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
If her age is 10 t + u (where t is the tens digit
and u is
the units digit) then reversing the digits gives 10
t. and the sum
is 11 t +
which is a multiple of 11. We know this has to be a square number.
Ong Xing Cong from Singapore sent in the following
11 t + 11 u = 11 x 11 = 121
t + u = 11
65 - 56 = 3 x 3
She is 65 years old.
The best solutions do not need trial and improvement methods and
they show that the answer or answers found are the only possible
answers. Knowing the digits add up to 11 ( t + u
= 11), you can also use (10 t + u ) - (10
u + t ) = 9 t - 9 u = 9(
t - u ) As this is also a square number you know
( t - u ) is either 1, 4, or 9. The solutions for
4 and 9 don't give whole number values for t in 0