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1 FERURY When 1 is divided by 5, the reminder is. nother wy to sy this is opyright 015 The Ntionl ouncil of Techers of Mthemtics, Inc. ll rights reserved. This mteril my not be copied or distributed electroniclly or in ny other formt without written permission from NTM. Theme: on t get md. Lern mod. Solve for ll integer vlues of n in which 3n (mod 5) nd 0 n 0. 4 Wht is the gretest negtive number tht is congruent to 9 mod 3? Suppose tht is equilterl, is right ngle, m = 30, nd = 10. Find the perimeter of qudrilterl (mod 5), red s 1 is congruent to modulo 5. Find if 37 (mod 8). Find the positive reminder when 17 is divided by 3. The multiplictive inverse (reciprocl) of number p is 1/p becuse the product p (1/p) equls 1, the identity element. Modulr rithmetic hs no frctions, but in some cses we cn still hve inverses. For emple, 3 5 = 15 1 (mod 7), so 3 nd 5 re multiplictive inverses modulo 7. Find n inverse of 3 modulo 17. Eplin why the number 3 does not hve multiplictive inverse in rithmetic modulo 6. How mny different reminders re possible when ny positive integer is divided by 6? How mny positive integers less thn 100 re congruent to 3 modulo 7? 1 3 Find the reminder when 13 0 is divided by 11. For four given numbers, the sum of the first two numbers is p, the sum of the second nd third numbers is q, nd the sum of the lst two numbers is r. Find n epression in terms of p, q, nd r for the sum of the first nd lst numbers E Three congruent circles ech contin the centers of the other two circles. The circles, two t time, intersect t three dditionl points. Find the rtio of the re of the tringle formed by these three dditionl points compred with the re of the tringle formed by the three centers F of the circles For given numbers p nd q, three of the following four epressions hve the sme vlue: p + q, p q, p /q, nd p q Find the bsolute vlue of the fourth epression. Solve for given tht nd b re two distinct positive numbers nd tht b log =. Find the lest positive integer congruent to modulo 7.

2 Solve for : 3 3 = 3 1 How mny positive integers less thn 10 stisfy the eqution (mod 4)? Point P is the center of the circle inscribed in with sides 6, 8, nd 10. Find the product of the three lengths: P P P Fctor 4 + 4y 4 into two polynomil epressions. Find ll positive integers m such tht 11 5 (mod m) Four circles re drwn, one centered t ech verte of squre. The circles ll contin the center of the squre. Find the rtio of the sum of the res of the overlpping portions of the circles to the sum of the res of the four circles. Prove tht in every right tringle with integrl side lengths, the product of the side lengths is multiple of 60. Pencils cost 16 ech. Sm bought n pencils nd pid with ectly d dollr bills, receiving 4 chnge. Wht is the mimum number of pencils he could hve purchsed if n 100? The Interntionl Stndrd ook Number (ISN) is unique numeric commercil book identifier. The tenth digit of ISN-10 is check digit. To find it, multiply the first digit by 10, the second by 9, the third by 8, nd so on until the ninth digit is multiplied by. The tenth digit dded to the sum of these nine products must result in 0 modulo 11. Find the check The ISN-13 check digit is found by multiplying ech of the first twelve digits by 1 nd 3 lterntely. The sum of these twelve products plus the check digit must be 0 modulo 10. Find the check digit for the ISN-13 number c. 1 digit, c, for c. 3 squre nd n equilterl tringle re drwn so tht the center of the squre lies on the tringle nd the incenter of the tringle lies on the squre. If the side of the squre mesures 1, wht is the mimum re of the tringle? In problem 6, wht is the minimum re of the tringle? Solve n 4 n+ 1 3 n/ n = Ntionl ouncil of Techers of Mthemtics, 1906 ssocition rive, Reston, V

