Cryptarithms are puzzles in which letters or symbols are substituted for the digits in an arithmetical calculation. Algebraic expressions might be regarded as cryptarithms of a sort, but algebra is not generally considered to be mathematically recreational. Cryptarithms have existed for centuries, and it is doubtful if it will ever be known when such puzzles were first devised. If a cryptarithm utilizes letters in place of the digits, and these letters form sensible words or phrases, the puzzle is termed an alphametic. J. A. H. Hunter coined the term in 1955.
S E N D M O R E --------- M O N E YDoes someone realy need a solution for this one?
T H I S I S A G R E A T T I M E ----------- W A S T E RRef: Unique solulion can be found in: AMM (1965) p316 E-1681
S A T U R N U R A N U S N E P T U N E P L U T O ------------- P L A N E T SAuthor: Willy Engren Ref: JoRM 12 (1979) 133-134
M A R S V E N U S S A T U R N U R A N U S ------------- N E P T U N EAuthor: Willy Engren Ref: JoRM 13 (1980/81) 293
H E A R T + E A R S + N O S E + T H R O A T = H E A L T HAuthor: Richard I. Hess Ref: JoRM 21 (1989) 61 Alpha 1680
H U R R A Y + H U Z Z A H = P U Z Z L E SAuthor: Richard I. Hess Ref: JoRM 21 (1989) 137 Alpha 1698
F A T H E R + M O T H E R = P A R E N TAuthor: David J. Porter Ref: JoRM 21:4 Alpha 1743
W I N T E R + I S + W I N D I E R + S U M M E R + I S = S U N N I E RAuthor: Brian Barwell Ref: JoRM 21:3 Alpha 1809
R E A S O N = (I T)(I S) + T H E R EAuthor: Robert B. Israel Ref: JoRM 21:3 Alpha 1731, (explained solution)
M A D * M A N = A S Y L U MAuthor: Robert B. Israel Ref: JoRM 21:4 Alpha 1751, (explained solution)
A L F R E D / E = N E U M A N A L F R E D * E = N E U M A NThese are two base-9 problems.
eve/did = .talktalktalk...There are 2 solutions here. One of them with a reduced fraction.
T H E S E T E A S E T I R E D ----------- R E A D E RAuthor: Hunter Ref: Joseph S. Madachy, Madachy's Mathematical Recreations, 1979, p185
A D A M A N D E V E --------- M O V E DRef: J. A. H. Hunter, Entertaining Mathematical Teasers, 1983, p93
S Q U A R E D A N C E ----------- D A N C E RThis problem has two solutions: one as an addition problem, the other as subtraction.
P I E R R E + E L L I O T T = T R U D E A URef: Dick Hess, Puzzles from around the world, 1997, Problem 6
A P P L E + G R A P E + P L U M = B A N A N ARef: Dick Hess, Puzzles from around the world, 1997, Problem 13
A P P L E + L E M O N = B A N A N AAuthor: Col. G. L. Sicherman
W E + W A N T + N O + N E W + A T O M I C = W E A P O NRef: Dick Hess, Puzzles from around the world, 1997, Problem 16
P E A N U T + T E E T H = C A R T E RSolve this alphametic in base 10 and base 9.
sqrt(ATOM) = A + TO + Mthis has two solutions
T E R R I B L E + N U M B E R = T H I R T E E NAuthor: Steven Kahan
Z E R O E S O N E S ----------- B I N A R YAuthor: Peter Macdonald, 1977
W R O N G W R O N G --------- R I G H Tthis has two solutions under the restriction of O = zero.
N O R A * L = A R O NRef: Martin Gardner, Mathematical Magic Show, 1977, Chap 15, Problem 2
A L P H A B E T L E T T E R S --------------- S C R A B B L ERef: G. Brandreth, The Puzzle Mountain, 1981, p61
N A I L O R L I N G E L ------------- F A C E O F FRef: rec.puzzles
A * (W E B T V) = D D D D D DRef: rec.puzzles
G A U S S R I E S E ----------- E U K L I DRef: Zweistein; 99 Logeleien von Zweistein, 1969, Problem 55
S I N ² + C O S ² = U N I T ERef: rec.puzzles
ALCOHOL + ALCOHOL + ... + ALCOHOL = HANGOVERhow many drinks do you need?
