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Bunuel
How many digits are in (8×10^14)(10×10^10)?

A. 24
B. 25
C. 26
D. 27
E. 28

Bunuel can you help me in this. Though I know the concept, still I am counting it as 27.

(8×10^14) ; 10^14 means fourteen "0s" and one "1" So total 15 digits ; 8 does not change the number of digits

(10×10^10) ; 10^11 means 12 digits

So 15+12 = 27

You cannot add like that. For example, 10 has 2 digits but 10*10 does not have 2 + 2 = 4 digits it has 3 digits.

You should do the following way: (8×10^14)(10×10^10) = 8*10^25 --> 10^25 has 26 digits. Multiplying it by 8 does not change the number of digits.

Answer: C.
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Bunuel
How many digits are in (8×10^14)(10×10^10)?

A. 24
B. 25
C. 26
D. 27
E. 28

Bunuel can you help me in this. Though I know the concept, still I am counting it as 27.

(8×10^14) ; 10^14 means fourteen "0s" and one "1" So total 15 digits ; 8 does not change the number of digits

(10×10^10) ; 10^11 means 12 digits

So 15+12 = 27

You cannot add like that. For example, 10 has 2 digits but 10*10 does not have 2 + 2 = 4 digits it has 3 digits.

You should do the following way: (8×10^14)(10×10^10) = 8*10^25 --> 10^25 has 26 digits. Multiplying it by 8 does not change the number of digits.

Answer: C.

Can we deduce that number less than 10 does not change the number of digits ever?
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Bunuel can you help me in this. Though I know the concept, still I am counting it as 27.

(8×10^14) ; 10^14 means fourteen "0s" and one "1" So total 15 digits ; 8 does not change the number of digits

(10×10^10) ; 10^11 means 12 digits

So 15+12 = 27

You cannot add like that. For example, 10 has 2 digits but 10*10 does not have 2 + 2 = 4 digits it has 3 digits.

You should do the following way: (8×10^14)(10×10^10) = 8*10^25 --> 10^25 has 26 digits. Multiplying it by 8 does not change the number of digits.

Answer: C.

Can we deduce that number less than 10 does not change the number of digits ever?

It won't if you have a*10^n, where a is a single digit integer. But for other cases it might not be true. For example, 3*400.
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Bunuel
QZ
How many digits are in (8×10^14)(10×10^10)?

A. 24
B. 25
C. 26
D. 27
E. 28

Bunuel can you help me in this. Though I know the concept, still I am counting it as 27.

(8×10^14) ; 10^14 means fourteen "0s" and one "1" So total 15 digits ; 8 does not change the number of digits

(10×10^10) ; 10^11 means 12 digits

So 15+12 = 27

You cannot add like that. For example, 10 has 2 digits but 10*10 does not have 2 + 2 = 4 digits it has 3 digits.

You should do the following way: (8×10^14)(10×10^10) = 8*10^25 --> 10^25 has 26 digits. Multiplying it by 8 does not change the number of digits.

Answer: C.

Can we deduce that number less than 10 does not change the number of digits ever?


Hi..
You can't take that as a rule..
It depends on other factors too..
8*10^2.... 8 will not effect
8*20^2.... 8 will effect...

Why you have gone wrong is that you have separated the 10s..
(8*10^14)(10*10^11)....
Although you have correctly solved 10*10^11, you have added an extra digit of 1 from 10^14..
10*10^11 could have meant 10 - 2 digits and 10^10- 10 zeroesand 1 digit for '1', so 10+1=11 digits
Total 2+11=13digits.. but is it so ...NO
It is 12 digits because 10*10^10=10^11, meaning 12 digits
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We have \((8*10^{14})(10*10^{10})\)

Since it is all multiplication, we can rearrange the terms as follows

\((8*10^{14}*10^1*10^{10})\)

which gives us

\(8*10^{25}\)

\(10^{25}\) gives us \(1\) followed by \(25\) zeroes or in total \(26\) digits.

Multiplying this by \(8\) does not change the number of digits.

Answer is Choice C
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You can't add like that, but if you wanted to take that approach, you will have:

(12 digit number)*(15 digit number)
= 10^11*10*14
= 10^25.
Including the 1 at the start you will have 26 digits.
AkshdeepS


Bunuel can you help me in this. Though I know the concept, still I am counting it as 27.

(8×10^14) ; 10^14 means fourteen "0s" and one "1" So total 15 digits ; 8 does not change the number of digits

(10×10^10) ; 10^11 means 12 digits

So 15+12 = 27
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