PUMaC 2011 · 团队赛
PUMaC 2011 — Team Round
题目详情
Team Round Time Limit: 25 minutes Maximum Possible Score: 81 The following is a mathematical Sudoku puzzle which is also a crossword. Your job is to fill in as many blanks as you possibly can, including all shaded squares . You do not earn extra points for showing your work; the only points you get are for correctly filled-in squares. You get one point for each correctly filled-in square . You should read through the following rules carefully before starting. • Your time limit for this round is 25 minutes, in addition to the five minutes you get for reading the rules. So make use of your time wisely. The round is based more on speed than on perfect reasoning, so use your intuition well, and be fast . • This is a Sudoku puzzle ; all the squares should be filled in with the digits 1 through 9 so that every row and column contains each digit exactly once. In addition, each of the nine 3 × 3 boxes that compose the grid also contains each digit exactly once. Furthermore, this is a super-Sudoku puzzle ; in addition to satisfying all these conditions, the four 3 × 3 boxes with red outlines also contain each of 1 , . . . , 9 exactly once. This last property is important to keep in mind – it may help you solve the puzzle faster. • Just to restate the idea, you can use the digits 1 through 9 , but not 0 . You may not use any other symbol, such as π or e or ε . Each square gets exactly one digit. • The grid is also a crossword puzzle ; the usual rules apply. The shaded grey squares are the “black” squares of an ordinary crossword puzzle. The white squares as well as the shaded yellow ones count as the “white” crossword squares. All squares, white or shaded, count as ordinary Sudoku squares. • If you obtain the unique solution to the crossword puzzle, then this solution extends to a unique solution to the Sudoku puzzle. • You may use a graphing calculator to help you solve the clues. The following hints and tips may prove useful while solving the puzzle. • Use the super-Sudoku structure described in the first rule; use all the symmetries you have. Remember that we are not looking for proofs or methods, only for correctly filled-in squares. • If you find yourself stuck on a specific clue, it is nothing to worry about. You can obtain the solution to that clue later on by solving other clues and figuring out certain digits of your desired solution. Just move on to the rest of the puzzle. • As you progress through the puzzle, keep filling in all squares you have found on your solu- tion sheet, including the shaded ones . Remember that for scoring, the shaded grey squares count the same as the white ones. Good luck! 1 Page 1 of 5
Figure 1: Good luck. Make sure you turn in the solution sheet, not this page! 1 Across • 1 Across. The following is a normal addition where each letter represents a (distinct) digit: GOT + T O + GO + T O = T OP . This certainly does not have a unique solution. However, you discover suddenly that G = 2 and P ∈ { / 4 , 7 } . Then what is the numeric value of the expression GOT × T O ? • 3 Across. A strobogrammatic number which reads the same upside down, e.g. 619 . On the other hand, a triangular number is a number of the form n ( n + 1) / 2 for some n ∈ N , th e.g. 15 (therefore, the i triangular number T is the sum of 1 through i ). Let a be the third i strobogrammatic prime number. Let b be the smaller number of the smallest pair of triangular numbers whose sum and difference are also triangular numbers. What is the value of ab ? • 6 Across. A positive integer m is said to be palindromic in base ` if, when written in base ` , its digits are the same front-to-back and back-to-front. For j, k ∈ N , let μ ( j, k ) be the smallest base- 10 integer that is palindromic in base j as well as in base k , and let ν ( j, k ) := ( j + k ) · μ ( j, k ) . Find the value of ν (5 , 9) . • 7 Across. Suppose you have the unique solution to this Sudoku puzzle. In that solution, let X denote the sum of all digits in the shaded grey squares. Similarly, let Y denote the sum of all numbers in the shaded yellow squares on the upper left block (i.e. the 3 × 3 box outlined 2 Page 2 of 5