H A P P Y - T I G E R = Y E A RSolve this under the conditions that
B O B B Y + L O V E D = J E W E L SRef: Ravi Narula, Brain Teasers, 1976, Jaiko Publ. House, p97
H E * H E = S H EAuthor: M. E. Larsen
B A * C B A = D C B ARef: Walter Lietzmann, Lustiges und Merkwürdiges von Zahlen und Formen, 11.Ed, p192
T H I S * I S = A P H I SThis one has three solutions.
J G D C H I F A B E ----------- B I B D E BRef: Sam Loyd, Cyclopedia of Puzzles, 1914, p238
E A R T H + A I R + F I R E + W A T E R = N A T U R EAuthor: Herman Nijon
T W O * T W O = S Q U A R ERef: H. E. Dudeney, Strand 78 (1929) 91, 208
NIIHAU ± KAUAI ± OAHU ± MOLOKAI ± LANAI ± MAUI ± HAWAI = 0Ref: Donald E. Knuth; The Art of Computer Programming, Section 7.2.1.2 Exc. 26.
N O R T H / S O U T H = E A S T / W E S TRef: Nob Yoshigahara; JoRM 27 (1995) 137
A B C B A = D * B E * B F F AThe factors shall be prime numbers.
A H H A A H / J O K E = H ARef: Ch. W. Trigg, Mathematical Quickies, 1985, Problem 19.
F I V E F I V E N I N E E L E V E N ----------- T H I R T YRef: Ken Russell, MENSA Magazine Oct. 1996, p28
V I N G T C I N Q C I N Q ----------- T R E N T EAuthor: Alan Wayne
E I N E I N E I N E I N ------- V I E RAuthor: Alan Wayne
S E C H S S E C H S ----------- Z W O E L FRef: Zweisteins Zahlenlogeleien, Insel it 1510, Problem 48
U N O + U N O + T R E E = C I N C OThis one has several solutions. The Solution
S I X + S I X + S I X = N I N E + N I N EAuthor: Alan Wayne
F O R T Y + T E N + T E N = S I X T YRef: Longley-Cook, New Math Puzzle Book, 1970, p54
T W O * T W O = T H R E EAuthor: H.E. Dudeney in the July 1924 issue of Strand Magazine
T H R E E = T W O + O N E + Z E R ORef: Richard L. Breisch; Recreational Math. Magazine 12 (Dec. 1962) 24
U N I T E D S T A T E S ------------- A M E R I C ARef: L. A. Graham; The surprise Attack in Mathematical Problems, 1968, Problem 9
F A R E S = F E E ²Solve this alphametic in base 6.
K Y O T O K Y O T O K Y O T O --------- T O K Y OWhich number system has a solution? Author: V. Dubrovsky, A. Shvetsov
U V + U V = U W UAuthor: Harry L. Nelson
X + X + X = X XSolution: 3 X = b X + X. Thus 2 X = b X and as X not equal 0 we have b = 2. Now X = 1.
X Y + X X = X Y XSolution: The last digit shows Y = 0. Dropping the last digit we again get 2 X = b X.
A B - B A = ARef: J. Lehmann, Mathe mit Pfiff, Manz Verlag 1977, p71 (P15: AB-BA=A)
X X X X X ----- * X X X X X X X X --------- X X X X XIn this remarkable cryptarithm, each digit is a prime (2, 3, 5 or 7). No letters or digits are provided as clues; but there is only one solution.
X X X * X X = X X X X X
x x x x 2 x ----- * x x x x x x x x 8 x ----------- x x 9 x 2 x
x x x x x ----- * x x x x x 4 --------- x x x 0 1
A T O M * A T O M = x x x x A T O MRef: Sphinx - Nov 1933 #167 - Page 167 (by M. Pigeolet) saying TOCK TOCK = xxxxTOCK.
