What is the least value of n such that the value of (8^n)(25^24) has

Question

What is the least value of n such that the value of (8^n)(25^24) has 49 or more digits?

A. 7
B. 14
C. 16
D. 32 
E. 48

gmatophobia · Accepted Answer

(8^n)*(25^{24})

(2^{3n})*(5^{48})

10^1 = 10 = 2  digits 
 10^2 = 100 = 3  digits 
 10^3 = 1000 = 4  digits

Therefore for 49 digits, we need  10^{48}

(2^{3n})*(5^{48}) = 10^{48}

Hence,  3n = 48

n = 16

Minimum value of n ⇒  n = 16

Option C

HarshaBujji · Answer

To make it simple & understand better try to choose lower n.o.

Here it is  2^3n * 5^48

2^3 * 5^3  will give 4 digit number.

In a similar way, 2^48 * 5^48 will give 49 digit number.

Hence 3n =48 => n=16

Kinshook · Answer

Asked: What is the least value of n such that the value of (8^n)(25^24) has 49 or more digits?

(4ˆ24 = 2ˆ48 = 8ˆ16) multiplied by (25ˆ24 = 100ˆ24) will provide (24*2 = 48) zeros at the end and overall 49 digits including 1 non-zero digit.

IMO C

KarishmaB · Answer

8^n * 25^{24} = (2^3)^n * (5^2)^{24} = 2^{3n} * 5^{48}

We need 49 or more digits so we should be able to make 48 zeroes. That will give us a number with 49 digits. 
We have 48 5s so we need 48 2s too i.e. 3n should be 48. Hence if n = 16, we will get 48 2s.

Answer (C)

GMAT695 · Answer

Please change the source from E-gmat to GMAT PREP ( FOCUS).  Bunuel

Bunuel · Answer

Done. Thank you very much!

hemanthPA · Answer

(8^n)(25^24) = (2^3*n) x (5^2*24) = (2^3n) (5^48)

2^1 x 5^1 = 10 – 2 digits
2^2 * 5^2 = 100 – 3 digits
2^3 * 5^3 = 1000 – 4 digits

We can say 2^n * 5^n = n+1 digits
So, for 49 digits, it is (2^48)(5^48) = 49 digits

3n = 48
n = 48/3 = 16

bumpbot · Answer

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What is the least value of n such that the value of (8^n)(25^24) has

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