red towards the top left). Concatenate X with Y in that order, and write that down. • 8 Across. For any n ∈ N such that 1 < n < 10 , define the sequence X , X , . . . by X = n , n, 1 n, 2 n, 1 and for r ≥ 2 , X is smallest number k ∈ N larger than X such that k and the sum n,r n,r − 1 of digits of k are both powers of n . For instance, X = 3 , X = 9 , X = 27 , and so on. 3 , 1 3 , 2 3 , 3 Concatenate X with X , and write down the answer. 2 , 2 2 , 4 • 9 Across. Find positive integers x, y, z satisfying the following properties: y is obtained by subtracting 93 from x , and z is obtained by subtracting 183 from y ; furthermore, x , y and z in their base- 10 representations contain precisely all the digits from 1 through 9 once (i.e. concatenating x , y and z yields a valid 9 -digit Sudoku answer). Obviously, write down the concatenation of x , y and z in that order. • 11 Across. Find the largest pair of two-digit consecutive prime numbers a and b (with a < b ) such that the sum of the digits of a plus the sum of the digits of b is also a prime number. Write the concatenation of a and b . • 12 Across. Suppose you have a strip of 2 n + 1 squares, with n frogs on the n squares on the left, and n toads on the n squares on the right. A move consists either of a toad or a frog sliding to an adjacent square if it is vacant, or of a toad or a frog jumping one square over another one and landing on the next square if it is vacant. For instance, the starting position F F F F F T T T T T has the position F F F F F T T T T T or the position F F F F F T T T T T as results of valid first moves. What is the minimum number of moves needed to swap the toads with the frogs if n = 5 ? How about n = 6 ? Concatenate your answers. • 15 Across. Let w be the largest number such that w , 2 w and 3 w together contain every digit from 1 through 9 exactly once. Let x be the smallest integer with the property that its first n m 5 multiples contain the digit 9 . A Leyland number is an integer of the form m + n for integers m, n > 1 . Let y be the fourth Leyland number. A Pillai prime is a prime number p for which there is an integer n > 0 such that n ! ≡ − 1 (mod p ) , but p 6 ≡ 1 (mod n ) . Let z be the fourth Pillai prime. Concatenate w , x , y and z in that order to obtain a permutation of 1 , · · · , 9 . Write down this permutation. 3 Page 3 of 5
• 19 Across. A hoax number k ∈ N is one for which the sum of its digits (in base 10 ) equals the sum of the digits of its distinct prime factors (in base 10 ). For instance, the distinct prime factors of 22 are 2 and 11 , and we have 2 + 2 = 2 + (1 + 1) . In fact, 22 is the first hoax number. What is the second? • 20 Across. Let a , b and c be distinct 2 -digit numbers satisfying the following properties: y x – a is the largest integer expressible as a = x = y , for distinct integers x and y . – b is the smallest integer which has three partitions into three parts, which all give the same product (which turns out to be 1200 ) when multiplied. – c is the largest number that is the sum of the digits of its cube. Concatenate a , b and c , and write down the resulting 6 -digit prime number. • 21 Across. Suppose N = a b c d is a 4 -digit number with digits a , b , c and d , such that N = 7 a · b · c · d . Find N . • 22 Across. What is the smallest number expressible as the sum of 2 , 3 , 4 , or 5 distinct primes? 2 Down • 1 Down. For some a, b, c ∈ N , let the polynomial 5 4 3 2 p ( x ) = x − 252 x + ax − bx + cx − 62604360 have five distinct roots that are positive integers. Four of these are 2 -digit numbers, while the last one is single-digit. Concatenate all five roots in decreasing order, and write down the result. • 2 Down. Gene, Ashwath and Cosmin together have 2511 math books. Gene now buys as many math books as he already has, and Cosmin sells off half his math books. This leaves them with 2919 books in total. After this, Ashwath suddenly sells off all his books to buy a private jet, leaving Gene and Cosmin with a total of 2184 books. How many books did Gene, Ashwath and Cosmin have to begin with? Concatenate the three answers (in the order Gene, Ashwath, Cosmin) and write down the result. • 3 Down. A regular octahedron is a convex polyhedron composed of eight congruent faces, each of which is an equilateral triangle; four of them meet at each vertex. For instance, the following diagram depicts a regular octahedron: 4 Page 4 of 5
