Choose two digits and arrange them to make two double-digit
numbers. Now add your double-digit numbers. Now add your single
digit numbers. Divide your double-digit answer by your single-digit
answer. Try lots of examples. What happens? Can you explain it?
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.
What is the sum of all the digits in all the integers from one to
No correct solution to this problem was originally received.
However, Mary of Birchwood Community High School gave a sound
argument that just needs some adaptation. Thank you Mary..
Firstly Mary considered the number of six digit numbers - this
10% of all six digit numbers start with a 5. So 90,000 six digit
numbers are of the form 5******
This leaves 810,000 numbers that do not start with a 5. How many
of these have a 5 as the second digit??
And so on.....
Here is a solution to this toughnut from Junwei of BHASVIC
Let the six digits number is abcdef, which a, b, c, d ,e, f
represent a digit respectively.
For a, neither 0 nor 5 could place in it, thus, 8 digits are
available here (1,2,3,4,6,7,8,9)
For b, c, d, e and f, they can't contain 5, hence, 9 digits are
available for them (0,1,2,3,4,6,7,8,9)
Therefore, the no. of six digits number which does not contain
any 5 is
8 * 9 * 9 * 9 * 9 *9 =472392 .