3 SOLUTIONS to clendr Looking for more lendr problems? Visit /clendr/defult.sp?journl_id= for collection of previously published problems sortble by topic from Mthemtics Techer. The Editoril Pnel of Mthemtics Techer is considering sets of problems submitted by individuls, clsses of prospective techers, nd mthemtics clubs for publiction in the monthly lendr. Send problems to the lendr editors. Remember to include complete solution for ech problem submitted. eprtment editors Mrgret offey, Thoms Jefferson High School for Science nd Technology, lendri, V 31; nd rt Klish, irector of the Institute of MERIT t SUNY ollege t Old Westbury The lrgest multiple of 8 tht is less thn 37 is 3, nd 37 3 = The reminder must be n integer in the set {0, 1,, 3, 4, 5} The smllest number to stisfy the condition is 3, nd the complete set my be written {3, 10, 17,, 94}. Ech number is of the form 3 + 7n nd must stisfy the inequlity n < 100. Thus, 0 7n < 97, implying tht 0 n < , 9, 14, 19. Test vlues to eclude 0, 1,, nd 3 (they do not mke the sentence true). Substitute 4 for n in the epression 3n nd then divide the result by 5 to find reminder of. ll numbers tht stisfy the congruence reltion must therefore be of the form 4 + 5k. Since 0 n 0, we hve k 0, resulting in 0 k 3. Thus, n = 4, 9, 14, nd ecuse the divisor, d, is 3, the only possible reminders, r, re 0, 1, or. We use the eqution n = qd + r, with n = 17 nd d = 3, to see tht q = 6 nd r = 1. (Tht is, 17 = ) The question sks for solution to the sttement 13 0 n (mod 11), where 0 n < 11. ecuse 13 is congruent to modulo 11, consider powers of. (onvince yourself, perhps using induction, tht 13 k k (mod 11), so tht we cn work with smller bse.) onsider 0 = ( 4 ) 5 = 16 5, written with smller eponent. ut (mod 11). Then, 5 5 = nd 5 3 (mod 11), so we hve = 45 nd, finlly, 45 1 (mod 11). lternte solution 1: Use 0 = ( 10 ) = 104. y pplying the divisibility rule for 11, we see tht 103 is multiple of 11. Therefore, (mod 11). lternte solution : 0 = ( 5 ) 4 = 3 4. Then 3 4 ( 1) 4 1 (mod 11). lternte solution 3: 13 0 = Since the bsolute vlue of the difference of the sum of the odd-positioned digits minus the sum of the even-positioned digits equls = 1, the reminder when dividing by 11 is p q + r. Let the four given numbers be, b, c, nd d. Then + b = p, b + c = q, nd c + d = r. Subtrct the second eqution from the first to get c = p q. Then dd to this result the third eqution to find tht + d = p q + r. lternte solution: The given informtion cn be written in mtri form s b c d = p q r. We need to write the dditionl row vector (1, 0, 0, 1) s liner combintion of the rows of the coefficient mtri. The first row plus the lst row minus the middle row gives the desired vector, so p + r q will equl + d. Tht is, b c d = p q r The numbers congruent to 9 modulo 3 cn be written s 9 + 3n. Using n = 1, we find the gretest negtive number to be ecuse 3 6 (mod 17), we re. 44 MTHEMTIS TEHER Vol. 108, No. 6 Februry 015 opyright 014 The Ntionl ouncil of Techers of Mthemtics, Inc. ll rights reserved. This mteril my not be copied or distributed electroniclly or in ny other formt without written permission from NTM.

