About POTW

Dear mathematics enthusiast,

We will be posting problems from the Singapore Mathematical Olympiad (SMO) archives. Each problem will be followed by its solution one week later to allow sufficient time for you to attempt it.

Your comments, including solutions are welcome. I may choose to not approve your comments right away so as not to give the game away.
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POTW142 solution

Answer to POTW142 is -1.

Rationalizing the denominators we have


Find the value of \dfrac{1}{1-\sqrt[4]{5}}+\dfrac{1}{1+\sqrt[4]{5}}+\dfrac{2}{1+\sqrt{5}}.

This is problem 4 from the 2014 SMO Junior Round 1.

POTW141 Solution

Answer to POTW141 is 2010.

The answer is not difficult to guess but since this is a round 2 problem, a rigorous solution is required.
The first thing to check is that the operation of erasing two numbers and then writing x+y+xy is a commutative and associative operation, i.e. the order of computation does not matter at all and we can assume the operation was done from \frac{1}{1} to \frac{1}{2} to … \frac{1}{k} to … \frac{1}{2010} in that order.

Next we can make a list to look for patterns:
1 + \frac{1}{2} + 1 \times \frac{1}{2} = 2
2 + \frac{1}{3} + 2 \times\frac{1}{3} = 3
3 + \frac{1}{4} + 3 \times\frac{1}{4} = 4

The pattern is clear and a straight forward induction argument suffices. We assume S_n = n where S_n is the value obtained from combining \frac{1}{1} to \frac{1}{2} to … \frac{1}{n}.

Assuming S_k = k, then we have
S_{k+1} = k + \frac{1}{k+1} + k \times \frac{1}{k+1} = k+1. Thus S_{k+1} = k+1 holds and our assumption holds for all n.


The numbers \frac{1}{1}, \frac{1}{2} , \ldots, \frac{1}{2010} are written on a blackboard. A student chooses any two of the numbers, say x, y, erases them and then writes down x+y+xy. He continues to do this until only one number is left on the blackboard. What is this number?

This is problem 2 from the 2010 SMO Senior Round 2.

POTW140 Solution

Answer to POTW140 is 5.

We first recall the fact that all primes except 2 is odd, and 2 cannot be the sum of two primes.
As the sum (similarly the difference) of two odd primes would be even, it forces the specific prime we are looking for to be of the form q-2 and r+2 for some primes q and r.

Hence we are essentially looking for three consecutive odd numbers which are all prime, i.e. p-2, p, p+2.

By considering the remainder of p modulo 3, one can check that at least one of p-2, p and p+2 must be divisible by 3. This means that p-2=3 and p=5 is the unique solution.


Find all primes that can be written both as a sum of two primes and as a difference of two primes.

This is problem 3 from the 2013 SMO Junior Round 2.

POTW139 Solution

Answer to POTW139 is 29.

If we let a and b be the solutions, by the sum and product of roots, we have
a+b = p and
ab= -580p = -4 \times 5 \times 29 \times p.

So p \mid ab \implies p \mid a or p \mid b. But a+b = p would then force p to divide both roots. Thus the only choices for p are 2, 5 or 29. The conclusion p=29 can then be deduced.


Let p be a positive prime number such that the equation x^2- px-580p=0 has two integer solutions. Find the value of p.

This is problem 27 from the 2003 SMO Senior Round 1.

POTW138 Solution

Answer to POTW138 is 1611.

n^4 +5n^2 + 9 \equiv n^4 - 1 \equiv (n-1)(n+1)(n^2+1) \pmod{5}.

If n \equiv 1, 2, 3, 4 \pmod{5}, then the above expression is \equiv 0 \pmod{5}. All integers except multiples of 5 satisfy the requirement.
The final answer is thus
2013 - \lfloor \frac{2013}{5}\rfloor = 2013 - 402 = 1611.


Find the total number of positive integers n not more than 2013 such that
n^4+5n^2+9 is divisible by 5.

This is problem 20 from the 2013 SMO Open Round 1.