H E * H E = S H E
O E E E E ------- * E O E E E O E ------- O O E EAll even digits are replaced with E and all odd with O. What was the multiplication?
E E O O O --------- * E O E O E O O --------- O O O O OAll even digits are replaced with E and all odd with O. What was the multiplication?
E O E E E O E ----------- * O O E E O E O E E E O O E ----------- E O O O E EAll even digits are replaced with E and all odd with O. What was the multiplication?
? ? ? ? ? ? 7 ? --------- * ? ? ? ? ? ? ? ? ? ? ? ? ? --------------- ? ? ? ? ? ? ? ?Only digit 7 is printed.
? ? 3 ? ? 3 ----- * 3 ? ? ? 3 ? ? ? 3 --------- ? ? ? ? ?Only digit 3 is printed.
T H I S I S --------- * x x T O O H A R D x ----------- x x x x x xRef: Joseph S. Madachy, Mathematics on Vacation, 1966, p192
P U Z Z L E x x x x x x x ------------- * x x x x x x x x W x x x x x x x O x x x x x W O R L D x x x x x L x x x x x x x D x x x x x x x x ------------------------- x P x U x Z x Z x L x E xRef: Puzzle World, Ed. Nob Yoshigahara and Richard Bozulich, Ishi Press, Summer 1992, Pilot Edition.
E S T M O D U S I N R E B U S ----------------- * x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x ----------------------------- x x x x x x x x I N R E B U SEach of the ten digits corresponds to at least one letter, so there is precisely one case where two different letters are replaced by the same digit. Fred. Schuh admits that the puzzle is far from easy.
x x x ) x x x x x x ( x x x x 0 x x ------- x x x x x 5 0 x ------- x x x x 4 x =====Ref: American Mathematical Monthly (1921) p37
x x ) x x x x 3 x ( x x x x.x x x x --- 4 x x x x x ----- x x x x --- x x x x --- x x x x x x =====Ref: Zweistein; 88 neue Logeleien von Zweistein, 1983, Problem 64
x x x ) 5 x x x x x x x x ( x x x x x x x x x x ----- x x x x x x x ----- 5 x x x x x ----- x 5 x x x x x x =======Ref: Zweistein; 88 neue Logeleien von Zweistein, 1983, Problem 19
x x x x 7 x ) x x 7 x x x x x x x ( x x 7 x x x x x x x x ----------- x x x x x 7 x x x x x x x x ------------- x 7 x x x x x 7 x x x x ----------- x x x x x x x x x x x 7 x x ------------- x x x x x x x x x x x x ===========Author: W. E. H. Berwick
x x x x 7 x x ) x x x x x x 7 x x x x x x x x x ( x x x 7 x x x x x x x x x x x x x ------------- x x x x x x x x x x x x x x x x --------------- x x x x x x x x x x x x x x x --------------- x x x x x x x x x x x x x x ------------- x x x x x x 7 x x x x 7 x x x --------------- x x x x x x x x x x x x x x x x --------------- x x x x x x x x x x x x x x x --------------- x 7 x x x x x x x 7 x x x x ------------- x x x x x x x x x x x x x x ------------- xRef: Fred. Schuh, The Master Book of Mathematical Recreations, Dover Publ. 1968, p318-320
x x x ) x x x x x x x x ( x 7 x x x x x x x ------- x x x x x x ----- x x x x x x x ------- x x x x x x x x =======Ref: American Mathematical Monthly 49 (1932) 489
x x x ) x x x x x x x x ( x x 8 x x x x x ------- x x x x x x x ------- x x x x x x x x =======Author: P. L. Chessin
x x x ) x x x x x x x x ( x x x x x x x x ------- x x x x x x x ------- x x x x 8 x x x =======Ref: W. Engel (Ed.), Mathematische Olympiade-Aufgaben, 1979, A.2.42
x x ) x x x x x x x x x x ( x x x x x 8 x x x x x ----- x x x x x x ----- x x x x x x ----- x x x x --- x x x x x x =====Ref: B. A. Kordemsky, The Moscow Puzzles, 1972, Problem 272 f
8 x x x ) 8 8 x x x x x x ( x x x x x x x x x --------- x x x x x x x x x x --------- x 8 x x x 8 x x x 8 --------- x x x x x x x x 8 8 =========Ref: Zeitschrift Archimedes, vor 1950
x x ) x x x x ( x x.x x x x x --- x x x x x ----- x x x x --- x x x x x x ----- x x x x ===Ref: Joseph S. Madachy, Mathematics on Vacation, 1966, p192
x x ) x x ( x.x x x x x x --- x x x x x x ----- x x x x x x ----- x x x x ===Ref: Zweistein, Logeleien für Kenner, 1974, Problem 7
x x x ) x x x x x x ( x x x x.x x x x x x x ----- x x x x x x ----- x x x x x x ----- x x x x x x ----- x x x x x x x x =======Author: A. Corrigan in Strand Magazine
x x x ) x x x x x x x x ( x x x x x x x x x ----- x x x x x x x ------- x x x x x x x ------- x x x x x x x x =======Determine the unique divisor.
