4 a b 1 1 c 1 d 3 e 2 f g 6 h i j k 7 l m n o 3 p q 5 r 2 s 4 t 3 3 u v 2

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1 Round Solutions Year 25 Academic Year //25. In the hexagonal grid shown, fill in each space with a number. After the grid is completely filled in, the number in each space must be equal to the smallest positive integer that does not appear in any of the touching spaces. You do not need to prove that your configuration is the only one possible; you merely need to find a configuration that satisfies the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.) Solution a b 1 1 c 1 d e 2 f g 6 h i j k 7 l m n o p q 5 r 2 s t u v 2 We label the cells a through v lexicographically as in the diagram to the left. The in row 1 must touch a 2, but the 2 can t be c because c already touches a 2. Therefore, a = 2. The must also touch a, so c =. Finally, b must be at least, but can t be greater than because there is no way to place a in a neighboring space. Therefore, b =. The 6 in the third row must touch a 5. That 5 cannot be l or m because those each neighbor the 5 in the fifth row. The 5 also cannot be d or h because there are not enough empty spaces for d or h to neighbor all of 1, 2,, and. Therefore, g = 5. Now the three vacant spaces touching g must have the numbers 2,, and. Since f touches a 2 and a, it must be the. Since d touches a, it must be the 2. This leaves l =. The progress so far is summarized in the diagram to the right. Consider the in the sixth row. It must touch a, but the only space surrounding it that does not touch a is v, so v =. The 2 in row 7 must touch a 1, so t = 1. Next, consider l =. It must touch a 1 and a 2, so m and q must have a 1 and 2 between them. But q touches a 1, so q = 2 and m = 1. To finish the right edge, consider the 5 in the fifth row. It must touch a, so r =. We can quickly determine that n = 2 and h = from here e h i j k 7 m n o p q 5 r 2 s t u v 2

2 Round Solutions Year 25 Academic Year Our progress so far is summarized in the diagram to the right. Next, the 7 in the center must touch the six numbers 1 to 6 once each. We already found the numbers 2 through 5, so k and p are 1 and 6 between them. Notice that regardless of which of the two is the 6, the remaining vacant space it touches (either j or s) must be a 5. In particular, this means each of j and s is either 5 or neighbors a 1, so neither can be a 1. Since the in row 7 must touch a 1, that 1 must be u. Then s = 5, since it now touches 1 through, but p e i j k o p s 1 u 2 The 2 in row 6 must touch a 1, so o = 1. All of the vacant spaces around i touch a 2, so i is at most 2. It is also at least 2 since it touches a 1, so i = 2. By similar reasoning, e is a and j is a. Since j is not a 5, the 6 must be p, and the 1 goes in k. This completes the grid. The only solution is the one shown below

3 Round Solutions Year 25 Academic Year //25. Let a 1, a 2, a,... be a sequence of positive real numbers such that a k a k+2 = a k for all positive integers k. If a 1 and a 2 are both positive integers, find the maximum possible value of a 201. Solution First we use the recurrence relation a k a k+2 = a k to compute a 201 in terms of a 1 and a 2, then we ll choose values of a 1 and a 2 that maximize a 201. The recurrence relation can be rewritten as Using this, we compute a k+2 = a k a k. a = a a 1 a = a + 1 a 2 = a 5 = a + 1 a = a 6 = a a = a 2 +1 a = a 2 + a 1 + 1, a 2 a 1 a 2 a 2 +a 1 +1 a 1 a a 2 +1 = (a 1a 2 + a 2 + a 1 + 1)(a 1 ) a 1 (a 1 a 2 )(a 2 + 1) a 1 +1 a a 2 +a 1 +1 = a 1 a 2 a 7 = a = a a a 1 +1 = a 2. 5 a 2 a 1 +a 2 +1 a 2 a 1 +a 2 +1 a 1 a 2 = a 1, = (a 1 + 1)(a 2 + 1)(a 1 ) (a 1 a 2 )(a 2 + 1) = a a 2, Since each term of the sequence depends only on the previous two terms and we have shown a 1 = a 6 and a 2 = a 7, we conclude that the sequence is periodic with period 5. Therefore, for any k, we have a k = a r, where r is the remainder when k is divided by 5. So a 201 = a = a 2 + a a 1 a 2 = 1 a a a 1 a 2. We are given that a 1 and a 2 are positive integers, so each term in this expression is at most 1 and the sum is at most. For a 1 = a 2 = 1, the sum is equal to. The maximum possible value of a 201 is from the sequence 1, 1, 2,, 2, 1, 1, 2,, 2,....

