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# Problems of the Week

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If $$N$$ is divisible by $$1,2,3,\ldots, 13$$, then $$N$$ must also be divisible by 14 and 15.

Using this same idea, what is the smallest integer $$M$$ such that the following statement is true?

If $$N$$ is divisible by $$1,2,3,\ldots,M$$, then $$N$$ must also be divisible by $$M+1,M+2,M+3,$$ and $$M+4$$.

Suppose there exists a triangle with integer side lengths such that its perimeter (in units) and area (in square units) have the same numerical value, $$p$$.

What is the sum of all possible values of $$p?$$

As shown in the diagram below, three overlapping circles with distinct radii define six points. The lengths of 5 line segments (in blue) connecting the points are known to us: 4, 6, 8, 3, 5.

What is the length of the remaining segment (in yellow), to 3 decimal places?

$\LARGE{ \require{enclose} \begin{array}{rll} \phantom{0}\ \mathrm{\large7} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} && \\[-1pt] \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \enclose{longdiv}{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} }\kern-.2ex \\[-1pt] \underline{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \phantom 0 \ \phantom 0 \ \phantom 0 } \\[-1pt] { \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{\large7} \ \phantom0 \ \phantom0 \ }\kern-.2ex \\[-1pt] \underline{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \phantom 0 \ \phantom 0 } && \\[-1pt] { \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} }\kern-.2ex \\[-1pt] \underline{ \mathrm{x} \ \mathrm{x} \ \mathrm{x} \ \mathrm{x} } && \\[-1pt] \mathrm{\large7} \end{array} }$

The above is a long division with most of the digits of any number hidden, except for the three 7's. Given that each of 0, 1, 2, ..., 9 was used at least once for the hidden digits, figure out all of the digits hiding and submit your answer as the value of the dividend (the 6-digit number being divided).


Details and Assumptions:

• Each $$\mathrm X$$ represents a single-digit integer.
• The leading (leftmost) digit of a number cannot be 0.

Find all integer solutions to the equation $$y^2=x^3-p^3$$, where $$p$$ is a prime number, $$y \neq 0$$, $$3 \nmid y$$, and $$p \nmid y$$. Enter your answer as $$\sum(x+p)$$.

Note: Actually, you are going to find integer points on a family of Mordell curves. A Mordell curve is an elliptic curve of the form $$y^2 = x^3 + k$$, where $$k$$ is a fixed non-zero integer.

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