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Difference between revisions of "2006 AIME I Problems/Problem 3"
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== Solution ==
 
== Solution ==
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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>\boxed{725}</math>.
 
  
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*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.
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Suppose the original number is <math>N = \overline{a_na_{n-1}\ldots a_1a_0},</math> where the <math>a_i</math> are digits and the first digit, <math>a_n,</math> is nonzero. Then the number we create is <math>N_0 = \overline{a_{n-1}\ldots a_1a_0},</math> so we get <cmath>N = 29N_0.</cmath> But <math>N</math> is <math>N_0</math> with the digit <math>a_n</math> added to the left, so <math>N = N_0 + a_n \cdot 10^n.</math> Thus, <cmath>N_0 = a_n\cdot 10^n = 29N_0</cmath> so <cmath>a_n \cdot 10^n = 28N_0.</cmath> The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number <math>10^n</math> is never divisible by <math>7,</math> so <math>a_n</math> must be divisible by <math>7.</math> But <math>a_n</math> is a nonzero digit, so the only possibility is <math>a_n = 7.</math> This gives <cmath>7 \cdot 10^n = 28N_0</cmath> or <cmath>10^n = 4N_0.</cmath> Now, we want to minimize ''both'' <math>n</math> and <math>N_0,</math> so we take <math>N_0 = 25</math> and <math>n = 2.</math> Then <cmath>N = 7 \cdot 10^2 + 25 = \boxed{725},</cmath> and indeed, <math>725 = 29 \cdot 25.</math> <math>\square</math>
  
 
== See also ==
 
== See also ==

Revision as of 15:23, 7 March 2014

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

Solution

Suppose the original number is $N = \overline{a_na_{n-1}\ldots a_1a_0},$ where the $a_i$ are digits and the first digit, $a_n,$ is nonzero. Then the number we create is $N_0 = \overline{a_{n-1}\ldots a_1a_0},$ so we get \[N = 29N_0.\] But $N$ is $N_0$ with the digit $a_n$ added to the left, so $N = N_0 + a_n \cdot 10^n.$ Thus, \[N_0 = a_n\cdot 10^n = 29N_0\] so \[a_n \cdot 10^n = 28N_0.\] The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number $10^n$ is never divisible by $7,$ so $a_n$ must be divisible by $7.$ But $a_n$ is a nonzero digit, so the only possibility is $a_n = 7.$ This gives \[7 \cdot 10^n = 28N_0\] or \[10^n = 4N_0.\] Now, we want to minimize both $n$ and $N_0,$ so we take $N_0 = 25$ and $n = 2.$ Then \[N = 7 \cdot 10^2 + 25 = \boxed{725},\] and indeed, $725 = 29 \cdot 25.$ $\square$

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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