The magic of 1001

Mike Rath
10th August 2016 at 18:29

Subject Genius, Mike Rath, The magic of 1001

I’ve been doing a little fun starter with small Year 5 groups recently.  I ask them to write down the number 91 then put the same single digit either side of it – like 3913 or 7917.  I then give out calculators and ask them to divide their number by 91.  The next step is to subtract 10 from their answer.   Finally, I ask them to divide their latest answer by the number they put at the front of their original 91 – and then I show them the number 11 that they finished with…hopefully!

Another way of ending the puzzle above is to divide by 91, subtract 10 and then simply say that they have a two-figure number consisting of the number they put in front followed by the number they had put at the back of their 91 originally.

This particular Year 5 seems to expect something strange from me all the time.  I assumed, for example, that they would gasp at the idea that all their original numbers would divide by 91, whatever they had put front and back - but no, no surprise at all!

Subject Genius, Mike Rath, The magic of 1001

The next thing I did was to ask them to start with a similar 4-digit number, but this time the number in the middle was 77.  This was divided by 77, then 10 was subtracted.  Then I asked them to divide by their original front number and produced the number 13 – which they had all finished on.

Can this be done with other numbers?  Well, since when I started with 91, I finished with 11 then, of course, if I start with 11 I will finish on 91.  Equally, if I start with 13, I will finish with 77.  The only other number I could start with is 07 – not just 7 because I need to create a 4-digit number – and then I would finish on 143.

By now, I am hopeful that you are wondering why I’ve headed this “the magic of 1001” – or, alternatively, you have worked it out.  91 × 11 = 77 × 13 = 7 × 143 = 1001 and, as far as I can work out, these will be the only numbers for which it will work.  I can’t put 143 in the middle because that would create a 5-digit number, and I want a 4-digit number.

This all started for me because I have long known the divisibility test for 7, 11 and 13.  In a 4-digit number, you take the first number from the remaining 3 and test the answer for divisibility by 7, 11 and 13.  So, for example, 8134 will divide by 7 because 134 – 8 = 126, and 126 divides by 7.  Similarly, 34125 will divide by both 7 and 13 because 125 – 34 = 91, which is 13 × 7.  Actually, in that case I guess it means that 34125 = 3 × 5³ × 7 × 13, since 34125 = 125 × (34 × 8 + 1) or 125 × 273, as 125 × 8 = 1000.

For a 6-digit number like most phone numbers, you subtract the smaller 3 from the larger 3 to test for 7, 11 and 13.  For example, 831 611 divides by 11 because 831 – 611 = 220, and 220 divides by 11.  Similarly, 357 578 divides by 13 because 578 – 357 = 221, and 221 divides by 13.

 

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