1/1/5. Alex is trying to oen lock whose code is sequence tht is three letters long, with ech of the letters being one of A, B or C, ossibly reeted. The lock hs three buttons, lbeled A, B nd C. When the most recent button-resses form the string, the lock oens. Wht is the minimum number of totl button resses Alex needs to be sure to oen the lock? The nswer is 9. This cn be chieved with the following sequence of resses: AAACCCBCCACBBCBACABCAABBBABAA There re = 7 different strings of three letters with ech letter being one of A, B, or C. All 7 of these strings er consecutively in the bove sequences of resses. Ech ress of the button corresonds to ttemting t most one more string, nmely the one formed by the revious three resses. No string cn be ttemted fter the first two resses. Therefore, the first time tht ll 7 strings cn be tried on the lock is fter 7 + = 9 resses of the button, so 9 is indeed the minimum.
/1/5. In the 5 6 grid shown, fill in ll of the grid cells with the digits 9 so tht the following conditions re stisfied: 1. Ech digit gets used exctly times.. No digit is greter thn the digit directly bove it.. In ny four cells tht form subgrid, the sum of the four digits must be multile of. 7 8 6 4 5 1 You do not need to rove tht your configurtion is the only one ossible; you merely need to find configurtion tht stisfies the constrints bove. (Note: In ny other USAMTS roblem, you need to rovide full roof. Only in this roblem is n nswer without justifiction ccetble.) Lbel the unknown cells with the vribles through v s shown to the b c d 7 e right. Using rule reetedly in the to two rows, we hve j k 4 l m b + 8 + c + g c + g + d + h d + h + 7 + i 7 + i + e + 6 (). 5 n o 1 q r s t u v The first equivlence bove gives us b +8 d+h (), nd the third equivlence gives us d + h e + 6 (). Thus b + 8 e + 6 (). Since 8 b 9 nd 6 e 9, we must hve tht (b, e) is either (8, 7) or (9, 8). Alying the technique bove reetedly gives us the following fourth rule: 4. If two columns re n even distnce rt, then ny djcent ir of cells in the first column hs the sme sum modulo s the corresonding djcent ir in the second column. Similrly, if two rows re n even distnce rt, then ny djcent ir of cells in the first row hs the sme sum modulo s the corresonding djcent ir in the second row. Using rule 4, we hve 5 + r 1 + u (). In rticulr, since u must be or 1, we must hve r be either or modulo. Now consider the sum of ll numbers of the bord. We hve three coies of every digit, so this sum is multile of. We lso know tht the six squres shown in the digrm to the right ll sum to multile of. Subtrcting these six squres out from the sum of the whole bord, we get tht 7 + e + + 4 + r + = 16 + e + r 1 + e + r (). b c d 7 e j k 4 l m 5 n o 1 q r s t u v But e must be 7 or 8 from bove, nd r must be or modulo, so the only ossibility is tht e = 8 nd r (). Hence (b, e) = (9, 8) nd lso u = 1.
Now the grid looks s shown. Next, note tht i () using the uer-right corner of the grid. So by rule 4 lied to the to two rows, we hve + f c + g 7 + i 1 (), 9 + 8 d + h 8 + 6 (). 9 c d 7 8 j k 4 l m 5 n o 1 q r or s t 1 v In rticulr, no column cn contin two 9 s, so ll three 9 s must be in the to row. Additionlly, by rule 4, c + d + 4 (mod ), so since one of c nd d must be 9, they both must be multiles of. Also is multile of since by rule 4, + 9 r + (). So, c, d, i, nd r re ll multiles of. At this oint, we hve shded the boxes tht we know must be multiles of. We know enough of the cells modulo for simle lictions of rules nd 4 to give us the vlue of ll of the cells modulo. For exmle, d + 7 + h + i + 1 + h + (), so h (). Reeting this round the grid gives us the following chrt: 9 7 8 8 1 6 1 4 5 1 1 1 1 1 9 c d 7 8 j k 4 l m 5 n o 1 q r s t 1 v We now just need to ly rules 1 nd reetedly to determine the exct vlues of ech cell. This is mostly cse of filling in from the to nd/or the bottom using rule, nd keeing trck of how mny times ech digit is used to ly rule 1. To strt, the only lce for the three s is the bottom row, nd the only lce for the remining 1 is the 4th column. The grid now looks s t right. The only numbers tht fit in the left column with the correct residues modulo re (, f, j) = (9, 7, 5), nd the only remining lce for the lst 8 is h. This mkes d = 9 nd c = 6. Then o = nd (q, v) = (5, ) re the only wys to lce the remining numbers tht re modulo, nd the rest of the grid esily fills to give us the finl nswer below. 9 9 6 9 7 8 7 8 4 8 6 5 7 4 6 5 4 1 1 5 1 9 c d 7 8 j k 4 l m 5 n o 1 1 q 1 v
/1/5. An infinite sequence of ositive rel numbers 1,,,... is clled territoril if for ll distinct ositive integers i, j with i < j, we hve i j 1. Cn we find territoril j sequence 1,,,... for which there exists rel number c with i < c for ll i? We will construct territoril sequence exists whose mximum vlue is. This llows us to choose c to be ny number greter thn. Consider the sequence ( i ) =, 1, 1,, 1 4, 4, 5 4, 7 4, 1 8, 8, 5 8, 7 8, 9 8, 11 8, 1 8, 15 8, 1 16, 16,.... Secificlly, 1 =, nd for ny ositive integer n, let k be the unique nonnegtive integer such tht k < n k+1, nd then n = (n k ) 1. Note tht ll the numbers in k the sequence re distinct. Suose we re given two ositive integers i, j with i < j. Let k be the unique nonnegtive integer such tht k < j k+1. Then j = t for some odd integer t. Since i k+1, we lso k hve i = u for some integer u with t u nd u not necessrily odd. Then k i j 1 > 1, k j s desired.
