2026 AIME II 详解
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所有题目均经美国数学协会（MAA）官方合法授权使用。
1.
求所有整数等差数列的第 项之和，这些数列的首项都等于 ，并且都包含 和 作为其中的项。
Find the sum of the th terms of all arithmetic sequences of integers that have first term equal to and include both and as terms.
小提示：
若公差为 ，则 和 都是项会迫使 同时整除 和 。
If the common difference is then and both being terms forces to divide both and
大提示：
公差必须是 的正因数；对所有这样的 ，把第 项 加起来。
The common difference must be a positive divisor of add up the th terms over all such
解答：
设公差为 。因为首项是 ，且 和 都出现，所以 整除 和 ，因此 整除 。为了从 到达 和 ，公差必须为正，所以 ；这些值也都可行，因为 和 都能被 整除，这使两个目标都落在数列中。
第 项为 ，所以所求和为
Let the common difference be Since the first term is and both and appear, divides and so divides The difference must be positive to reach and from so (and each of these works, since and are both divisible by which puts both targets in the sequence).
The th term is so the requested sum is
2.
下图是一排 个正方形组成的网格。每个正方形都有一条从左下顶点连到右上顶点的对角线。一只虫子沿线段从顶点走到顶点，永不经过同一条线段两次，并且在水平线段或对角线段上绝不从右向左走。设 为虫子从左下角 走到右上角 的路径数。图中粗线段展示了一条从 到 的路径。求 。
The figure below shows a grid of squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex, never traversing the same segment twice and never moving from right to left along a horizontal or diagonal segment. Let be the number of paths the bug can take from the lower left corner to the upper right corner One such path from to is shown by the thick line segments in the figure. Find
小提示：
因为虫子绝不向左走，它在 与 之间的每个竖条中恰好穿过一次，走底边、顶边或对角线。
Since the bug never moves left, it crosses the strip between and exactly once, along the bottom edge, the top edge, or the diagonal.
大提示：
竖直线段完全由这十次穿越的选择决定，所以每个正方形独立贡献 种选择。
The vertical segments are completely determined by the ten crossing choices, so each square independently contributes options.
解答：
令 ，。每次水平或对角移动都向右，所以虫子的 坐标从不减小；它会恰好穿过这 个竖条各一次，并在相应正方形的三条向右线段中选一条：底边、顶边或对角线。
这十个选择决定整条路径。每次穿越会以确定的高度到达（底边到达低处，顶边或对角线到达高处），也会从确定的高度离开（底边或对角线从低处离开，顶边从高处离开）。因此在每条竖直线上，虫子恰好在到达高度与离开高度不同时走那条竖直线段；每条竖直线段至多需要一次，所以不会重复线段。两端同理：虫子从 的低处开始，到 的高处结束，必要时使用端点处的竖直线段。反过来，每串选择都给出一条合法路径。
因此 ，且 。
Put and Every horizontal and diagonal move goes rightward, so the bug’s -coordinate never decreases, and it crosses each of the vertical strips exactly once, using exactly one of that square’s three rightward segments: the bottom edge, the top edge, or the diagonal.
These ten choices determine the whole path. Each crossing arrives at a definite height (bottom edge: low; top edge or diagonal: high) and departs at a definite height (bottom edge or diagonal: low; top edge: high), so at each vertical line the bug traverses the vertical segment exactly when the arrival and departure heights differ — and each vertical segment is needed at most once, so no segment repeats. The same applies at the ends: the bug starts low at and finishes high at using the end verticals if necessary. Conversely, every sequence of choices yields a valid path.
Therefore and
3.
设 是一个非凸五边形，内角满足 且 。已知 、、，并且点 、、 都在直线 的同一侧。还已知 是整数，，且五边形 的面积是 的整数倍。求 可能取值的个数。
Let be a nonconvex pentagon with internal angles and Suppose that and points and lie on the same side of line Suppose further that is an integer with and the area of pentagon is an integer multiple of Find the number of possible values of
小提示：
把 放在原点，并令 ； 和 处的直角使 与 竖直。
Put at the origin and the right angles at and make and vertical.
