According to the Journal of Irreproducible Results, any obtuse angle is a right angle! Here is

their argument.
Given the obtuse angle x, we make a quadrilateral ABCD with � DAB = x, and � ABC =
90◦, andAD = BC. Say the perpendicular bisector toDC meets the perpendicular bisector to
AB at P. ThenPA = PB andPC = PD. So the trianglesPADandPBC have equal sides
and are congruent. Thus � PAD = � PBC. But PAB is isosceles, hence � PAB = � PBA.
Subtracting, gives x = � PAD− � PAB = � PBC − � PBA = 90◦. This is a preposterous
conclusion – just where is the mistake in the “proof” and why does the argument break down
there?
5. Consider a rectangular array of dots with an even number of rows and an even number of
columns. Color the dots, each one red or blue, in such a way so that in each row half the
dots are red and half are blue, and also in each column half are red and half are blue. Now,
whenever two points of the same color are adjacent (in a row or column), join them by an edge
of that color. Show that the number of red edges is the same as the number of blue edges.
6. The squares of an infinite chessboard are numbered as follows: in the first row and first column
we put 0, and then in every other square we put the smallest non-negative integer that does not
appear anywhere below it in the same column or anywhere to the left of it in the same row.
What number will appear in the 1000th row and 700th column? Can you generalize?
7. Let’s agree to say that a positive integer is prime-like if it is not divisible by 2, 3, or 5. How
many prime-like positive integers are there less than 100? less than 1000? A positive integer
is very prime-like if it is not divisible by any prime less than 15. How many very primelike
positive integers are there less than 90000? Without giving an exact answer, can you
say approximately how many very prime-like positive integers are less than 1010? less than
10100? Explain your reasoning as carefully as you can.
8. Find a positive integer m such that
1
2m is a perfect square and
1
3m is a perfect cube. Can you
find a positive integer n for which
1
2n is a perfect square,
1
3n is a perfect cube and
1
5n is a
perfect fifth

Now now..... no using outside sources!! If you continue to do this, we will track your IP address to find out who you are and disqualify you...

Find a polynomial with integer coefficients for which √ √ is a root. That is find such that for some non-negative integer , and integers

and(√ √) .

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To answer the questions posed in the text, we will go through each question and provide explanations or solutions:

1. According to the Journal of Irreproducible Results, any obtuse angle is a right angle. The argument provided in the question is flawed. Let's analyze it step by step:
- Given the obtuse angle x, we construct a quadrilateral ABCD with ∠DAB = x and ∠ABC = 90°.
- The perpendicular bisector to DC intersects the perpendicular bisector to AB at point P.
- The key mistake in the argument lies in assuming that PA = PB and PC = PD. This assumption is not true in general, and without this assumption, the subsequent reasoning falls apart.
- Therefore, the conclusion that x = 90° is incorrect. The argument is invalid, and obtuse angles are not equivalent to right angles.

2. Consider a rectangular array of dots with an even number of rows and columns, where each dot can be colored red or blue, such that in each row half the dots are red and half are blue, and the same holds for each column. We want to show that the number of red edges is the same as the number of blue edges.
- Each colored dot has exactly one adjacent dot of the same color in its row and another adjacent dot of the same color in its column.
- When we join these adjacent dots by edges of the same color, every red dot contributes precisely one red edge, and every blue dot contributes precisely one blue edge.
- Since half the dots are red and half are blue in each row and each column, and each dot contributes exactly one edge, the number of red edges must be equal to the number of blue edges.

3. The squares of an infinite chessboard are numbered as follows: the first row and first column have the numbers 0, and then, in every other square, we put the smallest non-negative integer that does not appear anywhere below it in the same column or to the left of it in the same row. We want to find the number in the 1000th row and 700th column.
- To find the number at a given position, we need to consider the numbers in the row above and the column to the left of the desired position.
- The number in the 0th row and 0th column is 0.
- To find the number at position (i, j), where i is the row number and j is the column number, we look at the maximum numbers in the row above (i-1, j) and the column to the left (i, j-1), and take the minimum value not yet used.
- For example, to find the number in the 1000th row and 700th column, we consider the numbers in the 999th row (above) and the 699th column (to the left).
- The number at (1000, 700) will be the smallest non-negative integer that does not appear in the numbers below (in rows 1 to 999) or to the left (in columns 1 to 699).
- To generalize, we can use the same approach to find the number at any position (i, j) on the infinite chessboard, following the rules mentioned above.

4. To answer the question about the number of prime-like positive integers less than 100, less than 1000, and less than 90000, we need to define what it means to be "prime-like." A positive integer is prime-like if it is not divisible by 2, 3, or 5.
- To count prime-like positive integers, we can use the concept of prime factorization.
- Prime factorization involves finding the prime numbers that divide a given number and expressing it as a product of those primes.
- For example, to check if a number is divisible by 2, 3, or 5, we only need to check if it is divisible by any prime factors of these numbers.
- To count the number of prime-like positive integers less than 100, we can subtract the count of positive integers divisible by 2, 3, or 5, from the total count of positive integers less than 100.
- Similarly, we can apply the same logic to find the count of prime-like positive integers less than 1000 and less than 90000.
- To estimate the number of prime-like positive integers less than 1010 and less than 10100, we can use the same approach but with larger numbers.
- Keep in mind that finding an exact count may require more calculations, but this reasoning can provide an approximate answer.

5. To find a positive integer m such that 1/(2m) is a perfect square and 1/(3m) is a perfect cube, we need to analyze the properties of perfect squares and cubes.
- A perfect square is a positive integer that is the square of another positive integer (e.g., 4, 9, 16, 25...).
- A perfect cube is a positive integer that is the cube of another positive integer (e.g., 8, 27, 64, 125...).
- To find m, we need to find a positive integer that satisfies both conditions: 1/(2m) is a perfect square and 1/(3m) is a perfect cube.
- Let's consider the prime factorizations of 2m and 3m.
- For 1/(2m) to be a perfect square, the prime factors of 2m need to occur in even powers.
- Similarly, for 1/(3m) to be a perfect cube, the prime factors of 3m need to occur in multiples of three powers.
- By analyzing the prime factorizations, we can find a suitable value for m that satisfies both conditions.
- Additionally, we can try substituting different prime numbers in the prime factorizations to find other possible values for m that satisfy the conditions.

6. Following the same logic, let n be a positive integer such that 1/(2n) is a perfect square, 1/(3n) is a perfect cube, and 1/(5n) is a perfect fifth.
- We need to analyze the prime factorizations of 2n, 3n, and 5n.
- For 1/(2n) to be a perfect square, the prime factors of 2n need to occur in even powers.
- For 1/(3n) to be a perfect cube, the prime factors of 3n need to occur in multiples of three powers.
- For 1/(5n) to be a perfect fifth, the prime factors of 5n need to occur in multiples of five powers.
- By analyzing the prime factorizations, we can find a suitable value for n that satisfies all three conditions.
- Similarly, we can try substituting different prime numbers in the prime factorizations to find other possible values for n that satisfy the conditions.