When an unspecified heap of beads is divided into two equal piles
one bead is left over.
When the same heap is divided into three, four, five, or six, equal
piles, that remainder of one persists.
With this information can you say how many beads there were?
Is a solution even possible?
Is there more than one solution?
Was the result predictable?
Extend the enquiry by investigating other remainders, other
divisors, or even other rules for distribution.
Approaching a solution (off-screen planning is essential), one
layout could be:
- Column one: the number of beads. The first row has one, the
next two, and so on.
- Column two: the remainder after a division by two.
- Column three: the remainder after a division by three.
- Until all the required calculations are displaying their
Solutions occur wherever a row contains all ones.
Conditional formatting has been applied to automatically highlight
cells that hold a value of 1, but this is not necessary.
It is now immediately apparent that one solution is to have 61
beads, while scrolling down further reveals more solutions at 121,
181 and so on at intervals of 60.
Spared the tedious individual calculation we are free to ask: "why
intervals of 60"?
We might look for a relationship between this interval of 60 and
the divisors, 2, 3, 4, 5, 6, and notice that 60 is the lowest
common multiple of all these divisors. Is the lowest common
multiple was relevant to the problem?
This demonstration compresses the problem solving process from a
couple of hours to a few minutes. The spreadsheet is not the focus,
the problem is. The spreadsheet has merely produced helpful
numerical results. The mathematics happened when we were trying to
understand the problem, trying out some numbers manually, thinking
how a spreadsheet might be useful, planning what was needed and how
to organise the calculations in columns, interpreting the results,
making connections, trying to justify our insights to ourselves and
For more on this theme, see
Excel Investigation: More Beads