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Noppi, 2024/01/26 06:50
Problem 61¶
Cyclical Figurate Numbers¶
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:
| Triangle | $P_{3,n}=n(n+1)/2$ | $1, 3, 6, 10, 15, \dots$ |
| Square | $P_{4,n}=n^2$ | $1, 4, 9, 16, 25, \dots$ |
| Pentagonal | $P_{5,n}=n(3n-1)/2$ | $1, 5, 12, 22, 35, \dots$ |
| Hexagonal | $P_{6,n}=n(2n-1)$ | $1, 6, 15, 28, 45, \dots$ |
| Heptagonal | $P_{7,n}=n(5n-3)/2$ | $1, 7, 18, 34, 55, \dots$ |
| Octagonal | $P_{8,n}=n(3n-2)$ | $1, 8, 21, 40, 65, \dots$ |
The ordered set of three $4$-digit numbers: $8128$, $2882$, $8281$, has three interesting properties.
- 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).
- 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.
- This is the only set of $4$-digit numbers with this property.
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.
巡回図形数¶
三角数, 四角数, 五角数, 六角数, 七角数, 八角数は多角数であり, それぞれ以下の式で生成される.
| 三角数 | $P_{3,n}=n(n+1)/2$ | $1, 3, 6, 10, 15, \dots$ |
| 四角数 | $P_{4,n}=n^2$ | $1, 4, 9, 16, 25, \dots$ |
| 五角数 | $P_{5,n}=n(3n-1)/2$ | $1, 5, 12, 22, 35, \dots$ |
| 六角数 | $P_{6,n}=n(2n-1)$ | $1, 6, 15, 28, 45, \dots$ |
| 七角数 | $P_{7,n}=n(5n-3)/2$ | $1, 7, 18, 34, 55, \dots$ |
| 八角数 | $P_{8,n}=n(3n-2)$ | $1, 8, 21, 40, 65, \dots$ |
3つの4桁の数の順番付きの集合 (8128, 2882, 8281) は以下の面白い性質を持つ.
- この集合は巡回的である. 最後の数も含めて, 各数の後半2桁は次の数の前半2桁と一致する
- それぞれ多角数である: 三角数 ($P_{3,127}=8128$), 四角数 ($P_{4,91}=8281$), 五角数 ($P_{5,44}=2882$) がそれぞれ別の数字で集合に含まれている
- $4$桁の数の組で上の2つの性質を持つのはこの組だけである.
三角数, 四角数, 五角数, 六角数, 七角数, 八角数が全て表れる6つの巡回する4桁の数からなる唯一の順序集合の和を求めよ.
Noppi が2024/01/26に更新 · 1件の履歴