\((10^4 * 3.456789)^8=34567.89^8=(3456789*10^{-2})^8=3456789^8*10^{-16}\).

Answer: C.

Hope it's clear.
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\((10^4 X 3.456789)^8 = (10^4 X 10^{-6} X 3456789)^8 = (10^4 X 10^{-6} X 3456789)^8 = (10^{-2} X 3456789)^8\)

= \(3456789^8 X 10^{-16}\)

Answer is C

First of all split the terms (10^4)^8 * (3.456789)^8 --------> (10)^32 * (3.456789)^8

10^32 will have 32 zeros after 1.

(3.456789)^8 ----> Here we should understand 2 things 1) Any decimal when squared its decimal places doubles e.g. 1.2^2 = 1.44 2) x^2 means (x^2)^2)^2

So here when 3.456789 squared its decimal places would become 12. Squaring the resultant figure again would give 24 decimal places AND squaring the resultant figure for the final time would give 48 decimal places.

So we will have the figure as (x.abcd....48 times)*(1000000.....32 times). The 32 zeros would shift the decimal sign of x.abcd... to the right side by 32 places. So beyond decimal point we will have 48 - 32 = 16 places. Hence C
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In general, a fraction is something upon 10.
ex : 2.34 = 234/100
3.4 = 34/10
3.6789= 36789/10000

So when you square such numbers, you basiaclly square numerator and denominators both.
Say (2.34)^2 = (234/100)^2 = 234^2 / 100 ^2

so you end up having some number upon 10000, in this case.
Generally speaking you increase number of decimal digits by a multiplication factor of 2. If its cubing then multiplication factor is 3, and decimal will have 6 digits

So if you had one place of decimal, say 1.1 after squaring you shall have somenumber.2 places of decimal.

To our question
10^4 * 3.456789 becomes (34567.89)^8.

Sow we have (3456789/100)^8

or (3456789)^8 / 100^8

= some number / 10^16
or 16 places of decimal.

(34567.89)^8
Whenever you multiply a decimal by itself, you get the number of decimals times the power for the numbers on the right
0.2^2 = 0.04, 0.2^3=0.008
So, 2 numbers to the right of the decimal times power 8.
This is how I got the answer but if there's any flaw in the logic please let me know.

(10^4 * 3.456789)^8

Just resolve the power to move decimal point 4 digit ahead

= (34567.89)^8

There are 2 digits after decimal in the above term & the power is 8, so digits after decimal would be 8 x 2 =16 = Answer = C

Well, what I did was to do the multiplication in the brackets and end up with this: 3,4567.89

Now, we want to raise 3,4567.89 to the eighth power. Instead of doing this, I tried with 0.01. I noticed that 0.01^2, is adding two more digits: 0.0001.

So, ading 2 more digits until we reach the eighth power will lead to 14 digits. Plus the 2 we already have (.89) sums up to 16 digits. ANS C

i got to (34567.89)^8

now, we have 2 digits after the decimal point.
if we multiply two decimals, the number of digits after the decimal point are the sum of the digits after the decimal point of the decimals.
thus, we can easily eliminate A and B. C looks good, rest are way too much.

Since 10^4 x 3.456789 = 34567.89, (34567.89)^8 will have a total of 2 x 8 = 16 digits to the right of the decimal place.

Answer: C
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(10^4 * 3.456789)^8

Multiplying the inner part shifts the decimal point of 3.456789 to the right by 4 places

= (34567.89)^8

When we multiply out decimal numbers, the decimal places in the final result will be the total number of decimal places in each number being multiplied.

.89 equals 2 decimal places. So,2 decimal + 2 decimal +......(up to 8 times) or 2*8 = 16 decimal places .

There would be 16 decimal places or 16 digits to the right of decimal in final result.

Step 1: Power of 10 with a (+)Positive Integer Exponent

when Multiplying a Number by a Power of 10 with a (+)Pos. Integer Exponent, move the Decimal Point to the RIGHT as many Places as the (+)Pos. Integer Exponent


10^4 * 3.456789 = 34, 567.89



2nd) Rule: When you are Raising a Base that is a Decimal to a (+)Pos. Integer Exponent, the Number of Decimal Places will be =

(No. of Decimal Places in the Base) * ((+)Pos. Integer Exponent)



thus:

(34, 567.89)^8

(2 Decimal Places in the Base) * (8 Exponent) = Result will have 16 Decimal Places


Lastly, we know that none of the Digits at the END of the Decimal will be Trailing 0s because the Units Digit of (9)^x Power follows the Pattern:

[9 , 1 , 9 , 1]

the Last Digit in the Decimal will be 1, so all 16 Digits after the Decimal Point will be counted


Answer C

Thank you Narenn for the great explanation
This was exactly what I was looking for.