x x x ) x x x x x x x x x ( x x x x x x x x x ------- x x x x x x x ------- x x x x x x ----- x x x x x x x x ======= x x ) x x x x x x ( x x x x x x x --- x x x x x ----- x x x x x x ----- x x x x x x =====The six-digit quotient of the first must be equal the dividend of the second.
x x ) x x x x 0 x ( x x x x x x ----- x x x x x 1 ----- x x 2 x ===Ref: L. A. Graham; The surprise Attack in Mathematical Problems, 1968, Problem 9
x x ) x x x x 0 x ( x x x x x x ----- x x x x x 1 ----- x x 3 x ===Ref: Ch. W. Trigg, Mathematical Quickies, 1985, Problem 173
x x + x = xAuthor: Jaime Poniachik
X X X + X X X = X X XPut 1-9 once each to make the equation honest. Ref: Recreational Mathematics Magazine, No. 7 (Feb. 1962) 35-36
X X X --- + --- + --- = 1 X X X X X XPut 1-9 once each to make the equation honest. Ref: REC 8:5 (1993) 5-6 (Nob)
X X X --- * X X X X --- + X XPut 1-9 once each to make the equation honest.
X X X X X = k * X X X X XPut 0-9 once each to make the equation honest for k=7 and for k=9.
k : 2 3 4 5 6 7 8 9 s(k) : 48 6 8 12 0 1 16 3Ref: Angela Fox Dunn, Mathematical Bafflers, 1980, p99 (k=9)
X X X X X = k * X X X XPut 1-9 once each to make the equation honest for k=2..9.
X X X X X X X / X X X = 1984Put 0-9 once each to make the equation honest.
x x x x x x ----- * x x x x x x x x x --------- x x x x xNo digit occurs more than twice.
X O X O X * O X O X O = X O X O X O X O X OAll of the X's are different, as are all of the O's, as are all digits in the product (whose final O is 0).
B A C K E B A C K E ----------- K U C H E NRef: Zweisteins Zahlenlogeleien, Insel it 1510, Problem 47
V A T E R M U T T E R ----------- E L T E R NRef: J. Lehmann, Mathe mit Herz, Urania (1991), p82, problem 22
L A G E L E M G O ----------- S C H O E NAuthor: Ingo Althöfer
L A G E L I P P E ----------- S C H O E NAuthor: Torsten Sillke (but the Patriot is Udo Sprute)
A T L A N T I S = S C H L A M M + S C H L I C KStern Nr. 13 (1996)
Z W E I + B L I T Z = L I E B EStern Nr. 17 (1996)
Z W E I + B L I T Z = E R O T I KStern Nr. 18 (1996)
H A M M E L + H A E H N E = S T R E I TStern Nr. 18 (1996)
M U S I K + K U N S T = G E N U S SStern Nr. 35 (1996) (several solutions)
A T L A N T A = R U M M E L + P L E I T E NStern Nr. 37 (1996)
S U C H E N - M A C H T = S P A S S
N A M E N = S C H A L L - R A U C HTwo solutions but "schall" is unique.
