Question

find the sum of the first 64 digits of the decimal expansion of 4/7

Answer 1

Thanks for the question!

It is definitely a very interesting question to solve and do some brainstorming.

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We know, 4/7 = 0.571428 571428 571428 ..........

Also, 4/7 = 0.[571428]

So it is a recurring or a repeating decimal.

5+7+1+4+2+8=27.

We need uptil 64th digit, that means-----
the digits 5,7,1,4,2,8 will be repeated 10 times.

And addition of these digits give 5+7+1+4+2+8=27.

Hence, until 60th digit the sum is 27*10=270

The 61st digit is 5

The 62nd digit is 7

The 63rd digit is 1

The 64th digit is 4

So, we get 270+5+7+1+4 = 287

287 is the sum of the first 64 digits of the decimal expansion of 4/7

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Hope it helps and solves your query!!

Answer 2

The answer is given below :

We know that,

4/7 = 0.571428571... and (571428) will be repeating terms in the expansion.

Now, the number of digits in (571428) is 6.

Now, 64 = (6×10) + 4

That means to find the sum of the digits of the decimal expansion of 4/7 upto 64 terms, we need to add the sum of the first 4 digits (5, 7, 1, 4) with the 10 times the sum of the digits (5, 7, 1, 4, 2, 8).

Sum of the digits 5, 7, 1, 4, 2, 8
= 5 + 7 + 1 + 4 + 2 + 8
= 27

So, 10 times the sum
= 27 × 10
= 270

Now, the sum of the first four digits 5, 7, 1, 4 is
= 5 + 7 + 1 + 4
= 17

Therefore, the required sum of the first 64 digits in the decimal expansion of 4/7 is
= 270 + 17
= 287

Thank you for your question.