[KennethWilliams]

Hi Anupama,

Those illustrations in video 30 are to compare the Vedic and conventional methods. In fact the solution of quadratics shown in this lesson is for when the coefficient of x^2 is 1, and in the illustration it is 3.

We cover this fully in the Advanced course but since you ask let me explain.

The next step is -50 divided by 13, which gives -4 remainder 2.

The next step differs in that when we subtract the Duplex of -4 from 20 we have to treble it first so we have 20-48=-28 and -28 divided by 13 gives -3 remainder 11.

Since you may ask why we say -28 divided by 13 gives -3 remainder 11 rather than -2 rem -2, this becomes clear once you start to proceed to the following step.

[KennethWilliams]

That’s right Savita. I hope you have seen the full guidelines after lesson 26.

[KennethWilliams]

Hi Shikha,

There is a thread for week 8.

[KennethWilliams]

Seth, There must be plenty of such resources available on the internet and they can be adapted to whatever topic needs them.

[KennethWilliams]

Hi Amara,

Actually no-one talks about remainders as far a square roots are concerned. I was trying to answer your question as best as I could. So I have explained about the remainders for 5229 in my reply above.

I believe you can see your answers to the quizzes and which ones are wrong after the quiz is completed. For an explanation please post in the thread or contact me.

[KennethWilliams]

Thanks Preeti.

Yes, we can take a=2 rather than 3 and get the answer that way too.

We cover this subject more fully in the Advanced Course.

[KennethWilliams]

Hi Savita,

1. Yes, we go leftwards in pairs from the decimal point.

2. I think you slipped up on this line:  D(4)=16, 60-14=46, 46/12=3r10

The 14 should be 16.

[KennethWilliams]

Hi Savita,

It is a perfect square because when the the remainder 3 is attached in the usual place we find 309 at the end of the number.

And this is brought to zero by the remaining Duplexes: D(53) = 30 and D(3) = 9.

[KennethWilliams]

Hi Savita,

In the first practice only one figure is required, so for 3000 for example you would just give 50.

In the second practice two figures are required, so for 3000 you get 55.

So yes the answers are different but the 2nd is more accurate than the 1st.

[KennethWilliams]

Challenge Question 18

Missing Digits – Left to Right

Use vertically and crosswise to find the missing digits:

123 x ??? = 321??

[KennethWilliams]

Thanks for these answers – all correct

Challenge Question 17

There are certain whole numbers which divide into at least one of the elements of any perfect triple.

For example, 2 is such a number as all triples have at least one even element.

[We do not count triples that contain a zero, like 1,0,1]

Can you find them?

[KennethWilliams]

It gets a lot more complicated when you mix bases like this. When I looked into it I decided it was not worth the hassle.

I think there is an article or two on this though in the Online Journal in my website.

But I would suggest you use 8000 as base, not 7000, as then both numbers will be below a base  and your deficiency will be only 132.

[KennethWilliams]

It would be rather absurd to find 7868×9 using a base of 7000 as your deficiencies will both the large numbers, 868 and 6991.

[KennethWilliams]

If you mean the number of pages, it should be about 4 Savita.

[KennethWilliams]

Hi Preeti,

You are mixing your bases: 7000 and 10.
You would need to multiply the LHS by 7 only if you use 7000 as base for both numbers.

Good try though.

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