大提示：
和 处的 角迫使 且 ；鞋带公式把面积条件化为关于 的同余式。
The angles at and force and the shoelace formula turns the area condition into a congruence for
解答：
令 、，使五边形在直线 上方，并设 。 和 处的直角使 与 竖直：，，其中 。在 处，边 与向下的射线 成 角并进入五边形，所以 。类似地，在 处，边 与向下的射线 成 角，所以 ，其中 。比较坐标得 ，。此时 处的内角是反角 （角和 ），且 自动成立。
对 、、、、 使用鞋带公式，得到面积 能被 整除这一条件化为 ，也就是 。为了使 与 、 严格在直线 的同一侧，需要 。
所以 取 ，共有 个值。
Place and with the pentagon above line and write The right angles at and make and vertical: and with At the side makes a angle with the downward ray heading into the pentagon, so Similarly at the side makes a angle with the downward ray so where Matching coordinates gives and The interior angle at is then the reflex angle (angle sum ), and automatically.
The shoelace formula on gives area The condition that be divisible by reduces to that is, For to lie strictly on the same side of line as and we need
So runs over which is values.
4.
对每个正整数 ，令 为把通常写法 看作 进制数时的数值，其中 是大于 中最大数字的最小整数。例如，若 ，则 ，而 作为 进制数等于 ；因此 。求小于 且满足 的正整数 的个数。
For each positive integer let be the value of the base-ten numeral viewed in base where is the least integer greater than the greatest digit in For example, if then and as a numeral in base equals therefore Find the number of positive integers less than such that
小提示：
一位数在任何进制中读出的值都相同。对更长的数，比较位值：若 ，每个数字的权重都比通常写法中小。
One-digit numbers read the same in every base. For longer numerals compare place values: if each digit is worth less than in base ten.
大提示：
所以多位数解恰好是含有数字 的数；用补集计数。
So the multi-digit solutions are exactly the numbers containing a digit count those with complementary counting.
解答：
若 是一位数 ，则数字 在任何进制下的值都是 ，所以 ；全部 个一位正整数都可行。若 的数字为 ，其中 ，则总有 ；若 ，则 ，因为最高位满足 。所以多位数 满足 当且仅当 ，也就是 的某个数字等于 。
含有 的两位数有 到 ，以及 ，共有 个。含有 的三位数有 个，这是从所有三位数中减去不含 的数（首位为 到 ，其余位为 到 ）。
总数为 。
If has a single digit then the numeral has value in every base, so all one-digit numbers work. If has digits with then always, and if then because the leading digit satisfies So a multi-digit satisfies exactly when that is, when some digit of equals
Two-digit numbers containing a the numbers through plus for Three-digit numbers containing a subtracting the numbers with no (leading digit – others –).
The total is
5.
一个罐中有 个弹珠。每个弹珠不是红色就是蓝色，且每种颜色至少有 个弹珠。从罐中不放回地随机抽取 个弹珠时，恰好 个为红色的概率等于恰好 个为红色的概率。求使这件事可能的最小五个 值之和。
An urn contains marbles. Each marble is either red or blue, and there are at least marbles of each color. When marbles are drawn randomly from the urn without replacement, the probability that exactly of them are red equals the probability that exactly of them are red. Find the sum of the five least values of for which this is possible.
小提示：
用组合数写出两个概率， 和 ，然后约去公共因子。
Write both probabilities with binomial coefficients, and and cancel the common factors.
大提示：
方程化简为 ，所以 ；记得两种颜色都至少需要 个弹珠。
The equation reduces to so remember both colors need at least marbles.
解答：
设红弹珠有 个、蓝弹珠有 个，且 。条件为 。因为 且 ，约去公共因子得 也就是
因此 ，而 要求 ，所以 。 最小的五个选择为 ，对应 ，给出 。
和为 。
Say there are red and blue marbles, The condition is Since and cancelling gives that is,
So and requires so The five smallest choices are with giving
The sum is
6.
求所有实数 的和，使得以 为圆心、半径为 的圆至少在一点处与方程 的抛物线相切。
Find the sum of all real numbers such that there is at least one point where the circle with radius centered at is tangent to the parabola with equation
小提示：
圆心 在抛物线的对称轴上。把抛物线上的点写成 ，并研究到圆心的距离平方作为 的函数。
The center lies on the parabola’s axis. Write points of the parabola as and study the squared distance to the center as a function of
大提示：
相切恰好发生在半径垂直于抛物线的地方，也就是距离函数的临界点；不要忘记顶点，在那里两条曲线都有水平切线。
Tangency happens exactly where the radius is normal to the parabola, i.e. at critical points of the distance; don’t forget the vertex, where both curves have horizontal tangents.
