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Yes Anupama. Provided it is a 2-figure whole number.
It is equivalent to what we are doing – dividing the remainder by twice the first digit.
That is a great answer Preeti.
square root of 3 = a fifth of the square root of 75.Or, similarly: square root of 3 = a half of the square root of 12.
Another method is to note that since the square root of 300 is about 17 we can use that, i.e. have a divisor of 34.
Challenge Question 21
Noting that one element of a perfect triple is always divisible by 3 (see Challenge Question 19) how can the digit sum check for triples be simplified?
Hi Satya,
I have to say I have not noted this pattern before. It seems 1, 4, 7 are the only possible digits.
How that might be used or extended are open questions. Please keep looking – I will too.
In fact you have rather anticipated the next Challenge Question which I will post in a moment.
Hello Fabio,
Yes the reducing works as you say. Every time the quotient is reduced by 1 the remainder increases by the divisor.
So 12/2 = 6 rem 0 = 5 rem 2 = 4 rem 4 etc.
Good observations Satya.
So you could say triples these reduce to 9,7,7 and 9,4,4 and 9,1,1.
Are there triples that reduce to 9,2,2 or 9,3,3 etc?
And does that mean that all triples can fall into 9 classes?
Not quite Anupama.
6,8,10 is a perfect triple because 6^2 + 8^2 = 10^2.
The 2, 3, 5 property is only a property and does now prove ‘tripleness’ or non-tripleness.
Yes that’s right Amara. If you take the Ekadhika as 10 you will be working 1 figure at a time.
But if you take it as 1 you must work 2 figures at a time (as in lesson 32).
Try it.
Excellent Preeti. That is a really good way to do the division.
These little tweaks are simple to do and make the problem much easier.
Challenge Question 20
Finding the square root of 3 could be a bit tricky as the first figure is 1 and the divisor is therefore 2, i.e. quite small, leading to many subsequent reductions.
How can the square root of 3 be found more easily?
Challenge Question 19
How can the digit sum check be used to verify that 3 numbers form a perfect triple?
For example, I added two triples, got 77,36,85 and now wish to check this is a triple.
How can digit sums be used to verify that 77,36,85 is a triple?Well done Satya. That’s right, in fact 1, 2, 3, 4, 5 will always divide into at least one of the elements of a perfect triple.