4/1/5. Bunbury the bunny is hoing on the ositive integers. First, he is told ositive integer n. Then Bunbury chooses ositive integers, d nd hos on ll of the sces, + d, + d,..., + 1d. However, Bunbury must mke these choices so tht the number of every sce tht he lnds on is less thn n nd reltively rime to n. A ositive integer n is clled bunny-unfriendly if, when given tht n, Bunbury is unble to find ositive integers, d tht llow him to erform the hos he wnts. Find the mximum bunny-unfriendly integer, or rove tht no such mximum exists. Let M be the roduct of ll rime numbers less thn 14. We clim tht the mximum bunny-unfriendly integer is 1M. Note tht 1 = 11 6 is not rime, so ll of the rime divisors of 1M re less thn 1. First, we verify tht 1M is bunny-unfriendly. Suose, for ske of contrdiction, tht ositive integers nd d could be chosen so tht +1d < 1M nd gcd(+kd, 1M) = 1 for ll k 1. If there is rime less thn 14 tht does not divide d, then d 1 exists modulo. Choose k < such tht k d 1 (mod ). Then +kd is multile of, nd gcd( + kd, 1M), contrdiction. Therefore no such exists, nd thus d must be multile of M. But then + 1d > 1d > 1M, lso contrdiction. Thus, Bunbury will not be ble to find n nd d to use for his hoing, nd hence 1M is bunny-unfriendly. Next, we verify tht ll numbers greter thn 1M re bunny-friendly. Let n be n integer greter thn 1M, nd let be the lrgest rime divisor of n. We brek into cses bsed on the roerties of. Cse 1: < 14, so tht ll the rime divisors of n re less thn 14. Let x be the roduct of n s distinct rime divisors. Let = 1 nd d = x, nd note tht we hve tht gcd(n, 1 + kx) = 1 for ll integers k. Furthermore, 1 + 1x < 14x, which is t most n becuse n > n > 1. Thus, this choice of nd d roves tht n is bunny-friendly. x M Cse : > 14. We first consider = 1 nd d = n. If d is still multile of, then gcd(n, 1 + kd) = 1 for ll integers k. Furthermore, 1 + 1d = 1 + 1n < n. So this choice of nd d roves tht n is bunny-friendly. But if d is not multile of, then there is unique b with b < such tht 1 + bd is multile of, nd hence gcd(n, 1 + bd) =. For ll other k < with k b, we hve gcd(n, 1 + kd) = 1. Thus, if b > 1, choosing = 1 nd d = n still roves tht n is bunny-friendly, by the rgument bove. If b < 14, then we cn try to strt our hoing fter the bd multile of d. Tht is, we let = 1 + (b + 1)d. This works rovided tht + 1d = 1 + (b + 14)d < n.
So it is sufficient to hve 1 + 47d < n. If > 48, then this will lwys be the cse, becuse then 1 + 47d < 1 + 47n < n. 48 So our remining cse is when d is not multile of nd 47. In this cse we hve to try something different. Let = +d. Note tht gcd(n, +(k+1)d) = 1 for ll k 1. Furthermore, + 14d = + 14n = ( n + 14 ) n, but is very smll (recll tht n > 1M is very lrge reltive to < 48); in rticulr n n >, so < 1, nd hence n ( 1 + 14d < + 14 ) n = 15 n < n. Thus this choice of nd d roves tht n is bunny-friendly. In ll cses, ny n > 1M is bunny-friendly. Thus, 1M is the lrgest bunny-unfriendly integer.