解答：
配方得 ，所以令 时，抛物线上的点为 。圆心 在其对称轴上。圆在某点与抛物线相切，当且仅当两条曲线在该点有相同切线，也就是从圆心到该点的半径为抛物线的法线；这恰好出现在距离平方的临界点：
当 时，，所以 （这个圆在两个对称点与抛物线相切）。当 时，该点是顶点 ，到圆心距离为 。半径为 的圆与抛物线在那里都有水平切线，所以 也可行。
所求和为 。
Completing the square, so with the parabola is the set of points and the center lies on its axis. The circle is tangent to the parabola at a point exactly when the two curves share a tangent line there, i.e. when the radius to that point is normal to the parabola — which happens exactly at critical points of the squared distance
At so (the circle touches the parabola at two symmetric points). At the point is the vertex at distance where the parabola and the circle of radius both have horizontal tangent lines, so also works.
The sum is
7.
反复掷一枚标准公平六面骰。每次掷出 或 时，Alice 得到一枚硬币；每次掷出 或 时，Bob 得到一枚硬币；每次掷出 或 时，Carol 得到一枚硬币。Alice 和 Bob 都在 Carol 得到任何硬币之前各自至少得到两枚硬币的概率可写成 ，其中 和 是互质正整数。求 。
A standard fair six-sided die is rolled repeatedly. Each time the die reads or Alice gets a coin; each time it reads or Bob gets a coin; and each time it reads or Carol gets a coin. The probability that Alice and Bob each receive at least two coins before Carol receives any coins can be written as where and are relatively prime positive integers. Find
小提示：
按第一次出现 Carol 的掷骰是第 次来分类：之前的 次都是 Alice 或 Bob，且必须每人至少出现两次。
Condition on the first Carol roll being roll the rolls before it are Alice/Bob rolls that must include at least two of each.
大提示：
给定 次 Alice/Bob 结果，所有 种模式等可能，其中至多一个 Alice 或至多一个 Bob 的坏模式有 种。把相应级数求和。
Given Alice/Bob rolls, all patterns are equally likely, and of them have at most one Alice or at most one Bob. Sum the resulting series.
解答：
每次掷骰分别是 Alice、Bob、或 Carol 的概率都是 。事件成功当且仅当第一次 Carol 之前的结果中，Alice 至少出现两次且 Bob 至少出现两次。第一次 Carol 出现在第 次的概率为 ，在此条件下，前 次形成一个等可能的 Alice/Bob 字符串，共 种。当 时，坏字符串为至多一个 Alice 或至多一个 Bob，数量为 ，且没有字符串同时属于两类坏情形。因此
第一部分为 。第二部分中，，所以
因此 ，所以 。
Each roll is an Alice roll, a Bob roll, or a Carol roll, each with probability The event succeeds exactly when the rolls before the first Carol roll include at least two Alice rolls and at least two Bob rolls. The first Carol roll is roll with probability and given this, the first rolls form one of equally likely Alice/Bob strings. For the bad strings — at most one Alice, or at most one Bob — number and no string is bad in both ways. Hence
The first piece is For the second, so
Therefore and
8.