4 looking for number n such tht n 6 1 (mod 17). Use n = 3, which results in 3 6 = 18 1 (mod 17). lternte solution 1: Solve n 3 1 (mod 17) by tril nd error. lternte solution : onsider the congruence clss {1, 18, 35, 5, 69, 86, }. Which number is divisible by 3? The nswer is 69, the result of multiplying 3 by : 1. Let 1 be the lengths of the sides of the tringle formed by the centers of the circles:, E, nd F. The other three points,, nd form the endpoints of the dimeters of the circles nd re length. Since the tringles re similr nd the rtio of the sides is : 1, the rtio of their res is 4 : /10. Tke the log of both sides of the eqution nd pply the rules for logs: F log b log log = log log log = logb log loglog logblog = logb log logb log log = = 1 log 1 = 10 = 1/10 1. <soln 0 1> In tringle, the side opposite the 30 ngle is hlf the hypotenuse. Thus, equls 5, s do nd. The missing side length,, equls 513, so the perimeter is = 013. E ( ) ( ) 13. The options to consider in rithmetic modulo 6 re 0, 1,, 3, 4, nd 5. The products of ech of these nd the number 3 re 0, 3, 6, 9, 1, nd 15, respectively, equivlent to 0, 3, 0, 3, 0, nd 3 modulo 6, respectively. The number 1 is never result. theorem concerning modulr inverses sttes tht the only time p hs n inverse modulo m is when p nd m re reltively prime. This mens tht GF(m, p) = 1. When m nd p re reltively prime, integers nd b eist such tht m + bp = /. If p + q = p q, then q = 0, which is not vlid option becuse q is the denomintor of one of the epressions. Therefore, we must omit either p + q or p q s one of the three equivlent epressions. It must be tht p q = p/q, so q = 1, giving q = ± 1. If q = 1, then either p + 1 = p or p 1 = p, neither of which is possible. Therefore, q = 1. If we omit p + q, so tht p q = p q, we see tht p ( 1) = p( 1) p = 1/. Under these conditions nmely, tht q = 1 nd p = 1/ we hve p q = p/q = p q = 1/ nd p + q = ( 1) + ( 1/) = 3/. similr rgument follows if we omit p q from the three other epressions. Under this condition, in which q = 1 nd p = 1/, we found tht p + q = p/q = p q = 1/. This condition leds to the solution p q = 1/ ( 1) = 3/, the sme nswer When 014 is divided by 7 the reminder is 5. Thus, (mod 7). Evlute ( ) 014 mod 7 by tking dvntge of the simplifiction ( ) (mod 7). Thus, ( ) ( ) ( ) ( ) (mod7) , 3/. The solution = 0 my be pprent by inspection. Rerrnging ( 3 3) = 3( ), we find tht 0 = ( 3), so the second solution is t = 3/. 17. Four. Subtrct 3 from both sides of the given sttement to get 4 0 (mod 4). Thus, = 0 + 4n is solution. ut, = + 4n is lso solution. Notice tht subtrcting 3 from members of the congruence sttement is vlid step but tht dividing by is not. The four positive solutions less thn 10 re, 4, 6, nd Since the tringle side lengths re 6, 8, nd 10, the tringle is right tringle. rw the three segments from P to ech of the vertices. rop perpendiculr segments from P to the three sides of the tringle. Let the lengths of the segments be r. The sum of the res P + P + P equls the re of, so r + r + r = 6 8. Tht is, ( )r = 48 4r = 48 r =. P is the digonl of squre QPR, so its length is 1. We cn find the other two lengths by pplying the Pythgoren theorem: + 4 = P, so P = ; nd + 6 = P so P = 340 = 310. The product is = 80. Q R P S 19. ( + y y)( + y + y). The sum of perfect squres does not fctor over the rel numbers, but the difference does. Thus, we mke the given polynomil difference of perfect squres s follows: y + 4y 4 4 y = ( + y ) 4 y = ( + y y)( + y + y) 0. 1,, 3, 6. Since 11 5 mod m, 11/m = q + 5/m, where q is the integrl quotient. Multiply by m to get 11 = mq + 5 mq = 6. Therefore, the only positive vlues for m re the fctors of 6: 1,, 3, nd ll primitive triples cn be found using pirs of positive integers, m nd n, where m nd n re reltively prime nd ectly one of them is even. For m > n, the formuls m n, mn, nd m + n give ll primitive triples. To show tht the product of these three is lwys multiple of 60, show tht t lest one of them is divisible by 3, one by 4, nd one by 5. Since one of m or n is even, then mn 0 (mod 4). If m 0 (mod 3) or n 0 (mod 3), then mn 0 (mod 3). If neither m nor n is multiple of 3, then m ±1 (mod 3) nd n ± 1 (mod 3). Thus, m 1 (mod 3) nd n 1 (mod 3), Vol. 108, No. 6 Februry 015 MTHEMTIS TEHER 443

