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Difference between revisions of "1984 AHSME Problems/Problem 9"

(Created page with "==Problem== The number of digits in <math> 4^{16}5^{25} </math> (when written in the usual base <math> 10 </math> form) is <math> \mathrm{(A) \ }31 \qq...")
 
 
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==Solution==
 
==Solution==
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We can rewrite this as <math> 2^{32}5^{25} </math>. We can also combine some of the factors to introduce factors of <math> 10 </math>, whose digit count is simple to evaluate because it simply adds <math> 0 </math>s. Thus, we have <math> 2^{32}5^{25}=2^72^{25}5^{25}=2^710^25 </math>. We can see that this final number is <math> 2^7 </math> with <math> 25 </math> <math> 0 </math>s annexed onto it. <math> 2^7=128 </math>, which has <math> 3 </math> digits, so the entire number has <math> 25+3=28 </math> digits, <math> \boxed{\text{D}} </math>.
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We can rewrite this as <math> 2^{32}5^{25} </math>. We can also combine some of the factors to introduce factors of <math> 10 </math>, whose digit count is simple to evaluate because it simply adds <math> 0 </math>s. Thus, we have <math> 2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25} </math>. We can see that this final number is <math> 2^7 </math> with <math> 25 </math> <math> 0 </math>s annexed onto it. <math> 2^7=128 </math>, which has <math> 3 </math> digits, so the entire number has <math> 25+3=28 </math> digits, <math> \boxed{\text{D}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=8|num-a=10}}
 
{{AHSME box|year=1984|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 11:49, 5 July 2013

Problem

The number of digits in $4^{16}5^{25}$ (when written in the usual base $10$ form) is

$\mathrm{(A) \ }31 \qquad \mathrm{(B) \ }30 \qquad \mathrm{(C) \ } 29 \qquad \mathrm{(D) \ }28 \qquad \mathrm{(E) \ } 27$

Solution

We can rewrite this as $2^{32}5^{25}$. We can also combine some of the factors to introduce factors of $10$, whose digit count is simple to evaluate because it simply adds $0$s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$. We can see that this final number is $2^7$ with $25$ $0$s annexed onto it. $2^7=128$, which has $3$ digits, so the entire number has $25+3=28$ digits, $\boxed{\text{D}}$.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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