Let T be a regular octahedron of edge length 28 . What is the total surface area of T , rounded to the nearest integer? • 4 Down. Evaluate the value of the expression T 25 ∑ k, k = T +1 24 th where T denotes the i triangular number (the sum of the integers from 1 through i ). i • 5 Down. Suppose r and s are consecutive multiples of 9 satisfying the following properties: – r is the smallest positive integer that can be written as the sum of 3 positive squares in 3 different ways. – s is the smallest 2 -digit number that is a Woodall number as well as a base- 10 Harshad n number. A Woodall number is any number of the form n · 2 − 1 for some n ∈ N . A base- 10 Harshad number is divisible by the sum of its digits in base 10 . Concatenate r and s and write down the result. • 10 Down. For any k ∈ N , let ϕ ( k ) denote the sum of the distinct prime factors of k . Suppose p N is the largest integer less than 50000 satisfying ϕ ( N ) = ϕ ( N + 1) , where the common p p value turns out to be a meager 55 . What is N ? th • 13 Down. The n s -gonal number P ( s, n ) is defined as P ( s, n ) = ( s − 3) T + T , n − 1 n th th where T is the i triangular number (recall that the i triangular number is the sum of the i numbers 1 through i ). Find the least N such that N is both a 34 -gonal number, and a 163 -gonal number. • 14 Down. A biprime is a positive integer that is the product of precisely two (not necessarily distinct) primes. A cluster of biprimes is an ordered triple ( m, m + 1 , m + 2) of consecutive integers that are biprimes. There are precisely three clusters of biprimes below 100 . Denote these by, say, { ( p, p + 1 , p + 2) , ( q, q + 1 , q + 2) , ( r, r + 1 , r + 2) } , and add the condition that p + 2 < q < r − 2 to fix the three clusters. Interestingly, p + 1 and q are both multiples of 17 . Concatenate q with p + 1 in that order, and write down the result. • 16 Down. Find the least positive integer m (written in base 10 as m = a b c , with digits a , b , a c ), such that m = ( b + c ) . • 17 Down. Let X be a set containing 32 elements, and let Y ⊆ X be a subset containing 29 elements. How many 2 -element subsets of X are there which have nonempty intersection with Y ? • 18 Down. Find a positive integer K < 196 , which is a strange twin of the number 196 , in the 2 2 3 3 sense that K shares the same digits as 196 , and K shares the same digits as 196 . 5 Page 5 of 5
解析
Team Round Solutions 1 Across • 1 Across. Observe that the first digit of the 3 -digit number GOT changes when we add three 2 -digit numbers. Furthermore, one of the 2 -digit numbers starts with a 2 , and so we add at most 29 + 98 + 98 = 225 to GOT . This means T is either 3 or 4 or 5 . But then, T O is at most 59 , and so in fact, we add at most 59 + 59 + 29 = 147 to GOT . So in fact, T is either 3 or 4 . So GOT is something of the form 2 ∗ 3 or 2 ∗ 4 . Some calculator bashing now yields ( G, O, T, P ) = (2 , 6 , 3 , 1) , so that GOT × T O = 9468 . • 3 Across. The only digits which mean something meaningful upside down are 1 , 6 , 8 , 9 , 0 . A prime can end with only 1 out of this list, so all strobogrammatic primes start and end with 1 . Clearly, 11 is the first. Checking the three-digit numbers now give 101 and 181 as the next two, so a = 181 . Enumerating the triangular numbers yields 1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 etc. Clearly, 21 − 15 and 21 + 15 are both in the list, so b = 15 . Therefore, ab = 2715 . • 6 Across. This one is hard to solve, so leave it blank for now. • 7 Across. This is impossible to solve until you have all the solutions! • 8 Across. It is clear that X = 4 and X = 8 . A little more checking yields X = 512 , so 2 , 2 2 , 3 2 , 4 we are looking for 4512 . • 9 Across. You can possibly choose to leave this one blank as well, but if you are curious and have time, then you can figure it out with a bit of casework. Denoting x as 100 x + 10 x + x 1 2 3 and similarly for y and z as well, observe that we know { x , y , z } for i ∈ { 1 , 2 , 3 } all represent i i i different digits. From the first condition we get either 100 x + 10( x − 9) + ( x − 3) = 100 y + 1 2 3 1 10 y + y , 100( x − 1)+10 x +( x +7) = 100 y +10 y + y , or 100( x − 1)+10( x − 1)+( x − 3) = 2 3 1 2 3 1 2 3 1 2 3 100 y + 10 y + y , and similarly for the second condition. Now, doing a bit of casework yields 1 2 3 the unique solution (819 , 726 , 543) . • 11 Across. There are only so many pairs you have to check! Starting with the largest consec- utive pair of primes (89 , 97) , we easily see that the solution has to be (47 , 53) . • 12 Across. This can be solved using induction, but it is hard, so it’s better to leave it blank for now. • 15 Across. It is not too hard to see that x has to be 98 . Checking Leyland numbers one by one, observe that the list goes like 8 , 17 , 32 , 54 , . . . , and so y = 54 . Now, w is clearly a 3 -digit number, and since it’s a large 3 -digit number which is still 3 -digit when multiplied by 3 , it follows that the first digit of w is 3 and the second digit is less than 4 . Therefore, 2 w and 3 w start with 6 and 9 respectively, and so w cannot contain 6 , 9 , 5 , 4 , 8 (the last few because of this clue). Therefore, the second digit of w must be 2 or 1 . It’s now routine to check that w must be 327 . This leaves the digits 1 and 6 for z , and since it is a prime, it must be 61 . Therefore, ( w, x, y, z ) = (327 , 98 , 54 , 61) . 1 Page 1 of 4
