One of the essential skills at GCSE maths is the ability to check an answer for sense. This is a good general skill too and one of the few that can usefully be carried to later life.

Generally, good checking requires you to see another route to an answer, so at the level of arithmetic, one should check that your number is roughly the right size.

There is a very useful technique called the check-sum which I never saw included in the syllabus (specification in later life) but is used in the world of computing.  A checksum comes from adding the digits of a number, repeatedly. So 975468 has a checksum of 9+7+5+4+6+8 = 39, whose check sum is 12, whose checksum is 3. There are swifter ways to this, because you are throwing away all nines: 9 Ignore; 7+5 is 12 is 3; plus 4 and 6 is 13, which is 4, plus 8 is 12 which is 3.

Let's look at multiplication. 975468 (checksum 3) squared = 951537819024 (checksum 9).  Might this imply that an operation continues to work on the checksum? 

Choose some numbers of 6 digit length to multiply (e.g. a class acting as a group, independently picking numbers to see if this might be a rule. Any exception is either a mistake or it show s that the proposal (the checksum of a product is the product of the checksums) fails. Find the product (multiply the numbers) with a calculator, but work out the checksum without a calculator (which should be quicker).  

You will discover that the proposition works.

Might the checksum be useful elsewhere? I suggest you look at addition, by writing yourself a list of up to ten four digit numbers. 


the checksum for a sum is the same as the sum of the checksums

the checksums for each column of figures in a sum adds to the checksum of the result.

The checksum for all multiples of 3 is itself a multiple of 3.

There are short cuts to checksum; all 9s can be ignored, as can pairs that add to nine. The checksum for 1990990909099 is obviously 1. Less obvious, 182736450 has a checksum of 0, but so does 10807 02036 40500, since that too pairs off easily. Note that a digit sum of 9 is now declared to be the same as a digit sum of zero. Produce evidence to show that these are equivalent. Add one: 9+1=10 -> 1, 0+1=1. these are the same.

Theory: the checksum is the same as the number modulo nine, the same as the remainder after dividing by nine. A little fiddling with test examples shows that it is this remainder that is carried across any operation. Sixth formers should attempt a proof.

The checksum is a reliable way to check an answer, provided you are using all the digits available. Yes, you could have made balancing mistakes. Yes, swapping two digits in a dyslexic way will not affect the checksum of any single number (combination of digits) but each such error changes the checksum of the answer, suggesting you need many mistakes to produce a checksum the same as the original problem. Yes, in some problems a bucketful of missing zeroes won't help the checksum show that you are out in magnitude. This is why checking for the sense of your answer is always a good habit. The checksum is a very useful precision check.

As an arithmetic skill, something building familiarity with numbers and facility with these, I suggest that checksum is worth playing with. The extension that allows you to find the arithmetic error in a string of calculations without repeating the calculation (which tends to repeat whatever the error is) strikes me a particularly useful. Those who rely upon machinery for calculators need to be swindled (perhaps more than once) before they will exercise suitable care that figures are acceptably close. How could three identical drinks come to £10 exactly?  [10 is clearly not divisible by 3, with a checksum of 1]

Rapidly you can check that this is so, not merely for multiplication but for all operators (amazing). be aware that division and roots give problems with for recurring decimals, so you need to build in some adjustment for 'remainders'. Explore.

Using S(number) for digit sum repeated until a single digit result is found

S(1234) + S(5678) = S(10) + S(26) = 1 + 8 = 9.      S(11234+5678) = S(6912)=S(18)=9    OK

Subtraction copies since S(N) = S(-N)

Multiplication: S(1234x5678)=S(7006652) = 8 = S(1234) x S(5678) = 1x8                  OK

Division then clearly works for finite results, since S(N) = S(1000*N).

What about index?  S(3^5) = S(243) = 9, S(12^5)   = S(248832) = 9 , S(39^5) = S(90224199) = 9. So it works — 3=S(3)=S(12)=S(39), but it is not trivial.

This S function takes the remainder after dividing by 9, which is the same as adding digits (Sixth form, prove this). So (I say "so") there is a n equivalent function that takes the remainder after dividing by eleven. Call this function E(n) for any n that is integer or can be made integer by sufficient multiples of ten.[this isthe same as any number that can be written exactly in Standard Form]. So E(48)=4,  E(86)=11-2=9. To find E(4563)=9 one way is to add the odd digits from the right and subtract the other digits (3+5)-(6+4)=8-10 = -2 => 11- 2 = 9. So E(356789236) = (6+2+8+6+3)-(9+3+7+5)=25-24=1

[This process is the same as treating each pair of digits to subtraction, which may prove an easier task;   03 56 78 92 36 has paired differences of 3, 1, 1, -7=4, 3 summing to 12 mod 11 = 1.]

Try this out: E(235x468)=E(109980)= -9 mod11 = 2 = E(235) x E(468) = 4x6=24 mod11 = 2.

Can you prove that the 11-system works if the 9-system does? That is, can you prove that function E and S produce related results?

Can you show why it is that the checksum system works? If you can show it works for addition that is sufficient, I think, because you can cover subtraction by arguing that       S(-n)=S(n). For multiplication you can argue that it is compound addition and therefore you can proceed to showing that compound addition of the S function will amount to multiplication of S function results. The argument for division is described above, and though you can probably come up with a modifying technique tor  the checksum of a repeating decimal, I don't see how you can for an irrational. Reverse that; suppose the checksum works for square roots (which we assume true); then S(√121))=S(11)=2, but also S(√11)=S(√2), which has no value as yet, since we have not explained how S applies to infinite numbers. I wonder whether there is a relationship between the progressive digitsums of √11 and √2. One might write a computer program to find out.

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