Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 12"
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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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− | *[[Mock AIME 1 2006-2007/Problem 11 | Previous Problem]] | + | ==See Also== |
+ | *[[Mock AIME 1 2006-2007 Problems/Problem 11 | Previous Problem]] | ||
− | *[[Mock AIME 1 2006-2007/Problem 13 | Next Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 13 | Next Problem]] |
*[[Mock AIME 1 2006-2007]] | *[[Mock AIME 1 2006-2007]] |
Latest revision as of 18:14, 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.