等腰三角形 满足 。设 为 的内心。 与 的周长之比为 ，并且这两个三角形的所有边长都是整数。求 的最小可能值。
Isosceles triangle has Let be the incenter of The perimeters of and are in the ratio and all the sides of both triangles have integer lengths. Find the minimum possible value of
小提示：
设 、。内切圆与 相切于其中点，所以 ；海伦公式把它化简为 。
With and the incircle touches at its midpoint, so Heron’s formula simplifies this to
大提示：
令 为最简形式；周长比变为 并由整除性确定 。
Set in lowest terms; the perimeter ratio becomes and divisibility pins down
解答：
设 、，则 。内切圆与 相切于中点（从 出发的切线长为 ），所以 。由海伦公式，，因此 周长条件为 。
因为 是有理数，令 为最简形式。于是 迫使 能被 整除；写 ，得到 、，以及 。周长条件中的 完全约去： 因为 ，可知 能被 整除；而且 必须为偶数，因为若 为奇数，左侧两个因子均为奇数，右侧却为偶数。写 并化简，得 。由 ，左侧两个因子都与 互质。若有素数整除 ，右侧会迫使它也整除左侧乘积，产生矛盾；因此 。此时 的唯一正整数解为 。
所以 、、，它为整数当且仅当 能被 整除。取 得到 的边长 ，以及 的边长 ，其周长 与 的比确为 。因此 的最小可能值为 。
Let and so The incircle touches at its midpoint (tangent length from is ), so By Heron’s formula, and therefore The perimeter condition is
Since is rational, write in lowest terms. Then forces to be divisible by writing gives and The perimeter condition then loses entirely: Since we get that is divisible by and must be even, since for odd both factors on the left are odd while the right side is even. Writing and simplifying, Both factors on the left are coprime to (as ). If a prime divided the right side would make it divide their product, a contradiction; hence Now has the unique positive solution
So and which is an integer exactly when is divisible by Taking gives with sides and with sides whose perimeters and are indeed in ratio The minimum possible is
9.
设 表示无穷和 的值。求小于或等于 的最大整数除以 所得的余数。
Let denote the value of the infinite sum Find the remainder when the greatest integer less than or equal to is divided by
小提示：
把每个 展开为关于 的等比级数，并收集相同的 的幂： 的系数会数出 的因数个数。
Expand each as a geometric series in and collect equal powers of the coefficient of counts the divisors of
大提示：
乘以 后，只有 会影响最后三位，而剩下的尾项在 和 之间。
After multiplying by only affect the last three digits, and the leftover tail is between and
解答：
每一项为 ，所以对 求和并收集指数 ，得到 其中 是 的因数个数。因此 ，其中 。
由 、、、，尾项开头为 ，又因为 ，余下各项贡献小于 。所以 ，且
模 时，所有 的项都是 的倍数，只剩 。因为 、、且 ，余数为 。
Each term is so summing over and collecting the exponent where is the number of divisors of Hence with
From the tail starts and since the remaining terms contribute less than So and
Modulo every term with is a multiple of leaving Since and the remainder is
10.
设 中的点 在 上，且 平分 。设 为经过 且在 处与线段 相切的圆。令 和 分别为 与线段 和 的交点。已知 、，并且 、、、 都是正整数。求 的最大可能值。
Let be a triangle with on such that bisects Let be the circle that passes through and is tangent to segment at Let and be the intersections of with segments and respectively. Suppose that and all of and are positive integers. Find the greatest possible value of
小提示：
在 处相切给出 和 的幂： 与 ；角平分线给出 。
Tangency at gives the powers of and and the bisector gives
大提示：
写 、；那么 是整数，而 和 为整数迫使 能被 整除，同时 给出 的上界。
Write then is an integer, and integrality of and forces to be divisible by while bounds
解答：
因为 在 处与 相切，由点 的幂得 ，由点 的幂得 。角平分线定理给出 ，所以 、，其中 是正整数。于是 所以 ，。
为使 和 都为整数，需要 能被 整除，也就是 。此时 迫使 ，且 。当 时，，并且 、、、 都是正整数；边长 也形成合法三角形，因为 。
的最大可能值为 。
Since is tangent to at the power of gives and the power of gives The angle bisector gives so and where is a positive integer. Then so and
For and to be integers we need to be divisible by that is, Then forces and At with all positive integers, and the sides form a valid triangle since
The greatest possible value of is
11.
求最大的整数 ，使三次多项式 的根为 、、，其中 、、 是复数，并且 恰好有七个不同的可能值。
Find the greatest integer such that the cubic polynomial has roots and where and are complex numbers, and there are exactly seven different possible values for
小提示：
固定这三个根的平方根 ；可能的和为 ，这八个表达式成对为 。
Fix square roots of the three roots; the possible sums are eight expressions that pair off as
大提示：
恰好出现七个值时，有一个符号组合为 ；四个表达式 的乘积等于根的对称函数中的 。
Exactly seven values occur when one combination is the product of the four expressions equals in the roots’ symmetric functions.
