of n!, should be distributed across two terms (A and B) such that, A - B = 1, and AB * 4 = n! + 1.
119. Continuing on 118, prove that either A or B contains 2^({k_{1}} - 2), where k_{1} is the exponent of 2 in n! (meaning, its prime number expansion).
120. Continuing on 119, prove that if equal number of prime numbers are segregated into A and B, a solution to Brocard's problem is not possible. (Use 113).
121. Continuing on 117, if prime numbers within same interval, are factored such that equal number of them are in A and in B, then give an estimate of their ratio, in terms of e (Reduce the ratio as much as possible).
122. Prove that for large n, the smallest term, in the prime number expansion of n! is always not lesser than (n + 1) / 2.
123. MRB constant is given by , \sigma_{n=1} (-1)^{n} * 1/(n^{n}) (till infinity). Prove that, for any first n terms of the MRB constant (denoted by MRB(n)), (MRB(n))^{n!^{n}} is a rational number.
124. Prove that, any sub-sequence of terms for the MRB constant, is not a transcendental number.