Both DS and DD methods are simple checks to quickly verify your answer without having to do the calculation again.

The only way to be 100% sure that you have the right answer is to do the calculation again.

The only time that the DS method will fail is if the Digit Sum of the incorrect answer matches the Digit Sum of the Right answer.

7 x 6 = 42 (DS is 6).

So 15 x 16 should have DS 6 on both sides.

However if you calculate incorrectly and the incorrect answer also has then digit sum 6 then the DS method will fail.

The probability of that happening is approximately 10% of the time. It is exactly around 11.11% of the time. The remaining 88.88% of the time (approximately 90%) it should correctly identify the mistake.

How do we calculate this probability?

The incorrect answers digit sum can be between 1 to 9. The digit sum will identify the error 8 out of 9 times. But 1 out of 9 times it will match with the correct answer digit sum.

So 8/9 = 88.88% (approximately 90%) the DS method will identify the error. But 1/9 times it will fail.

In the example 15 x 16, the DS method will fail if the incorrect answer has DS 6. But it won’t fail if the answer has DS between 1 to 5 and 7 to 9.

DD method has a similar error rate. However since the methods use different computations usually what fails in one method should be caught in the other method. However there are going to be exceptions where both methods fail.

Both these methods are meant to be simple quick checks and nothing is going to replace doing the calculation again which is the only way I know of to get 100% accuracy.

Hope this clarifies.

What kind of mental math do you do in school?

The reason I am asking is to know whether the book and the course has everything that you do in school.

Will include additional content based on your feedback.

You make the number smaller by dividing by prime numbers first 2, 3, 5, 7 and 11. 

Mostly it will reduce the number unless its a prime number or a multiple of a prime number greater than 11

506 divided by 2 gives 253. 

253 cannot be divided by 3,5 or 7. How do I know its not divisible so quickly? I use the divisibility test.

A number is divisible by 2 if the last number is even. 
A number is divisible by 3 if the sum of the digits is divisible by 3
A number is divisible by 5 if the last digits are 0 or 5. 
For 7 you need to actually divide to check. 
So we try 11. 

253 divided by 11 gives 23
23 is also a prime number so we can’t reduce it further. 

This is the only way to reduce it.
 
You can combine the prime factors together if you want to reduce steps. 
For example, the prime numbers of 506 is 2, 11 and 23. 
We can combine 2 and 11 to get 22.
So we are left with 22 x 23. 

Hope this helps

Hi Toni,

Any goal you hope to achieve by inculcating this new habit? Like a specific grade or something?

Hi Toni

What do you spend time learning? Is it school or university work or some new skill?

Abhishek

The method is the same for negative and positive numbers. There is absolutely no difference.

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Unfortunately we cannot discount the course any further to be fair to people who paid full price.

Hi Jaggari,

The first digit left of the decimal is odd and you alternate between odd and even.

For example, let us take 7324.

So the first digit left of the decimal is 4. So it becomes Odd (O)
O
7 3 2 4

Now you alternate between Odd (O) and Even (E)
E O E O
7 3 2 4

Hope this clarifies.