5 so m n 0 (mod 3). In similr fshion, if m 0 (mod 5) or n 0 (mod 5), then mn 0 (mod 5). If m is not multiple of 5, then m is congruent to ±1 or ± modulo 5; thus, m 1 (mod 5) or m 4 (mod 5). Equivlently, m ±1 (mod 5). Similrly, n ±1 (mod 5). When m nd n re congruent modulo 5 (both re either 1 or 1), their difference, m n, is congruent to 0 modulo 5. When one is 1 nd the other 1, the sum, m + n, is congruent to 0 modulo 5. Therefore, every Pythgoren triple includes numbers tht re multiples of 3, 4, nd 5, nd the product is multiple of The tenth digit is the dditive inverse of the sum mod 11. Multiply nd dd the following: 10(0) + 9(9) + 8(7) + 7(7) + 6(3) + 5(0) + 4(4) + 3(5) + (6) = 47. Then we need to find the vlue of c tht stisfies 47 + c 0 (mod 11). Since 47 5 (mod 11), use c = Multiply ech of the twelve known digits by 1 nd 3 lterntely nd then dd the products: 1(9) + 3(7) + 1(8) + 3(0) + 1(9) + 3(7) + 1(7) + 3(3) + 1(0) + 3(4) + 1(5) + 3(8) = 15. The check digit, c, must be dded to this result to get multiple of 10, so c = (p )/(p). Since there re four congruent circles nd four identicl overlpping regions, the rtio of the sum of the res is the sme s the rtio of pir of res. Let the rdius of ech circle be 1; then its re is p. n overlpping region, seen shded in the digrm, hs n re determined by finding the re of sector OP, subtrcting the re of OP, nd then doubling tht result. Since PO is 90, the re of the overlpping region is (p/4 1/) = p/ 1. Thus, the rtio we seek is (p/ 1)/p = (p )/(p) Since ech pencil costs 16, the totl number of cents spent is 16n. Sm used d dollr bills nd got bck 4, so he spent 100d 4 cents. Set these epressions equl nd solve for integer solutions: 100d 4 = 16n. Possible solutions cn be found using spredsheet, but modulr rithmetic gives nother pproch. First, simplify the eqution to 5d 1 = 4n. It follows tht 5d 1 4n (mod 5), which reduces to 4 4n (mod 5), giving n = 6 s potentil solution. (When n = 6, we hve d = 1.) Since we wnt the mimum number of pencils less thn or equl to 100, find numbers equivlent to 6 modulo 5. These re = 31, 56, nd 81. The lrgest possible number is 81. Reders might consider vrition of this pproch using the slope of the line whose eqution is (5 1)/4 = y The side length, s, of the tringle determines the re of the tringle since = s 13/4. The mimum ltitude,, of the tringle mimizes the side of the tringle since s = 13/3. side of the tringle must contin the center of the squre, K, nd the tringle s ltitude to tht side must contin the incenter of the tringle. Thus, to mimize, we wnt the incenter of the tringle to be s fr from K s possible. This occurs when the incenter is t corner of the squre,. The side length of the squre is 1, so the distnce K is 61. ut K is 1/3 of the ltitude. Thus, the ltitude is 181, nd side of the tringle is 116. The re of the tringle is ( ) = s 3/4= 1 6 3/4 = K R of the tringle. We do so by plcing verte of the tringle t the center of the squre nd plcing the incenter of the tringle t midpoint of the side of the squre. This distnce is 6, which is /3 of the ltitude of the tringle. The ltitude of the tringle is 9 so side is 613, nd the re is R T ( 6 3) 3/4= n = 5. The right side equls 9. Use the rules of eponents to simplify the left side of the eqution. The numertor is = n 4 1/ n 1/4 5n 13 /4 = ( + ) = ( ), 1/ 1/ n 4 n+ 1 n 4 1 n+ 1 nd the denomintor is Then S ( ) 3 n/ ( n )/ n 1 =. (5n 13)/4 ( 4n 4)/4 9 =. Thus, (9n 9)/4 = 9, nd n = 5. P S O T To minimize the re of the tringle, minimize the distnce from the center of the squre to the incenter 444 MTHEMTIS TEHER Vol. 108, No. 6 Februry 015

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