5/1/5. Niki nd Kyle ly tringle gme. Niki first drws ABC with re 1, nd Kyle icks oint X inside ABC. Niki then drws segments DG, EH, nd F I, ll through X, such tht D nd E re on BC, F nd G re on AC, nd H nd I re on AB. The ten oints must ll be distinct. Finlly, let S be the sum of the res of tringles DEX, F GX, nd HIX. Kyle erns S oints, nd Niki erns 1 S oints. If both lyers ly otimlly to mximize the mount of oints they get, who will win nd by how much? (We use the common nottion tht [P QR] denotes the re of tringle P QR.) We clim tht the gme will end with S = 1. In rticulr, for ny X tht Kyle chooses, we will show tht Niki cn choose her oints so tht S 1, nd if X is the centroid of ABC, then the best tht Niki cn do is S = 1. Note tht none of the rgument below deends on how tringle ABC is drwn, so how Niki drws ABC is irrelevnt. In the digrm t right, S is the sum of the shded res. Consider the sum of the res [AF I]+[BEH]+ [CDG]. Notice tht this sum counts ech white region once but ech shded region twice: for exmle, [HIX] is counted in both [AF I] nd [BEH]. Therefore, [AF I] + [BEH] + [CDG] = 1 + S, F G A X H I nd hence C E S = [AF I] + [BEH] + [CDG] 1. We will show tht for ny X Kyle chooses, Niki cn choose her oints such tht D B [AF I] + [BEH] + [CDG] 4, which imlies tht S 1. Let X be Kyle s chosen oint. Let, b, nd c be the lengths of BC, AC, nd AB, resectively, nd let h A, h B, nd h C be the distnces from X to BC, AC, nd AB, resectively. We will focus on Niki s choices of F nd I to minimize [AF I]. Let r = AI nd s = AF. On the one hnd, [AF I] = [AF X] + [AIX] = 1 (rh B + sh C ). On the other hnd, since [ACB] = 1, we hve F s A h b h c r X I [AF I] = [AF I] [ACB] = rs sin F AI bc sin CAB = rs bc. C B
Combining the bove two equtions, we hve Set x = [AF I]. By the AM-GM inequlity, [AF I] = 1 (rh B + sh C ) = rs bc. ( ) x = 1 (rh B + sh C ) rh b sh C. But rs = 1 bc(rh B + sh C ) = bcx by ( ), therefore Squring nd solving for x gives x bch B h C x. x bch B h C, with equlity if nd only if rh B = sh C. In the equlity cse, ( ) gives rh B = rs bc, so s = bch B, nd similrly r = bch C. Thus, Niki minimizes [AF I] by choosing F nd I so tht r = bch C nd s = bch B. She will ble to mke these choices rovided tht bh B 1 nd ch C 1. So, if Kyle icks X such tht h A, bh B, nd ch C re ll less thn 1, then Niki cn choose her oints so tht [AF I] + [BEH] + [CDG] = bch B h C + ch C h A + bh A h B. Let s disose of the contrry cse first: suose Kyle icks X such tht (without loss of generlity) h A 1, or equivlently h 1. Niki cn choose F nd I so tht F I CB. Then by similrity, nd using the fct tht the height from A to BC is, we hve F G A H X I ( [AF I] = h ) ( 1 ) = 1 4. Furthermore, Niki cn ick G nd H very close to A so tht [BEH] + [CDG] is s close to 1 s she wishes (in the extreme cse, G = H = A nd [BEH] + [CDG] = 1), in rticulr, Niki cn get [BEH] + [CDG] < 1. Thus, Niki cn chieve 1 [AF I] + [BEH] + [CDG] < 1 4 + 1 1 = 4. Now for the remining cses: let = h A, q = bh B, nd r = ch C, nd ssume tht Kyle icks X such tht, q, r re ll less thn 1. Then we hve [AF I] + [BEH] + [CDG] = qr + r + q. C E D B
But we lso hve the generl inequlity (qr + r + q) 1 ( + q + r), with equlity if nd only if = q = r. (This is corollry of the rerrngement inequlity q+qr+r +q +r, which itself follows from the inequlity ( q) +(q r) +(r ).) On the other hnd, so Niki s choices result in + q + r = [BCX] + [CAX] + [ABX] = [ABC] =, [AF I] + [BEH] + [CDG] = qr + r + q 1 ( + q + r) = 4, nd this inequlity is strict unless = q = r. Since Kyle wnts to choose X to mximize the mount of re tht Niki must choose, he wnts to force the inequlity to be n inequlity by choosing X such tht = q = r. This mens tht h A = bh B = ch C, which mens tht X is the centroid of ABC. In ll cses, Niki cn ick oints such tht S = [AF I] + [BEH] + [CDG] 1 1, nd Kyle cn force her into S = 1 by icking X to be the centroid of ABC. Thus, with otiml ly, Niki will score nd Kyle will score 1. c 1 Art of Problem Solving Foundtion