4 Round Solutions Year 25 Academic Year //25. Let A 1 A 2 A... A 20 be a 20-sided polygon P in the plane. Suppose all of the side lengths of P are 1, the interior angle at A i measures 108 degrees for all odd i, and the interior angle at A i measures 216 degrees for all even i. Prove that that the lines A 1 A 9, A 2 A 12, A A 15, A A 18, and A 6 A 20 all intersect at the same point. A 9 A 10 A 11 A 12 A 8 A 7 A 6 A 5 A A A 2 A 1 Solution A 20 Let O be the center of P. Let B 2i be the point on the line segment A 2i O such that A 2i B 2i has length 1. For each i, draw segment A 2i B 2i and B 2i B 2i+2. The resulting figure is shown below left, with the line segments A 2 A 8, A A 10, A 5 A 1, A 6 A 16, and A 7 A 19 in gray. A 1 A 1 A 15 A 16 A 18 A 19 A 17 A 11 A 9 A 8 A 7 A 10 B 10 A 12 B 12 A 1 A 1 B 8 A 6 B 6 O B B 1 B 18 B 16 A 5 B 2 B 20 A A A 2 A 20 A 19 A 1 We claim that the 10 small pentagons are regular unit pentagons. By symmetry we may check only A 2 A A B B 2, so consider only this pentagon for now. First note that A 2 B 2 bisects a 216 angle so A 2 = 108. Since A 2 = A = A = 108 and B = B 2, the pentagon is equiangular. By construction, edges have length 1, so all 5 edges have length 1 and the pentagon is regular. In particular, since the pentagon is regular, the line segments B 2i B 2i 2 have length 1. By symmetry, the inner decagon is also equiangular so it must be regular. Specifically, each angle of the decagon has measure 1. A 18 A 16 A 15 A 17 We now claim that A 2 A 8, A A 10, A 5 A 1, A 6 A 16, and A 7 A 19 all intersect at B 6. Since B 6 lies on A 6 O by definition, A 6 A 16 necessarily passes through B 6. If A 2 A 8 passes through B 6, then A A 10 passes through B 6, because A A 10 is the reflection of A 2 A 8 through A 6 A 16. Similarly, if A 5 A 1 passes through B 6, then so does A 7 A 19. So to solve the problem, it suffices to show that A 2 A 8 and A 5 A 1 both contain B 6. Lines B B 6 and A 2 A 8 are both perpendicular to A 5 A 15 so are parallel (and hopefully equal). The point B 6 lies on A 2 A 8 if and only if these lines are equal. This occurs if and only if B B 6 A 8 = 180. To prove this, we first write B B 6 A 8 = B B 6 B 8 + B 8 B 6 A 8. Since B B 6 B 8 is the interior angle of a regular decagon, its measure is 1. Further, B 6 B 8 A 8 is isosceles with vertex angle 108, so the base angle is B 8 B 6 A 8 = 6. This forces B B 6 A 8 = = 180,

5 Round Solutions Year 25 Academic Year and A 2 A 8 passes through B 6. Similarly, B 6 B 12 is parallel to A 5 A 1, so A 5 A 1 contains B 6 if and only if B 6 B 12 A 1 = 180. To prove this, first write A 9 A 8 A 7 A 6 A 5 A A B 6 B 12 A 1 = B 6 B 12 B 10 + B 10 B 12 A 12 + A 12 B 12 A 1. Since A 12 B 12 A 1 is the smaller angle formed by a diagonal and a side in a regular pentagon, its measure is 6. Since B 10 B 12 A 12 is an interior angle of a regular pentagon, its measure is 108. Therefore B 6 B 12 A 1 = B 6 B 12 B = B 6 B 12 B A 11 A 10 B 10 A 12 B 12 B 6 B 8 O B B 2 B 20 B 1 B 18 B 16 A 1 A 1 A 18 A 16 A 15 A 17 A 2 A 1 A 20 A 19 To find the measure of B 6 B 12 B 10, we view it as an interior angle of the quadrilateral B 6 B 8 B 10 B 12. The sum of the interior angles of a quadrilateral is 60. Two of the interior angles of the quadrilateral are interior angles of a regular decagon, and therefore have measure 1. The other two angles are also equal to one another, since they are reflections of each other about the line A 9 A 19. So we have Dividing by 2 gives B 6 B 12 B 10 = 60. B 6 B 12 A 1 = B 6 B 12 B = 180. therefore A 5 A 1 passes through B 6. All 5 lines meet at B 6.

