 # Trachtenberg and Arthur Benjamin Method

Great, kinma Here are some questions i need to find the best way to do in my mind im confused in trachtenberg and arthur benjamin method:

26757*991=?
how do you get result when you see √15129 ?

For the second one -
√15129

First step - I divided this in two parts first part and second part and the second part should have 2 numbers.
151 | 29

Second step -In first part , I find a nearest square for number 151 .
Here - (12^2 = 144 )is the nearest square.

Third step - In second part , I see the last digit and find which number square has the last digit 9 (In 1 to 9 number square.)
Here 9 is the last digit in 29.

And there are two numbers 3 & 7 that squares digit endes with 9.

Final step - So we have two answers.

123 & 127

And we have to find which one is the correct answer.
So we apply 5 rule square method and find the square of middle no.
Here - 125 is the middle no.

So - 125^2

12×13 | 25
15625 and it is greater than 15129

So the correct answer is 123.

I only described this in detail otherwise it is not a lengthy process , very easy to do.
You have to practise more , if you interested in speed calculation.

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26757 × 991 = ?

I don’t know the fastest way to solve the first question.
If I have to solve this , I do this like

26757 × 1000 = 26757000 (because 991 is 1000 - 9)
26757 × 9 = 240813

26757000 - 240813

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@Rajadodve786 Thank you dear for taking part here but lte me complete my question package:
26757 / 991: what do you do when you see it
and about the last one this:√333333333 which method do you use to solve?
and best regards

If you want to make it even easier on yourself and basically just shift around the decimal point, 991 is also…

1,000 - 10 + 1
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you made it too hard man,
it was main question in ucimac courses

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As is already answered, I would also use either 1000-9 or 1000 - 10 + 1

Raja’s method is solid if you know the answer is an integer.

My method is similar at first, but this also works for non integer answers:

Like Raja I would start wth (12 * 10)^2 = 14400 as a ball park figure.
The difference of 14400 - 15129 is about 5%.
5% / 2 = 2.5%

12 plus 2.5% is 12.3.

So I revise (12 * 10)^2 = 14400 to 123^2 = 15129.
123^2, I calculate using cross multiplication with 12 and 3.
The 12 is actually 120, but I position the numbers in my head automatically: 12^2 | 2 * 12 * 3 | 3^2

I round 991 up to 1000 and remember that for every 1000 I subtract, I have to add 9.
Start by subtracting 26*1000. Remainder is 757. Add 26*9 = 234.
757+234 = 991. The remainder is exactly one more time 991. So correct the 26 to 27.

I count the number of 3s. 9 in total. I first round it to 333 (million).
18*18 = 324, which is the closest integer that squared is lower than 333.
The difference between 324 and 333 is 9.
9/2 = 4.5.

I add 1.4% to 18,000 to get 18,180 + 72 = 18,252.

Then I realise that my answer can be improved upon.
9 is actually 9.333333, so we need a little more than 4.5: 4.6666665
And my 333 is actually 333.333333.
So we get: 4.6666665 / 333.333333.
In my mind 4.6666665 is 4 2/3 = 14/3 and 333.333333 = 1/3 (disregarding the decimal point).
Dividing by 1/3 = multiplying by 3.
14/3, multiplied by 3 = 14.
Putting the decimal point back and 14 becomes 1.4%.

My original answer of 18,000 plus 1.4% stays.
A calculator gives 18,257.412…
We are 5 off, which is close enough for me.

If you want to get an exact answer with all the decimals, check out my post about the duplex method:

Tbh; too much work for me.
If my answer is exact in the first 4 decimals, this is fine for me and my purposes.

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This is an excellent app for practicing maths tables.
And some mathematical calculation problems.

I personally used this a lot in childhood.

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