statement.html

problem 61 · 38 lines

<div class="problem_content" role="problem">
<p>Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:</p>
<table><tr><td>Triangle</td>
<td> </td>
<td>$P_{3,n}=n(n+1)/2$</td>
<td> </td>
<td>$1, 3, 6, 10, 15, \dots$</td>
</tr><tr><td>Square</td>
<td> </td>
<td>$P_{4,n}=n^2$</td>
<td> </td>
<td>$1, 4, 9, 16, 25, \dots$</td>
</tr><tr><td>Pentagonal</td>
<td> </td>
<td>$P_{5,n}=n(3n-1)/2$</td>
<td> </td>
<td>$1, 5, 12, 22, 35, \dots$</td>
</tr><tr><td>Hexagonal</td>
<td> </td>
<td>$P_{6,n}=n(2n-1)$</td>
<td> </td>
<td>$1, 6, 15, 28, 45, \dots$</td>
</tr><tr><td>Heptagonal</td>
<td> </td>
<td>$P_{7,n}=n(5n-3)/2$</td>
<td> </td>
<td>$1, 7, 18, 34, 55, \dots$</td>
</tr><tr><td>Octagonal</td>
<td> </td>
<td>$P_{8,n}=n(3n-2)$</td>
<td> </td>
<td>$1, 8, 21, 40, 65, \dots$</td>
</tr></table><p>The ordered set of three $4$-digit numbers: $8128$, $2882$, $8281$, has three interesting properties.</p>
<ol><li>The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).</li>
<li>Each polygonal type: triangle ($P_{3,127}=8128$), square ($P_{4,91}=8281$), and pentagonal ($P_{5,44}=2882$), is represented by a different number in the set.</li>
<li>This is the only set of $4$-digit numbers with this property.</li>
</ol><p>Find the sum of the only ordered set of six cyclic $4$-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.</p>
</div>