6 Round Solutions Year 25 Academic Year //25. An infinite sequence (a 0, a 1, a 2,...) of positive integers is called a ribbon if the sum of any eight consecutive terms is at most 16; that is, for all i 0, i=k a i + a i a i A positive integer m is called a cut size if every ribbon contains a set of consecutive elements that sum to m; that is, given any ribbon (a 0, a 1, a 2,...), there exist nonnegative integers k l such that l a i = m. Find, with proof, all cut sizes, or prove that none exist. Solution We claim that m is a cut size if and only if m is a positive multiple of 16. First we show that any positive integer m that is not a multiple of 16 is not a cut size. We do this by constructing counter-examples based on the highest power of 2 that divides m. If m is odd, then m is not a sum of terms from the ribbon 2, 2, 2, 2, 2, 2, 2, 2,.... Thus any cut size is even. If m is an odd multiple of 2 (meaning m 2 (mod )) then m is not a sum of terms from the ribbon, 1,, 1,, 1,, 1,.... Specifically, any sum of an even number of consecutive terms from this sequence is a multiple of and any odd number of terms gives an odd sum. Therefore any cut size is a multiple of. Likewise, the ribbon 5, 1, 1, 1, 5, 1, 1, 1,... shows that any cut size must be a multiple of 8 and the period 8 ribbon 9, 1, 1, 1, 1, 1, 1, 1,... shows that any cut size must be a multiple of 16. Now we must show that for any m that is a multiple of 16, every ribbon contains a consecutive set of terms that sums to m. Let some m = 16n be given. Consider first all ribbons a 0, a 1,... such that each set of 8 consecutive terms gives a sum of 16: a i + a i a i+8 = 16. We call such a ribbon maximal. Given a maximal ribbon, the sum of the first 8n terms is a 0 + a a 8n = 16n = m

7 Round Solutions Year 25 Academic Year showing that m can be obtained as a sum of consecutive terms from this ribbon. Now we are left to show that m can be achieved as a sum of consecutive terms for any ribbon that is not maximal. For any i let s i = a 0 + a a i be the sum of the first i + 1 terms. Since the a i are all positive integers, s i is a strictly increasing sequence of positive integers. We wish to find i, j such that s j s i = m, which means that a i+1 + a i a j = m. Since our ribbon is not maximal, there exists some k such that s k+8 s k < 16. Consider the set T = {s k, 1 + s k, s k }. Since s k+8 < 16 + s k, the nine partial sums s i, s i+1,..., s i+8 are members of T. Next consider U = {m + s k, m s k,... m s k }. We will show that at least 8 of these values are also partial sums. Let s b be the largest partial sum smaller than s k + m. This means s b+1 is an element of U. Since we know so s b+1 through s b+8 are all elements of U. s b+8 s b = a b+1 + a b a b+8 16 s b s b < m s k Now we know that T contains at least 9 partial sums from our sequence and U contains at least 8 partial sums. By the Pigeonhole Principle, one of the sixteen sets {s k, m + s k }, {1 + s k, m s k },... {15 + s k, m s k } contains two partial sums. These two give us the desired s i, s j with difference m, so the proof is complete.

8 Round Solutions Year 25 Academic Year //25. For any positive integer b 2, we write the base-b numbers as follows: (d k d k 1... d 0 ) b = d k b k + d k 1 b k d 1 b 1 + d 0 b 0, where each digit d i is a member of the set S = {0, 1, 2,..., b 1} and either d k 0 or k = 0. There is a unique way to write any nonnegative integer in the above form. If we select the digits from a different set S instead, we may obtain new representations of all positive integers or, in some cases, all integers. For example, if b = and the digits are selected from S = { 1, 0, 1}, we obtain a way to uniquely represent all integers, known as the balanced ternary representation. As further examples, the balanced ternary representation of the numbers 5,, and 25 are: 5 = (1 1 1), = ( 1 0), 25 = ( ). However, not all digit sets can represent all integers. If b = and S = { 2, 0, 2}, then no odd number can be represented. Also, if b = and S = {0, 1, 2} as in the usual base- representation, then no negative number can be represented. Given a set S of four integers, one of which is 0, call S a -basis if every integer n has at least one representation in the form n = (d k d k 1... d 0 ) = d k k + d k 1 k d d 0 0, where d k, d k 1,..., d 0 are all elements of S and either d k 0 or k = 0. (a) Show that there are infinitely many integers a such that { 1, 0, 1, a + 2} is not a -basis. (b) Show that there are infinitely many integers a such that { 1, 0, 1, a + 2} is a -basis. Solution Given a set S of integers with one of them 0, we will say a nonzero integer n is representable using S if we can write n in the form (d k d k 1... d 0 ) = d k k + d k 1 k d d 0 0, where d k, d k 1,..., d 0 are all elements of S and d k 0. Part a: We will show that if a + 2 is a multiple of, we do not get a -basis. Any value of a 1 (mod ) will force a+2 so there are infinitely many such S. Let b = a + 2 and suppose that S = { 1, 0, 1, b} is a -basis. Since b is representable using X, b = (d k d k 1... d 0 ) = d k k + d k 1 k d d 0 0,

