Am I thinking correctly ?

What is the sum of digits of (9999999999)^3 ?

Is the answer 180 ?

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

how did u find that out ??

Sivaramakrishnan Sivakumar - 6 years, 10 months ago

For example take (99)^3. In this sum of the number (99) is 9+9=18. Now take the cube ,which is 970299. In this the sum (9+7+0+2+9+9) is 36. I used this in the question. (9999999999)^3 has 10 nines. Therefore the sum is 90. Hence the sum of the digits of (9999999999)^3 is 180.

Sharath Chandar - 6 years, 10 months ago

take 9 and 9^3 , Your logic doesn't work !

Sivaramakrishnan Sivakumar - 6 years, 10 months ago

@Sivaramakrishnan Sivakumar – It works.. 9^3= 729. 7+2+9= 18

Sharath Chandar - 6 years, 10 months ago

@Sharath Chandar – oh..ya... then i will use a computer to check !

Sivaramakrishnan Sivakumar - 6 years, 10 months ago

@Sivaramakrishnan Sivakumar – You may not be able to!!

Sharath Chandar - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$