Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 12"
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==Solution== | ==Solution== | ||
The digit-removal condition is equivalent to the statement <math>k = 4\cdot10^n + m</math> where <math>10^n > m</math> and <math>n \geq 1</math>. Thus <math>14m + 1 = 4\cdot 10^n + m</math> so <math>13m = 4\cdot 10^n - 1</math> and <math>m = \frac{4 \cdot 10^n - 1}{13}</math>. It's easy to see that this value of <math>m</math> is small enough, so all we need to check is that it is an [[integer]]. That happens if and only if 13 is a [[divisor]] of <math>4\cdot 10^n - 1</math>, so <math>4\cdot 10^n \equiv 1 \pmod{13}</math> and multiplying by <math>4^{-1} \equiv 10 \pmod{13}</math> we have that <math>10^n \equiv 10 \pmod {13}</math> Certainly <math>n = 1</math> is a solution. All we need is the order of 10 <math>\pmod {13}</math>. Now <math>10^2 = 100 \equiv 9 \pmod{13}</math> so <math>10^3 \equiv 90 \equiv -1 \pmod{13}</math>, <math>10^6 \equiv 1 \pmod{13}</math> and the order of 10 mod 13 is 6. Thus, we get one value of <math>m</math> each time <math>n = 6j + 1</math>. There are <math>335</math> such values of <math>n</math> which fall in the required range. | The digit-removal condition is equivalent to the statement <math>k = 4\cdot10^n + m</math> where <math>10^n > m</math> and <math>n \geq 1</math>. Thus <math>14m + 1 = 4\cdot 10^n + m</math> so <math>13m = 4\cdot 10^n - 1</math> and <math>m = \frac{4 \cdot 10^n - 1}{13}</math>. It's easy to see that this value of <math>m</math> is small enough, so all we need to check is that it is an [[integer]]. That happens if and only if 13 is a [[divisor]] of <math>4\cdot 10^n - 1</math>, so <math>4\cdot 10^n \equiv 1 \pmod{13}</math> and multiplying by <math>4^{-1} \equiv 10 \pmod{13}</math> we have that <math>10^n \equiv 10 \pmod {13}</math> Certainly <math>n = 1</math> is a solution. All we need is the order of 10 <math>\pmod {13}</math>. Now <math>10^2 = 100 \equiv 9 \pmod{13}</math> so <math>10^3 \equiv 90 \equiv -1 \pmod{13}</math>, <math>10^6 \equiv 1 \pmod{13}</math> and the order of 10 mod 13 is 6. Thus, we get one value of <math>m</math> each time <math>n = 6j + 1</math>. There are <math>335</math> such values of <math>n</math> which fall in the required range. | ||
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| + | ==See Also== | ||
*[[Mock AIME 1 2006-2007 Problems/Problem 11 | Previous Problem]] | *[[Mock AIME 1 2006-2007 Problems/Problem 11 | Previous Problem]] | ||
Latest revision as of 17:14, 3 April 2012
Problem
Let 
be a positive integer with first digit 4 such that after removing the first digit, you get another positive integer, 
, that satisfies 
. Find the number of possible values of between 
and 
.
Solution
The digit-removal condition is equivalent to the statement 
where 
and 
. Thus 
so 
and 
. It's easy to see that this value of is small enough, so all we need to check is that it is an integer. That happens if and only if 13 is a divisor of 
, so 
and multiplying by 
we have that 
Certainly 
is a solution. All we need is the order of 10 
. Now 
so 
, 
and the order of 10 mod 13 is 6. Thus, we get one value of each time 
. There are 
such values of 
which fall in the required range.
