Is there an efficient way to work out how many factors a large number has?
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
Helen made the conjecture that "every multiple of six has more
factors than the two numbers either side of it". Is this conjecture
Keep sending us YOUR OWN alphanumerics and we'll publish them in
collections from time to time. The following two came from Jonathan
Gill, St Peter's College, Adelaide, Australia.
There is a one-to-one correspondence between digits and letters,
each letter stands for a single digit and each digit is represented
by a single letter. How many different solutions can you find?
Ling Xiang Ning(Allan) form Tao Nan School, Singapore, who
solves many of are hardest problems, has sent 7 solutions to
CARAVAN and 88 solutions to AUSTRALIAN. Is this all there are? Here
is one solution to each.
Soh Yong Sheng, age 12, also from Tao Nan School, Singapore has
sent this solution for.
and there are al lot more.
We have the following solutions from Allan Ling (Tao Nan School,
Singapore): For the equation
T has to be 9 or 0, in order for it to satisfy T+A=A. However if
T=0, it is impossible, as H+0 is not L. So T has to be 9.
The following are the possible sums (total 59):
Jonathan also proved that the following alphanumeric does not
work, that is it cannot have any solutions. Well done Jonathan.
If it was an alphanumerics then H = 0 to satisfy 0 + S = S, but
then H cannot be zero, otherwise C + 0 (H) = C and not R. We know
that C and R cannot both represent the same number therefore
cannot be made into an alphanumeric.