• 19 Across. A brute force check in this case works. We get the second hoax number to be 58 . • 20 Across. It is well known that a = 16 (in fact, it is the only integer expressible as both y x 4 2 x and y for distinct x and y ). For b , factor 1200 = 2 · 3 · 5 . A simple check gives you 39 = 24 + 10 + 5 = 25 + 8 + 6 = 20 + 15 + 4 . Finally, since c cannot contain the digits 5 , 8 , 1 , 6 , 3 , 9 (because of the previous clue), it follows that c is a 2 -digit number using two digits from { 2 , 4 , 7 } . Since the whole thing is prime, c ends with 7 . Checking 27 and 47 , it is clear that only 27 is the sum of the digits of its cube 19683 . So the 6 -digit prime number must be 163927 . 7 • 21 Across. If d ≥ 4 , then d already has 5 digits. Therefore, d ∈ { 1 , 2 } . But obvious d 6 = 1 , 3 because then abc can be at most 9 = 729 , which is not 4 digits. Therefore, d = 2 or d = 3 . If d = 3 , N is a multiple of 2187 . There are only four 4 -digit multiples, and it is easy to check that they do not satisfy the conditions. So d = 2 . The multiples of 128 that end in 2 are the 4 7 (mod 5) multiples. A quick check now yields 6912 = 6 · 9 · 1 · 2 . • 22 Across. If you have time, brute force this! If not, you can leave this clue blank. 2 Down • 1 Down. This can be solved with enough time by factoring 62604360 cleverly, and finding which combinations add to 252 . If you try it that way, a good first step may be to figure out what the one-digit factor is. If all else fails, don’t fill this in yet. • 2 Down. This is just a simultaneous equations problem! Denoting by G , A and C the original number of books, you get G + A + C = 2511 , 2 G + A + C/ 2 = 2919 , 2 G + C/ 2 = 2184 . Solving this system yields ( G, A, C ) = (864 , 735 , 912) . √ 2 • 3 Down. Each face of T is an equilateral triangle of side 28 , so its area is 3 · 28 / 4 ≈ 339 . 48 . Therefore, the total surface area of the whole figure is 339 . 48 ∗ 8 ≈ 2716 square units. This gives the result. • 4 Down. Observe that T − T = 25 . Therefore, the given sum is just T + 1 + T + 2 + 25 24 24 24 . . . + T + 25 = 25 · T + T = (24 · 25 · 25) / 2 + (25 · 26) / 2 = 7500 + 325 = 7825 . 24 24 25 • 5 Down. Even if s is hard to figure out, r is not; take all the multiples of 9 in order, and check 2 2 2 2 2 2 2 2 2 the property. At 54 , we get 54 = 3 + 3 + 6 = 7 + 2 + 1 = 5 + 5 + 2 . Therefore, s must be 63 , so we are looking for 5463 . • 10 Down. This is pretty hard, so leave it blank for now. • 13 Down. This is also fairly hard, so leave it blank. 2 Page 2 of 4
• 14 Down. The multiples of 17 hint gives it away. (33 , 34 , 35) is clearly a cluster, as 33 = 3 · 11 , 34 = 2 · 17 , 35 = 5 · 7 . So, p + 1 = 34 . Now, 85 , 86 , 87 is also a cluster, so q = 85 . So we are looking for 8534 . a • 16 Down. We have 100 a + 10 b + c = ( b + c ) . Since this is the least positive integer, we may as well try a = 2 ( a = 1 would not work, because then the number of digits don’t add up). If 2 200 + 10 b + c = ( b + c ) , then the square has to be either 225 , 256 or 289 . Clearly the last one fits the bill, and so we are looking for 289 . • 17 Down. These 2 -element subsets will have intersections of size either 1 or 2 with Y . The number of ways to have a size- 1 intersection is to pick something from Y and something from X − Y , which is a total of 29 · 3 . Similarly, the number of ways to have a size- 2 intersection is to choose two things from Y , which is a total of 29 · 28 / 2 . Therefore, we get 493 possible subsets. • 18 Down. This may be a bit tricky to check with a calculator, so leave this out for now! At this stage, your crossword looks as follows. Note that we have assumed that you have not filled in the hard clues above, or even the ones which you could have afforded to skip. Figure 1: Halfway done! After this, just use the Sudoku structure (keep the extra symmetries in mind!) to fill in the rest of the blanks to obtain the final solution. 3 Page 3 of 4
Figure 2: The complete solution 4 Page 4 of 4