解答：
三次多项式的根为 。固定它们的平方根 ；那么 在八个表达式 中取值，这些表达式分成四对 。一般情况下八个值都不同。由于三次多项式的常数项为 ，所选三个平方根的乘积非零，所以每个 都非零。若两个非相反的选择发生重合 ，它们必有两个符号不同，从而迫使某个 ，其中 ，这会使八个值减少到至多六个。因此，恰好出现七个值，当且仅当某个选择满足 （其相反选择也给出同一个值 ），并且没有其他退化情形。
这个条件等价于下式为零： 其中 是多项式的根， 是它们的初等对称函数。由韦达定理， 且 ，所以 ，即 ，其根为 和 。
当 时，三次多项式分解为 其互异的根为 ，，和 。可选择平方根 ，，和 ；此时 。八种符号选择使 出现两次，另外六个非零值为 ，，，所以恰好出现七个和。最大的这种整数是 。
The roots of the cubic are Fix square roots of them; then ranges over the eight expressions which come in four pairs Generically all eight are distinct. The product of the three chosen square roots is nonzero because the cubic’s constant term is so every is nonzero. A coincidence between choices that are not opposite must differ in two signs and forces for some which collapses the eight values to at most six. So exactly seven values occur precisely when one choice satisfies — its opposite is then the same value — and no further degeneracies occur.
That condition is the vanishing of where are the roots and their elementary symmetric functions. By Vieta’s formulas and so i.e. with roots and
For the cubic factors as with distinct roots and Choose their square roots as and then The eight sign choices give twice and the six distinct nonzero values so exactly seven sums occur. The greatest such integer is
12.
有一个四面体，它有两个等腰三角形面，边长分别为 、、，还有两个等腰三角形面，边长分别为 、、。四面体的四个顶点都在一个以 为球心的球面上，四个面都与一个以 为球心的球相切。距离 可写成 ，其中 和 是互质正整数。求 。
Consider a tetrahedron with two isosceles triangle faces with side lengths and and two isosceles triangle faces with side lengths and The four vertices of the tetrahedron lie on a sphere with center and the four faces of the tetrahedron are tangent to a sphere with center The distance can be written as where and are relatively prime positive integers. Find
小提示：
边长 与 的边彼此相对，另外四条边都是 。把这两条相对边放在互相垂直的水平轴上；两个球心都在竖直对称轴上。
The edges are and opposite each other plus four of Place those two edges on perpendicular horizontal axes; both centers lie on the vertical symmetry axis.
大提示：
这两条边在竖直方向相距 。在对称轴上分别求两个球心： 到各顶点等距， 到两个相邻面等距。
The two edges sit apart vertically. Find each center on the axis: equal distances to vertices for equal distances to two adjacent faces for
解答：
四个面的边长总多重集为 ，且每条边属于两个面，所以四面体 有相对边 和 ，另外四条边都等于 。把各点放置为 这是相容的，因为 。该构型关于 和 对称，所以两个球心都在 轴上。
对 ，令到 与到 的距离相等，得 ，所以 。对 ，面 的平面为 ，面 的平面为 ，等距条件给出 并且由两个镜面对称可知此点到四个面的距离都相等（距离为 ）。
因此 ，已经最简，所以 。
The four faces have side multiset and each edge lies on two faces, so the tetrahedron has and as opposite edges and the other four edges equal to Place which is consistent since The configuration is symmetric under and under so both centers lie on the -axis.
For equating distances to and gives so For face has plane and face has plane so equal distances require and by the two mirror symmetries this point is equidistant (at distance ) from all four faces.
Therefore which is in lowest terms, so
13.
若两个有限整数集合 和 满足以下条件，就称它们为表亲：
• 和 的元素个数相同，
• 和 不相交，并且
• 可以把 的元素与 的元素配对，使每一对中的两个元素恰好相差 。
例如， 与 是表亲。已知集合 恰好有 个表亲。求集合 可能拥有的最少元素个数。
Call finite sets of integers and cousins if
• and have the same number of elements,
• and are disjoint, and
• the elements of can be paired with the elements of so that the elements in each pair differ by exactly
For example, and are cousins. Suppose that the set has exactly cousins. Find the least number of elements the set can have.
小提示：
每个元素都必须移动 到 外的值，并且映射要一一对应。若某个元素的两个邻居都在 中，它就无法移动，所以 由长度为 或 的连续整数块组成。
Each element must move by to a value outside injectively. An element with both neighbors in is stuck, so consists of blocks of or consecutive integers.