9 Round Solutions Year 25 Academic Year where d k, d k 1,..., d 0 are all elements of X and d k 0. Choose some representation with k minimal. Notice that b b a (mod ) but our expression for b also gives b (d k d k 1... d 0 ) d 0 (mod ). The only choice for d 0 in S that is congruent to 2 modulo is d 0 = b. Then removing the final digit we get b = (d k d k 1... d 0 ) = (d k d k 1... d 1 ) + b. Isolating b gives b = (d k d k 1... d 1 ). However, this gives us a representation for b with fewer digits, contradicting the minimality of k. So there could not have been a representation of b, and { 1, 0, 1, a + 2} is not a -basis. Part b: We claim that S b = { 1, 0, 1, b 2} is a -basis for every positive integer b. Since b 2 is of the form a + 2, this will suffice. For a given n, let d be the element of S b such that n d (mod ). Note that if then n d = d k k + d k 1 k d d 0 n = d k k+1 + d k 1 k + + d d d. Therefore, Lemma 1: If n d is representable by S b, then n is representable by S b. The number n d will often be smaller than n, suggesting that we may be able to use the fact that n is representable when n is small to prove that every n is representable. We first explore when n d < n. Assume that n 0. If d = 1, 0, or 1, then n d n + 1 n. However, if d = c 2 then n d n + (c 2) which could be larger than n. In the case that n c, we have c n so n d n + (c 2) n + c n + n < n. In summary, we see that:

10 Round Solutions Year 25 Academic Year Lemma 2: If n b then n d < n. Suppose that we know that every n with 0 < n < b is representable by S b. Then we can prove that every n is representable by strong induction on n as follows. Suppose that n is representable when n N, and suppose that n = N + 1. If n d = 0 then n is in S b so is representable. Assume not. By the first lemma, n is representable if n d is representable. Either 0 < n d < b n d, in which case is representable by assumption, or n d b, in which case n d n d < n, by our second lemma, so that is representable by our inductive hypothesis. So to prove that every n is representable, it suffices to prove that every n with 0 < n < b is representable. As an example, the b = 1 case is follows since values of n for which 0 < n < are 1 and 1, and these values are in S 1. For values of n such that n is representable by S 1 with k b digits, we can use such a representation to construct a representation of n by S b as follows. Suppose that where d 0, d 1,..., d k 1 S 1 and k b. Then n = d k 1 k 1 + d k 2 k d d 0 0 n = ( d k 1 ) k 1 + ( d k 2 ) k ( d 1 ) 1 + ( d 0 ) 0 We want to use this to represent n by S b. Consider a new sum, e k+b 1 k+b 1 + e k+b 2 k+b e 0 0 where e i is defined as follows. If i < k then { di if d e i = i { 1, 0, 1} b 2 if d i = 2 If k i < b then we set e i = 0. When i b, let { 0 if di b { 1, 0, 1} e i = 1 if d i b = 2 Then if d i 2, Furthermore, if d i = 2, we get e i+b i+b + e i i = 0 + ( d i ) i = ( d i ) i. e i+b i+b + e i i = ( 1) i+b + ( b 2) i = 2 i = ( d i ) i

11 Round Solutions Year 25 Academic Year as well. Therefore k 1 n = ( d i ) i i=0 Each e i is in S b so we are done. i=0 k 1 = e i+b i+b + e i i = i=0 k+b 1 e i i. We now use mathematical induction to prove that every n with n < b can be represented by S 1 using b or fewer digits. The b = 1 case is that ±1 may be represented with a single digit. This is true since these are indeed in S 1. Now we suppose that the claim is true for b and prove it is true for b + 1. That is, if n d is representable using b digits, then n is representable using b + 1 digits. So to prove the claim, it suffices to show that if n < b+1 then n d < b. If d { 1, 0, 1} then n d is the nearest multiple of to n. If d = 2 then n d is the next lower multiple of. So replacing n by n d rounds n to the nearest multiple of, rounding down to break ties. Suppose that n < b+1. Since n is an integer and b+1 is one more than a multiple of, 1 b+1 n b+1 1. When we round all elements in this interval to the nearest multiple of we find This means b+1 1 b n d b+1. n d b 1. making n d < b b+1 n d. So if n <, then n < b is representable by S 1 using no more than b digits. < b. This proves that every n with Since every n with n < b is representable by S 1 using at most b digits, every such n is representable in S b and this gives us the base cases for our induction. c 201 Art of Problem Solving Foundation