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.
This is an engaging context in which to reinforce rules of divisibility and challenge students to reason mathematically and work systematically.
If students are unfamiliar with the standard divisibility tests for two, three, four and six, Dozens provides a good starting point. The article Divisibility Tests also provides a useful reference to explain where the rules come from.
Introduce the game, and hand out some digit cards so that students can play in pairs.
Once they have had the chance to play a few times:
"Has anyone come up with any useful strategies that they'd like to share? Are there advantages to going first, or going second?"
Discuss any strategies that have emerged (including a discussion of divisibility rules if necessary).
"Has anyone's game used all ten digits?" (very very very unlikely!)
"What's the most digits that anyone's game has used so far?" (probably around 5, 6 or 7)
"I'd like you to stop playing against each other now and instead work together to make the longest number you can so that as you place each digit it follows the rules of the game."
As students are working, you could stop them for the occasional mini-plenary to share any examples of long numbers that have been found so far, and ask the students who found them to explain the strategy they used to find it.
The task of finding a ten digit number that satisfies the rules could be left as a simmering activity (a long-term challenge that students can keep returning to when time allows).
How do you establish that a number is divisible by 2, by 3, by 4, by 5...?
This problem need not stop at a 10-digit number! Students can be challenged to come up with the longest number which fits the rule, by relaxing the need to use particular digits.
For example, is it possible to find a 15-digit number where
1. the number formed by the first 2 digits from the left is divisible by 2
2. the number formed by the first 3 digits form the left is divisible by 3
How far can they get? A 20-digit number....?
A more challenging extension is to prove that there is only one ten-digit number using the digits 0-9 which satisfies the rule.
Those who are not confident with factors and multiples could play the Factors and Multiples Game as well as Dozens beforehand.