F I N D E + D I E S E = K I N D E R
M A C H * M A L = L A U T E R
K E I N E + A N G S T = S I G R I DRef: Zeit Magazin, Nr 47, 15. Nov. 1996, p51, Logelei
I N A * I S T = S U E S S
V I E L E N : D A N K = K A I V E E V --------- A V U E A L K O --------- I N U N I N U D ------- K
D E I N E + B E I N E = S U P E RThis alphametic has two solutions but the sum remains equal. What number is SUPER?
N I E M A N D : Z U = H A U S E N I H ------- I M A I N S ------- Z I N U D M ------- N H D N H D =====
K U S S + L U S T = O L A L ARef: Zeit Magazin, Nr 23, 30. Mai 1997, p42, Logelei
D I E * Z E I T = R A E T S E L
D E N K E R + G E R N E = L O G E L NRef: Zeit Nr 20, 12. Mai 1999, Logelei
S E K T * B I E R ----------------- T B T S R B E R A S S E K T I ------------- S A E U F E RRef: Zeit Magazin, Nr 23, 30. Mai 1997, p42, Logelei
S C H W I M M T : E I N = F I S C H S N E I --------- I C I I E I N --------- C E T M C H F T --------- C M W M H C S C --------- H C C T N S C W ------- E M SRef: Zeit Magazin, Nr 35, 22. Aug. 1997, p42, Logelei
U N S E R + O S C A R = S C H L A U
K L U G E + L E U T E = L O G E L NRef: Zeit Magazin, Nr 35, 22. Aug. 1997, p42, Logelei
J E D E R + L I E B T = B E R L I NThis alphametic has two solutions, but EIN BIER is unique.
W A I G E L + S P A R E = E I F R I GRef: Zeit Magazin, Nr 3, 10. Jan. 1997, Logelei
W E R + I S T + Z W E I = S T E I NThis alphametic has two solutions,
+ C E G E F e c + C c B A D B c + C E G A F D c --------------- c c d d B f f COne frequency is false and must be corrected!
ABC * DA = BEFA + * - GHHA + GJE = GFDC ------------------ GJEJ + HJJJ = FFDCRef: M Sändig, Hobbymatik, 1984, Titelpage
ABCD : EFG = HC - * + FHA - IG = FJD ------------------ AHBE - ACFE = FAJRef: Hör Zu, 17/1972
ABC * DE = CFGH + * - JDHJ + DGC = JGKK ------------------ JEDK + EBAH = FAGHRef: Hör Zu, 38/2001
AB * CDC = BABE * + - AFD + GHJE = GGDC ------------------ HFKH + GJEF = KDAHRef: Hör Zu, 10/1995
EGA * AH = DGFG * + - LL + EFFC = EFDL ------------------ ABCD + EFGH = KLBKRef: Hör Zu, 00/1995
ABAC : DEF = GB - + * HEGF - HBID = HED ------------------ EKBI - HKGI = ICDCRef: Hör Zu, 45/2002
ABBC : DDE = DF + + * AGFG - AHIB = KKG ------------------ KGEH + ACKE = FCIIRef: Hör Zu, 29/2003
ABCD + DCEF = GHFI - : + AHGE + AIC = AKGF ------------------ KEG * KI = EBKBRef: Hör Zu, 31/2003
ABCD + EBFG = FDBH : - + AE + EBII = EBAE ------------------ AHC * FG = KFGKRef: Hör Zu, 37/2003
ABCD - EABF = GDGA : - - HII + EGKD = EFCK ------------------ IH * EDH = HABFRef: Hör Zu, 51/2003
ABCD + EFCF = BGHE - : + AAIF + GDE = AEDC ------------------ FCH * GK = KHHIRef: Hör Zu, 01/2004
ABBC : DE = FGD - + * FBHI - FGKD = FD ------------------ EHGI - FBCG = CCEKRef: Hör Zu, 05/2004
ABCD + EFGG = HEBG : - + CI + EHKI = EFCH ------------------ CKB * GH = IKFDRef: Hör Zu, 07/2004
ABCD : EE = EFD + + * GHEH - GIKC = IF ------------------ HHDI + GICF = DBCARef: Hör Zu, 53/2004
ABCD + EFGH = KLBK : - + LL + EFFC = EFDL ------------------ EGA * AH = DGFGRobert Israel solved it with a Maple Program.
