In turn 4 people throw away three nuts from a pile and hide a
quarter of the remainder finally leaving a multiple of 4 nuts. How
many nuts were at the start?
The nth term of a sequence is given by the formula n^3 + 11n . Find
the first four terms of the sequence given by this formula and the
first term of the sequence which is bigger than one million. Prove
that all terms of the sequence are divisible by 6.
Try to move the knight to visit each square once and return to the starting point on this unusual chessboard.
Luke of Madras College and Ling Xiang
Ning, Raffles Institution, Singapore, sent us answers to the
question "How many divisors does $n!$ have?"
Luke gave more detail:
To understand the reason for this, think
of each divisor being expressed as a product of prime factors. The
number of possible divisors is found by working out the number of
ways of choosing the prime factors of the divisor. The prime $p_1$
may not occur as a factor of the divisor, or it may occur to the
power 1 or 2 or 3 or any power up to at most $a$, hence there are
$(a+1)$ possibilities for a divisor to contain $p_1$ as a factor.
Similarly there are $(b+1)$ possibilities for a divisor to contain
$p_2$ as a factor and $(c+1)$ possibilities for a divisor to
contain $p_3$ as a factor and so on. We multiply these numbers of
possibilities to find the total number of possibilities.