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How many digits does the product contain?

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by Vemuri » Wed Apr 29, 2009 9:11 am
How many digits does the product of 4^12 and 5^23 contain?
A. 12
B. 13
C. 23
D. 24
E. 35

I do not have the OA. I hope there is a easy way to solve this problem.
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by scoobydooby » Wed Apr 29, 2009 9:34 am
4^12 * 5^23
=(2^2)^2 * 5^23
=2^24 * 5^23
=(2*5)^23 *2
=10^23 *2

10^23 gives 23 zeroes
10^23 *2 therefore has 2+23 zeroes
=24 digits

hence, D
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by Vemuri » Wed Apr 29, 2009 9:56 am
scoobydooby wrote:4^12 * 5^23
=(2^2)^2 * 5^23
=2^24 * 5^23
=(2*5)^23 *2
=10^23 *2

10^23 gives 23 zeroes
10^23 *2 therefore has 2+23 zeroes
=24 digits

hence, D
Hey soobydooby,

That's a very good solution.
Last edited by Vemuri on Wed Apr 29, 2009 10:05 pm, edited 1 time in total.
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by dumb.doofus » Wed Apr 29, 2009 12:37 pm
Vemuri wrote:
scoobydooby wrote:4^12 * 5^23
=(2^2)^2 * 5^23
=2^24 * 5^23
=(2*5)^23 *2
=10^23 *2

10^23 gives 23 zeroes
10^23 *2 therefore has 2+23 zeroes
=24 digits

hence, D
Hey soobydooby,

That's a very good solution. If 10^23 has 23 zeros, multiplying this number with 2 should not change the number of zeros right? You seem to have hurried up in the last part. I think the answer should be 23 digits, i.e. C.
I think it is correct.. it should be 24.
10^5 = 100000 There are 5 zeros..

so 10^23 will have 23 zeros.. also we have 1.. so altogether 24 digits..
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Isn't that C?

by g2000 » Wed Apr 29, 2009 4:58 pm
10^23 *2

I'm just curious the number of zeroes. It seems it's 23 zeroes, doesn'it?

If multiply by 2, that only changes the coefficient of 10^23.

For instance,
10^2 * 5
= 100 * 5
= 500
That's not 5000!?! (the extra mysterious one!?!)
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by DeepakR » Wed Apr 29, 2009 5:45 pm
Thanks Scoobydooby for the clear solution:

I think the answer should be 24. For eg.) 10^1 * 2= 20 2 digits and now 10^2 * 2= 200 we get 3 digits i.e the power of 10 + 1. So in case of 10^3 * 2 = 2000 we have power of 10 =3 and hence 3 + 1 = 4 digits.

Similarly for 10^23 * 2 we'll have 24 digits.

-Deepak
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