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Philoman's Question Archive new (1-10) (11-20) (21-29) |
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Suppose you go to a bookstore and buy 11 copies of the same book. After coming home, in how many ways can you place these 11 identical books on 3 bookshelves (considering that each shelf can hold 11 books at a time)?
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Let's represent the books by "0"s and let's use "1"s as seperators. What do I mean? For example,
0010000100000
0000010000100
are two different arrangements that can be done. "1"s seperate the "0"s into three different groups in the same way that you should group 11 books into 3 shelves. As all books are exactly same, there is no problem in this representation. Now, the problem is simplified; in how many ways can you arrange 11 "zero"s and 2 "one"s. This can be calculated easily with repeated permutation or combination. Both would give the same result. The result equals 13! / (2! * 11!) = 78.
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Suppose you go to a bookstore and buy 11 different books. After coming home, in how many ways can you place these 11 different books on 3 bookshelves (considering that each shelf can hold 11 books at a time)?
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Consider that you are a book. You have three choices to go and be placed. You have 10 other friends and all of them are in the same position as you are. As each book has 3 choices and there are 11 different books, the result equals the product of eleven threes, in other words, 3^{11}.
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Prove that if a number is not divisible by any of the prime numbers upto its square root, then it is another prime number.
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All integers can be grouped into two: composite or
prime. An integer is either composite or prime, there is no third possibility. Composite
numbers can be expressed as product of prime factors; every composite number has at least
two prime factors. At least one prime factor of a composite number must be less than the
its square root. To prove this, let's suppose that a composite number has prime factors
that are greater than its square root only. Since there has to be at least two prime
factors and each are greater than the square root of the composite number, their product
is greater than the composite number itself. So, it is not possible such a possible
number.
(To express it mathematically, let's do this:
x = a * b (where a and b are prime numbers)
a > Öx
b > Öx
Then, a * b > x, which is impossible. If there are more factors, then the
product would be even greater; so, it doesn't make a difference.)
If a number that has no prime factors less than its square root can not be
composite, then it has to be prime (since we have said there is no third possibility).
QED.
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Suppose you are given
lengths of 3, 4 and 5 units on a plane and you are asked to construct a length of 1,4
units using only compasses and a ruler (which you are not allowed to use to measure
lengths of course). (Hint: little trigonometry is needed.) ---- In this example it is easier to work on the figure. But before starting, let's complete the trigonometric part of the question that I gave as a hint. If you construct a 3, 4, 5 triangle, it is a right triangle; the triangles ABH and AHC are 3, 4, 5 right triangles where |BH| = |HC| = 3, |AH| = 4 and |AB| = |AC| = 5. I think you will not have any difficulty in constructing the triangle ABC. Now, let's say m(BAH) = m(HAC) = x. Then, sin(x) = 0,6 and cos(x) = 0,8. Since sin(2x) = 2 * sin(x) * cos(x), sin(2x) is equal to 0,96 and knowing sin(2x), we calculate cos(2x) = 0,28. (Note that m(BAC) = 2x). Suppose you have a right triangle with interior angles of 2x and 90-2x and a hypotenuse of 5 units in length. Then using the sine and cosine values, we can conclude that one of the right sides of such a triangle will be 1,4 units in length. Now we have an angle of 2x already drawn, that is BAC, and the only problem is to draw the right triangle with a hypotenuse of 5. To do that, measure |AC| with the compasses and draw an arc taking A as center. Then, measure |BC| with the compasses and draw a second arc taking B as center. It is C' where the two arcs intersect. Note that |CC'| is perpendicular to |AB|. If you look at the final figure carefully, you may see that ADC is the triangle that I've described before, the one with interior angles of 2x and 90-2x and hypotenuse of 5. So, |AD| = 1,4. |
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Assuming that n and k are integers, prove that a prime number in the form of 3n+1 can also be expressed in the form 6k+1. (Example: 31 = 3 * 10 + 1 and 31 = 6 * 5 + 1)
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Suppose you have a prime number p in the form of 3n + 1, where n is either an odd or an even integer. Let's say n is odd, or in other words n = 2k + 1 where k is an integer. Substituting for n, we get p = 3 (2k + 1) + 1 = 6k + 4 which is obviously divisible by 2. In other words, n can not be an odd number if we want p to be a prime number. Since n has to be an even number in order to make p a prime number, substituting n = 2k, we get p = 6k + 1. So, all prime numbers that are in the form 3n + 1 are also of the form 6k + 1.
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Using intersection points of 5 lines, how many triangles could be formed at maximum?
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First of all, we need to get the maximum number of intersection points in order to form the maximum number of triangles with them. To do this, we assume that no three lines intersect at the same point. With that assumption, the number of intersection points would equal C(5,2) where C represents combination and it is equal to 10. So, we can have 10 intersection points at maximum. Now let's calculate the number of triangles that could be formed with 10 points. It is equal to C(10,3), which you can easily calculate as 120. However, there is something important to be considered. Some of those points that we are using to form triangles are collinear. Three collinear lines do not form a triangle, they form a line. There are four collinear points on each of the 5 lines which we used at the beginning. So, considering any one of the lines, we must exclude C(4,3) = 4 possible triangles which are in fact not triangles but lines. Having 5 lines in total that makes 20 triangles that are not triangles in fact. So, the number of triangles that could be formed at maximum is 120-20=100.
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Show that if f is both even and odd, f(x)=0 for every x in the domain of f.
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For all x: If f is an even function, then f(x) = f(-x); also, if f is an odd function, then f(x) = -f(-x). Using these two equalities f(-x) = -f(-x), which is only possible if f(-x) = 0. Then for all x, f(x) = f(-x) = -f(-x) = 0.
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Show that f(x) = (x-a)^{2}(x-b)^{2} + x is equal to (a+b)/2 for some value of x.
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First, define a new function h(x) = f(x) - (a+b)/2. Now, h(a) = f(a) - (a+b)/2 = a - (a+b)/2 = (a-b)/2 and h(b) = f(b) - (a+b)/2 = b - (a+b)/2 = (b-a)/2. Therefore h(a) and h(b) have opposite signs and for some value c between a and b h(c) = 0. Then, as h(c) = 0, f(c) - (a+b)/2=0 and therefore f(c) = (a+b)/2.
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Fixed Point Theorem: Suppose that function f is continuous on the closed interval [0,1] and that 0 £ f(x) £ 1 for every x in [0,1]. Show that there must exist a number c in [0,1] such that f(c) = c (c is called fixed point of f).
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To begin with, if f(0) = 0 or f(1) = 1, the hypothesis of the theorem is satisfied. If these two possibilities are omitted, let us define a new function h(x) = f(x) - x. Then h(1) = f(1) - 1 < 0 since f(1) < 1. Similarly, h(0) = f(0) - 0 > 0 since f(0) > 0. Then, according to to the Intermediate Value Theorem (it could be applied because both f(x), x and therefore h(x) is continuous on [0,1]), h assumes zero for some value of x between 0 and 1. since h(c) = 0 at some point x = c, f(c) - c = 0 and finally f(c) = c for some c between 0 and 1.