Is there an efficient way to work out how many factors a large number has?
Choose any 3 digits and make a 6 digit number by repeating the 3
digits in the same order (e.g. 594594). Explain why whatever digits
you choose the number will always be divisible by 7, 11 and 13.
Find the number which has 8 divisors, such that the product of the
divisors is 331776.
If the five digit number A679B, in base ten, is divisible by 72
then A = 3 and B = 2. The number is 36792.
This is how Jamie from Hethersett High School
Norfolk explained his method.
To be divisible by 72 the number must be divisible by 8 and 9 as
8 x 9 = 72. I don't know a rule for finding if a number is
divisible by 8, but I realised that I could look for a number
divisible by 4. The rule for this is that if the last two digits of
a number are divisible by 4 then the whole number is divisible by
4. In the nineties (as, in the question, 9 is the second from the
last digit) only 92 and 96 are divisible by 4. Bearing these
numbers in mind I brought the rule about a number being divisible
by 9 into play. To be divisible by 9 the sum of all the digits in
the number added together must be divisible by 9. These are the
sums I did.
A + 6 + 7 + 9 + 2 = 24 + A. To make this up to a number
divisible by 9, A would have to be 3.
A + 6 + 7 + 9 + 6 = 28 + A. To make the number divisible by 9, A
would have to be 8. However when 86796 is divisible by 4 and 9 but
not by 8.
You may like to try a similar problem
sent by John of Madras College: ```A512B is divisible by 65,
find A and B''.
Well done Sonya, Genevieve, James, and Angela
of Hethersett High School, Norfolk; Claire, Kim, Joanna, Rhona,
David, Elisabeth, Nicola, Colin, Ross, Myles, and Helen from Madras
College, Scotland; Danielle from The Mount School York; and Chin
Siang and Xinxin of Tao Nan School, Singapore for your excellent
The following solution came from Stephen.
We have been learning BBC BASIC since September and have just
learnt the for-next loop. We used this to write a programme, shown
below, to solve this problem.
10 FOR A = 1 TO 9
20 FOR B = 0 TO 9
30 IF (10000*A + 6790 + B)/72 = INT((10000*A + 6790 + B)/72)
THEN PRINT A,B
40 NEXT B
50 NEXT A
So A = 3 and B = 2.