大提示：
只隔一个缺失整数的块会连成链：一条含 个单点块的链恰好有 个像，不同链的选择相乘。分解 ，使总大小最小。
Blocks separated by a single missing integer chain together: a chain of singleton blocks has exactly images, and separate chains multiply. Factor to minimize total size.
解答：
一个表亲 是某个映射的像，该映射把每个 送到 或 ，落在 外，并且是单射。若 ，那么 无处可去，所以 的每个最大连续块长度只能是 或 。一个双点块 被迫映到 ，而一个单点块 可选择 或 。只有当两个块之间恰好隔一个整数时，它们才可能争抢同一个值，因此把块分成若干链：链内相邻块之间的间隔恰好为一。在一条链中，唯一一致的模式是“前 个块向左移，后面的块向右移”，因为某块向右而下一块向左会发生碰撞；双点块同时起到左移与右移的作用，迫使切换恰好在那里发生。因此一条含 个单点块的链产生 个不同的像；含一个双点块的链恰好产生 个；含两个双点块的链产生 个。不同模式给出不同的集合 ，而不同链的选择彼此独立，所以表亲个数等于所有纯单点链的 的乘积。
需要 ，同时最小化元素个数 （含双点块的链只会浪费元素）。把一个合数因子 替换成两个因子 会严格降低代价，因为 。所以最优方案使用质因数分解： 可由五条分别含 个单点块的链实现，也就是若干段相隔一个整数的单点块，彼此放得足够远。
最少元素个数为 。
A cousin is the image of an injection sending each to or landing outside If then has nowhere to go, so every maximal block of consecutive elements of has size or A double block is forced to map to while a singleton chooses or Two blocks can fight over a value only when exactly one integer separates them, so group blocks into chains: consecutive blocks with gaps of exactly one. Within a chain the only consistent patterns are “the first blocks shift left and the rest shift right,” since a block choosing right and its successor choosing left would collide; a double block acts as both left and right, forcing the switch to happen exactly at it. Hence a chain of singletons produces distinct images, a chain containing one double produces exactly and a chain with two doubles produces Distinct patterns give distinct sets and choices in different chains are independent, so the number of cousins is the product of over the all-singleton chains.
We need while minimizing the element count (chains with doubles only waste elements). Replacing a composite factor with the two factors strictly lowers the cost, because So the optimum uses the prime factorization: realized by five chains of singletons — runs of every-other integer — placed far apart.
The least possible number of elements is
14.
对整数 和 ，若 为奇数且 为偶数，定义 ；否则定义 。求正整数序列 ，，，， 的个数，使得 且 其中运算从左到右进行；也就是说， 表示 。
For integers and let if is odd and is even, and otherwise. Find the number of sequences of positive integers such that and where the operations are performed from left to right; that is, means
小提示：
运行值始终与运行和奇偶性相同，所以一项被减去当且仅当它是偶数，且它之前各项之和为奇数。
The running value always has the same parity as the running sum, so a term is subtracted exactly when it is even and the sum of the terms before it is odd.
大提示：
被减去的项总和必须为 。前缀和奇偶性只在奇数项处改变，所以按奇数项把序列分段（奇数项个数为 、 或 ），逐案计数。
The subtracted terms must total Prefix-sum parity flips only at odd terms, so split the sequence at its odd terms (there are or ) and count each case.
解答：
因为 ，前 步后的运行值与 奇偶性相同。所以 被减去，当且仅当 为偶数且前缀和 为奇数；最终值等于 减去被减去项总和的两倍。我们要数的是 的组成，其中位于奇数前缀和位置的偶数项总和恰好为 。前缀奇偶性恰好在奇数项处翻转，因此奇数项共有 个（总和为偶数），而被减去的项正是第 个奇数项与第 个奇数项之间的偶数项；这些 个“奇区间”中的偶数项总和为 ，其他 个区间中的偶数项总和为 ，其中 是所有奇数项之和。
令 为用 个有序区间填入若干偶数项，使总和为 的方法数。一个区间就是把 写成偶数部分的组成，也就是 的组成：当 时 ，且 ；卷积给出下面需要的值：、、，以及 。把 写成 个奇数部分的组成数为 。
对 和 分类：当 时， 分别给出 、、以及 。当 时， 给出 和 。当 时， 给出 。总数为 。
Since the running value after steps has the same parity as So term is subtracted exactly when is even and the prefix sum is odd, and the final value is minus twice the total of the subtracted terms. We must count compositions of in which the even terms sitting where the prefix sum is odd total exactly The prefix parity flips exactly at odd terms, so the odd terms come in (the total is even), and the subtracted terms are precisely the even terms lying between the st and th odd terms; these “odd stretches” must hold even terms totaling while the other stretches hold even terms totaling where is the sum of the odd terms.
