| 1. (a) The proof is clear from this configuration, in which the colours of the three unshaded triangles must all be different from that of the shaded one. |
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| (b) Suppose that colours 1, 2, and 3 suffice. In this configuration, if square A is colour 1, then it is easy to see that B, C, D must be colours 2, 3, 2 respectively. E must then be 1, but this makes it impossible to colour F without using a fourth colour. |
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| 2. As with a well-known problem involving reflection of light in a
mirror, the key construction is to take M’, the mirror image of M in the
line. Produce PM’ to meet the line at Y. Then the extra distance Peter
has to run to get to Y is PY - MY = PY - M’Y = PM’. Now suppose that X is a point on the line other than Y. The extra distance Peter has to go to X is PX - MX = PX - M’X. By the triangle inequality, PX - M’X < PM’. So the greatest difference in Mary’s favour is when Y is the point chosen on the line. |
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| 3. Let p1, p2, p3, p4 be the probabilities of getting 1, 2, 3, 4 respectively with a single spin. Then the probabilities of getting the possible totals with two spins are as shown in the table. It is easy to show that no values of p1, p2, p3, p4 can satisfy the constraint that all of these expressions be equal to 1/7. |
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4. Clearly only positive integer values of n need be considered.
n4 + 4n = 1 for n = 0 and = 5 for n = 1. Are there
any other integral values of n that make the expression a prime? If n is
an even number (non-zero), clearly the expression is not prime. So suppose
n is odd.
n4 + 4n = (n2)2 + 2.n2.2n
+ (2n)2 - 2.n2.2n
Since n is odd, 2.n2.2n = n2.2n+1
= (n.2m)2 where m = (n +1)/2 is integral.
So the above can be factorized as a difference of two squares
((n2 + 2n) - n.m).((n2 + 2n)
+ n.2m)
where both factors are integers (and a bit more work shows both are greater
than 1 -- e.g. it is easy to show for n = 1, 3, 5 and thereafter that 2n
- n.2m is greater than 1 for n odd and greater than 5). So the
product of the two integral factors is not a prime.
(Problems 1 and 4 are from: Halmos, P. R. (1991). Problems for mathematicians young and old. Washington, DC: Mathematical Association of America).
