MHB Max $m$ for $(\frac{1}{11^m}\prod_{i=1000}^{2014}i)\in N$

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To determine the maximum value of \( m \) such that \( \frac{1}{11^m} \prod_{i=1000}^{2014} i \) is an integer, the focus is on the prime factorization of the product \( \prod_{i=1000}^{2014} i \). The product contains all integers from 1000 to 2014, and the number of factors of 11 in this range must be calculated. The maximum \( m \) is found by counting the multiples of 11, 121, and higher powers of 11 within the range. The final result reveals the largest \( m \) that satisfies the condition. The solution confirms that \( m \) is determined by the total count of 11s in the prime factorization of the product.
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$(\frac{1}{11^m}\prod_{i=1000}^{2014}i)\in N$
please find max($m$)
 
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Albert said:
$(\frac{1}{11^m}\prod_{i=1000}^{2014}i)\in N$
please find max($m$)

we need to find how many numbers between 1000 and 2014 are divisible by $11 ,11^2, 11^3$ so on

$\lfloor\dfrac{999}{11}\rfloor = 90$
$\lfloor\dfrac{2014}{11}\rfloor\ = 183$

$\lfloor\dfrac{999}{11^2}\lfloor\ = 8$
$\lfloor\dfrac{2014}{11^2}\rfloor\ = 16$

$\lfloor\dfrac{999}{11^3}\lfloor\ = 0$
$\lfloor\dfrac{2014}{11^3}\rfloor = 1$

so number of numbers divisible by 11 is 93 by $11^2$ is 8 and $11^3$ is 1

so highest poser is 93 + 8 +1 = 102
or m = 102
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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