Difference between revisions of "2006 Alabama ARML TST Problems/Problem 11"
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==Solution== | ==Solution== | ||
| − | Now either <math>5^k</math> starts with 1, or <math>5^{k+1}</math> has one more digit than <math>5^k</math>. From <math>5^0</math> to <math>5^{2005}</math>, we have | + | Now either <math>5^k</math> starts with 1, or <math>5^{k+1}</math> has one more digit than <math>5^k</math>. From <math>5^0</math> to <math>5^{2005}</math>, we have 1402 changes (since 1 increases 1402 times to become 1403 after <math>5^{2005}</math>), so those must not begin with the digit 1. <math>2006-1402=\boxed{604}</math> |
==See also== | ==See also== | ||
| + | {{ARML box|year=2006|state=Alabama|num-b=10|num-a=12}} | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Latest revision as of 22:11, 12 April 2012
Problem
The integer 
has 1403 digits, and 1 is its first digit (farthest to the left). For how many integers 
does 
begin with the digit 1?
Solution
Now either starts with 1, or 
has one more digit than . From 
to 
, we have 1402 changes (since 1 increases 1402 times to become 1403 after ), so those must not begin with the digit 1. 
See also
| 2006 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 10 |
Followed by: Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