Let be the number of ways to fill ordered stretches with sequences of even terms totaling One stretch is a composition of into even parts, i.e. of for and convolving gives the values needed below: and Compositions of into odd parts number
Casework on and for give and For give and For gives The total is
15.
求满足下列性质的有序 元组 的个数：
• 对所有 都有 。
• 是 的倍数。
• 是 的倍数。
Find the number of ordered -tuples having the following properties:
• for all
• is a multiple of
• is a multiple of
小提示：
在模 下工作，其中 、。七个指标三元组 构成一个 Fano 平面：每一对位置恰好在一个三元组中。
Work modulo with and The seven index triples form a Fano plane: every pair of positions lies in exactly one triple.
大提示：
按取值为 的位置集合 分类：一个乘积只有在它的三元组避开 时才保留下来，且每个保留下来的乘积按其三元组中 的个数贡献 。
Classify by the set of positions holding a a product survives only if its triple misses and each surviving product is according to how many ’s its triple holds.
解答：
在模 下工作：取值 为 ，取值 为 。因为 的差覆盖模 的每个非零剩余类恰好一次，七个三元组 是位置集合上的 Fano 平面的直线：每一对位置在恰好一条直线上，任意两条直线恰好交于一点。设 为取值为 的位置集合，。一个乘积项保留下来当且仅当它对应的直线避开 。若 是该直线上取值为 的项数，它的贡献就是 ；线性条件则限制其余 个 的和模 等于 。
对 分类。：全为 的元组可行，得 个。：单个 不可能和为 ，没有解。：没有直线保留下来；两个非零项必须一个为 、一个为 ，有 个。：三个 和为 只有全相同，且三个非零位置不能成一条直线，否则其乘积为 ，所以有 个。：四个 必须二二分开；若 不是一条直线，则恰有一条直线避开它（破坏乘积和），而若 是一条直线，则没有直线避开它，得到 个。：五个 必须四个同号、一个异号；恰有两条直线避开 ，它们交于点 并覆盖五个位置，其乘积相消当且仅当唯一的少数值不在 上，所以有 个。：六个 和为 时要么全相同，要么正负各三个；避开 的四条直线两两交于六个非零位置，且四个直线乘积之积为 ，因此需要恰好两条负直线。全相同时有 或 条负直线；若有三个 ，把位置看作这四条直线形成的 的边，一条直线为负当且仅当它在所选 条边中度数为奇；在 个三边子集中，恰有 条三边路径给出两个奇度数，所以有 个。：七个 需要两个或五个 ，它们分别产生 或 条负直线，但 需要 ，没有解。
总数为 。
Work modulo entries are and entries are Because the differences of hit every nonzero residue mod exactly once, the seven triples are the lines of a Fano plane on the positions: every pair of positions lies on exactly one line, and any two lines meet in exactly one point. Let be the set of positions holding a and A product term survives exactly when its line avoids If is the number of entries equal to on that line, its contribution is and the linear condition constrains the values to sum to mod
Casework on the all-s tuple works: a single can’t sum to none. no line survives; the two nonzero entries must be a and a three s sum to only if all equal, and the three nonzero positions must not form a line, else its product is four s must split two and two; exactly one line avoids a non-line (spoiling the sum), while a line is avoided by no line: five s must go four and one; exactly two lines avoid meeting at a point and covering the five positions, and their products cancel exactly when the lone minority value avoids six s sum to if all equal or three of each; the four lines avoiding pairwise meet in the six nonzero positions, and since the product of all four line-products is we need exactly two negative lines. All-equal gives or negative lines; for three ’s, viewing positions as edges of on the four lines, a line is negative exactly when it has odd degree in the chosen -edge set, and exactly the three-edge paths (of the subsets) give two odd degrees: seven s need two or five ’s, which make or lines negative respectively, but needs none.
The total is