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JMO mentoring scheme answers
Charitable Trust
November 2011 paper
1
Ans : 7
2
Ans : 4
The worst case is if she picks three different colours on the first three selections. The fourth sock is
bound to be one of the three colours.
3
Ans : 6937
9 (unit numbers) + 2 ´ 90 (numbers in the tens) + 3 ´ 900 (numbers in the hundreds) + 4 ´ 1012.
4
Ans : 8 values
Let the digits be x, y, z : (100x + 10y + z) - (100z + 10y + x) = 100(x - z) + (z - x) = 99(x - z).
Assuming that 1 £ z < x £ 9, this gives 8 possible values for x - z, namely 1, 2, ..., 8.
So there are 8 possible answers, namely 99k with k = 1, 2, ..., 8.
5
Ans : 2
If Katrina turns round to face each corner in turn, the numbers can appear in three orders, d being
opposite a different number in each case. Choosing u, v, x, y as the distances to the edges so that
d² = x² + y², we can write down equations using Pythagoras’s rule.
We don’t need to find the values of u, v, x, y individually nor even their square values.
We find that d² is one of 49 + 81 - 36, 81 + 36 - 49, or 36 + 49 - 81 and must be the last one as the
others do not yield integer answers.
6
If two numbers are even, then 4 is a factor. Otherwise, five numbers must be odd and one even,
so one of the numbers is 9, and so 9 is a factor.
A more general method arises from thinking this way. There are 4 prime numbers between 1 and 10.
The product of these four with 1 itself as a fifth number can not be divided by a square number greater
than 1. Including a sixth number will add a further factor which is composed with one of these prime
numbers again.
Can you generalise the question?
7
Ans : 3Ö3.
Preliminary : any 30°, 60°, 90° triangle has sides in the ratio 1 : Ö3 : 2 which can be worked out by
splitting an equilateral triangle with sides of length 2 into two equal right-angled triangles.
Learn this!
Either (i) : because ÐDCE = 60 ° and because CD : CE = 1 : 2, DCDE is right-angled at D and DAEF
is congruent to DCDE. Thus AF = 4, BF = 2 and so DBFD is also congruent to these two triangles.
Or (ii) : if M is the midpoint of BC so that ÐAMB = 90 °, then DAEF is similar to DBMA.
Because AE : BM = 2 : 3, AF = 2 ´ AB ¸ 3 = 4. Again BF = 2 and DAEF, DBFD, DCDE congruent.
Finally : the quickest way to proceed from here is to see that DDEF is equilateral and that it has sides
of length DE = Ö12 = 2Ö3. We can now calculate the height of DDEF from the same ratio 1 : Ö3 : 2
or by using Pythagoras's rule. It comes to 3. Thus the area of DDEF = 3 ´ 2Ö3 ¸ 2.
A more extended calculation would find the area DCDE + DAEF + DBFD and subtract from the area
of DABC. Other routes can also be taken to get this result.