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PostPosted: Wed May 13, 2020 5:56 pm 
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Hi. I wonder if anyone can help please. We had this question:

Which of the following is divisible by 23?
a) 111, b) 293, c)1,495, d)938, e)1,235

Are there any hints or 'tricks' to answer this type of question? Like, I know to see if a number is divisible by 3 you add all the digits up and then if that number is divisible by 3 then yes it is. I wondered if something similar might apply here?

Or, is it that you have to do it individually/estimate?

I hope that makes sense! Thank you.


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PostPosted: Mon May 18, 2020 4:42 pm 
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It turns out that there is a trick to check if a number is divisible by 23.

To determine if a number is divisible by 23, take the last digit and multiply it by 7. Then add that to the rest of the number. If the result is divisible by 23, then the number is divisible by 23. If its still not clear, repeat the process.

a) 111 -> 1 x 7 +11 = 18. No :(
b) 293 -> 3 x 7 + 29 = 21 + 29 = 50. No :(
c) 1495 -> 5 x 7 + 149 = 35 + 149 = 184 Not sure :? . Repeat the process. 184 -> 4 x 7 + 18 = 28 + 18 = 46. Yes! :D


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PostPosted: Mon May 18, 2020 7:05 pm 
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Location: Solihull, West Midlands
That's a good one! Now scurrying off to prove how it works.....


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PostPosted: Tue May 19, 2020 9:29 am 
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I've no idea how it works. If you can prove how I'd be very impressed! :)

(BTW, do you know how the one for 3 and 9 works? i.e. why is the whole number divisible by 3 or 9 if the sum of its digits are?)


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PostPosted: Tue May 19, 2020 11:38 am 
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OK for the 23 here goes

A number is divisible by 23 if and only if we can add or subtract more multiples of 23 to/from it. And any multiple of a number divisible by 23 is also divisible by 23 (lots of hidden background stuff about prime numbers here)

We can rewrite any number as 10n + m where n and m are integers
- (so for example given the number 276 n=27 and m=6)

The "trick" given involves creating a new number by adding n (the "tens and hundreds") to 7m

So we have two numbers A (our original) which is also writable as 10n + m

and B (our new one) which is n + 7m

Now if B is divisible by 23, so is 10 times B which is 10n + 70m

But what's the difference between 10B and our original number ? Subtracting :
10B-A =10n +70m-10n-m = 69m

So the difference is 69 times an integer, which (as 69=3 x 23) is divisible by 23

Now if B is divisible by 23, so is 10B, so is (10B - 69m ) which is A

Well it makes sense to me anyway though I'm sure there's a neater proof!


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PostPosted: Tue May 19, 2020 2:00 pm 
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Well done! That sort of makes sense even if it does make my brain hurt a bit! :D

What about the divide by 3 and 9 rule? Is that any simpler to explain?


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PostPosted: Tue May 19, 2020 3:29 pm 
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Location: Solihull, West Midlands
Surferfish wrote:

What about the divide by 3 and 9 rule? Is that any simpler to explain?


I'm sure it is, and I suspect it will be something to do with 9 being one less than 10 which is the basis of our decimal number system.

Actually to save my brain (!) here is a good explanation via the wonders of Google...

https://www.facebook.com/notes/math-dud ... 580234021/


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PostPosted: Sat May 23, 2020 7:15 pm 
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This is brilliant. Thank you so much! I'll add it to my list to teach my DD :)

Thanks!


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