Difference between revisions of "2006 AIME I Problems/Problem 3"
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| − | The number can be represented as <math>10^na+b</math>, where <math> a </math> is the leftmost digit, and <math> b </math> is the rest of the number.* We know that <math>b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na</math>. Thus <math> a </math> has to be 7 since <math> 10^n </math> can not have 7 as a factor, and the smallest <math> 10^n </math> can be and have a factor of <math> 2^2 </math> is <math> 10^2=100. </math> We find that <math>b=25</math>, so the number is <math>725</math>. | + | The number can be represented as <math>10^na+b</math>, where <math> a </math> is the leftmost digit, and <math> b </math> is the rest of the number.* We know that <math>b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na</math>. Thus <math> a </math> has to be 7 since <math> 10^n </math> can not have 7 as a factor, and the smallest <math> 10^n </math> can be and have a factor of <math> 2^2 </math> is <math> 10^2=100. </math> We find that <math>b=25</math>, so the number is <math>\boxed{725}</math>. |
*It is quite obvious that <math>n=2</math>, since the desired number can't be single or double digit, and cannot exceed <math>999</math>. From <math>100a+b=29b</math>, proceed as above. | *It is quite obvious that <math>n=2</math>, since the desired number can't be single or double digit, and cannot exceed <math>999</math>. From <math>100a+b=29b</math>, proceed as above. | ||
Revision as of 16:52, 8 March 2012
Problem
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 
of the original integer.
Solution
The number can be represented as 
, where 
is the leftmost digit, and 
is the rest of the number.* We know that 
. Thus has to be 7 since 
can not have 7 as a factor, and the smallest can be and have a factor of 
is 
We find that 
, so the number is 
.
- It is quite obvious that
, since the desired number can't be single or double digit, and cannot exceed 
. From 
, proceed as above.
, since the desired number can't be single or double digit, and cannot exceed 
. From 
, proceed as above.See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
