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I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.

  1. What is the largest integer $k$ such that the following equation is an integer?

$$\frac {1001\times 1002 \times ... \times 2008} {11^k}$$

(A) $100$ (B) $101$ (C) $102$ (D) $103$ (E) $105$

The first way I did it is calculate the multiple of $11$ in $1001$ to $2008$

$\lfloor (2008-1001) \div 11 \rfloor$ ($91$ multiples of $11$) + $\lfloor (2008-1001) \div 121 \rfloor$ ($8$ multiples of $121) + 1$ ($1331$ is $11^3)= 100$

Are there any mistakes or miscalculations? Are there any more effective ways to find the value of $k$?

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  • $\begingroup$ Hint: product in denominator is $2008!/1000!$, and use this result $\endgroup$ – Michael Galuza Jul 25 '15 at 8:56
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It is almost correct. The number of multiples of 121 and 1331 are correct. However there are 92 multiples of 11 between 1001 and 2008. To see this, note that 1001 is a multiple of 11 since the alternating sum of the digits is 0.

Now calculate

$$\lfloor \frac{2008-1002}{11} \rfloor = 91$$

This means that there are 91 multiples of 11 between 1002 and 2008. With the number of multiples of 121 and 1331 you were lucky you were correct. For example, for 121 you should have searched for the smallest multiple of 121 bigger than 1001 which is 1089. Now have $$\lfloor \frac{2008-1098}{121} \rfloor = 7$$

So the answer is 92+8+1=101.

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We have to use the Polignac Legendre formula twice.

The answer is: $\lfloor\frac{2008}{11}\rfloor+\lfloor\frac{2008}{121}\rfloor+\lfloor\frac{2008}{1331}\rfloor-(\lfloor\frac{1000}{11}\rfloor+\lfloor\frac{1000}{121}\rfloor+\lfloor\frac{1000}{1331}\rfloor)$

So $182+16+1-(90+8+0)=101$

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