1. How many positive integers have square less than 107?
2. The squares of a chessboard have side 4. What is the circumference of the largest circle that can be drawn entirely on the black squares of the board?
3. What is the remainder on dividing 1234567 + 891011 by 12?
4. Given the real number k, find all differentiable real-valued functions f(x) defined on the reals such that f(x+y) = f(x) + f(y) + f(kxy) for all x, y.
5. A road has constant width. It is made up of finitely many straight segments joined by corners, where the inner corner is a point and the outer side is a circular arc. The direction of the straight sections is always between NE (45o) and SSE (157½o). A person wishes to walk along the side of the road from point A to point B on the same side. He may only cross the street perpendicularly. What is the shortest route? 
6. The real-valued function f(x) is defined on the reals. It satisfies |f(x)| ≤ A, |f ''(x)| ≤ B for some positive A, B (and all x). Show that |f '(x)| ≤ C, for some fixed C, which depends only on A and B. What is the smallest possible value of C?
Solutions
Problem 1
How many positive integers have square less than 107?
Answer
3162
Solution
I find this a somewhat baffling question (in the sense of why on earth would anyone set a question like this!). There are algorithms for calculating the square root, but no one is likely to remember them. So one has to resort to kludges. Probably most people remember enough squares to know that 312 = 961, 322 = 1024, so the answer is 31xx. We have 31002 = 9610000, which is 390000 short. We have 31xy2 = 31002 + 6200·xy + xy2. So we want roughly xy = 3900/62 = about 63. So we try 3162 and 3163.
Problem 2
The squares of a chessboard have side 4. What is the circumference of the largest circle that can be drawn entirely on the black squares of the board?
Answer
radius = √40, circumference = 2√40 π
Solution
Consider the arc of the circle that lies in a particular black square. Its endpoints must be vertices of the square (otherwise the continuation of the arc would enter a white square). Suppose the two vertices are opposite (A and B in the diagram below). Then the two vertices of the next square along the circle cannot be opposite or the circle would pass through three collinear points (so they must be B and C in the diagram). Hence the center lies on the the perpendicular bisectors of AB and BC, so it must be O. That works, as shown in the diagram above.
If no square on the circle has opposite vertices on the circle, then it is easily seen that the circle must pass through the four vertices of a square and so is smaller than the case above. (Of course, if the circle is contained in a single square, then it is smaller still.)
Problem 3
What is the remainder on dividing 1234567 + 891011 by 12?
Answer
9
Solution
1234 = 1 mod 3, so 1234567 = 1 mod 3. 89 = -1 mod 3, so 891011 = -1 mod 3. Hence 1234567 + 891011 = 0 mod 3. 1234 is even, so 1234567 = 0 mod 4. 89 = 1 mod 4, so 891011 = 1 mod 4. Hence 1234567 + 891011 = 1 mod 4. So it is 9 mod 12.
How many positive integers have square less than 107?
Answer
3162
Solution
I find this a somewhat baffling question (in the sense of why on earth would anyone set a question like this!). There are algorithms for calculating the square root, but no one is likely to remember them. So one has to resort to kludges. Probably most people remember enough squares to know that 312 = 961, 322 = 1024, so the answer is 31xx. We have 31002 = 9610000, which is 390000 short. We have 31xy2 = 31002 + 6200·xy + xy2. So we want roughly xy = 3900/62 = about 63. So we try 3162 and 3163.
Problem 2
The squares of a chessboard have side 4. What is the circumference of the largest circle that can be drawn entirely on the black squares of the board?
Answer
radius = √40, circumference = 2√40 π
Solution
Consider the arc of the circle that lies in a particular black square. Its endpoints must be vertices of the square (otherwise the continuation of the arc would enter a white square). Suppose the two vertices are opposite (A and B in the diagram below). Then the two vertices of the next square along the circle cannot be opposite or the circle would pass through three collinear points (so they must be B and C in the diagram). Hence the center lies on the the perpendicular bisectors of AB and BC, so it must be O. That works, as shown in the diagram above.
If no square on the circle has opposite vertices on the circle, then it is easily seen that the circle must pass through the four vertices of a square and so is smaller than the case above. (Of course, if the circle is contained in a single square, then it is smaller still.)
Problem 3
What is the remainder on dividing 1234567 + 891011 by 12?
Answer
9
Solution
1234 = 1 mod 3, so 1234567 = 1 mod 3. 89 = -1 mod 3, so 891011 = -1 mod 3. Hence 1234567 + 891011 = 0 mod 3. 1234 is even, so 1234567 = 0 mod 4. 89 = 1 mod 4, so 891011 = 1 mod 4. Hence 1234567 + 891011 = 1 mod 4. So it is 9 mod 12.
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Swedish Mathematical Society
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