ABCDE - FGEE = AHBDE : - - DI * HDB = BKDB -------------------- CDE + FKCD = KDGDRef: Lehmann
ATU + IAS = IITE - - : NEG : IOG = E ----------------- PAU - NS = PPARef: Boris A. Kordemsky, Köpfchen muss man haben, 1975, P241
HEB + EUS = LFS - - : WUB : EH = US ----------------- EHB - ESU = RARef: Boris A. Kordemsky, Köpfchen muss man haben, 1975, P241
ABC * DC = EAEF * + - BF + AEGG = AEBF ------------------ FFEE + AEDC = DBHGRef: Torsten Sillke
AB * CD = EFCB * + - EC + GBB = GEC ------------------ EBDB + GCD = EDBDRef: Torsten Sillke
ABC * BB = ADBC * + - E + ACC = ACE ------------------ FGEC + ABB = ABCBRef: Torsten Sillke
ABC : C = DE - * + AE + AF = DD ------------------ BGA - FHC = IJRef: J. Petigk, Mathematik in der Freizeit, 1998, p151
A + B = CD + + + CE + F = CF --------------- CA + CC = DERef: J. Petigk, Mathematik in der Freizeit, 1998, p159
ABC - DBB = EFG + + + ABB - ECC = BBB ----------------- DBEB - HBB = IBCRef: Jürgen Köller, Symbolrätsel 2
AB * BC = DAD * : * EC : F = F ----------------- BCD * G = BBEARef: Jürgen Köller, Symbolrätsel 4
HJ + AD = DF + + + BG + EC = DF ---------------- AC + GE = HEKRef: Trend-Nüsse 1981, p52
ISE + EID = ALK - + - RAS - KL = RUM ---------------- ERR + MDR = LLMRef: Zeit Magazin 19/1972
ABC - DEF = EG + - : HIE : DAH = H --------------- FBD : HB = DFRef: Eckstein, Symbolrätsel, www.eckstein.de
AB + C = AC + + + AD + E = FB ---------------- FD + G = ECRef: W. N. Bolchowitinow, B. I. Koltowoi, I. K. Lagowski; Spass für freie Stunden, 1980, problem 194 A
AB * A = CCD - * + AE : F = G ---------------- B * CF = CHERef: J. Lehmann, Mathe mit Pfiff, Manz Verlag 1977, p72, problem 25
E * K = BE + * : B : N = N ---------------- DN + DN = NBRef: Walter Lietzmann, Lustiges und Merkwürdiges von Zahlen und Formen, 11.Ed, p192
AB * C = CD - + : E : B = F ---------------- F + G = HRef: Walter Lietzmann, Lustiges und Merkwürdiges von Zahlen und Formen, 11.Ed, p192
AB * C = DEB + + - AC * B = FBB ---------------- GE - FE = CERef: Walter Lietzmann, Lustiges und Merkwürdiges von Zahlen und Formen, 11.Ed, p192
AB * B = CDD : * : CB : B = E ---------------- E * D = BDRef: Walter Lietzmann, Lustiges und Merkwürdiges von Zahlen und Formen, 11.Ed, p192
AB * C = DE * + - F * F = AC ---------------- GH + AI = CHRef: Walter Lietzmann, Lustiges und Merkwürdiges von Zahlen und Formen, 11.Ed, p192
ABB : C = DB * + - E * A = C ----------------- FC : GA = BRef: Walter Lietzmann, Lustiges und Merkwürdiges von Zahlen und Formen, 11.Ed, p192
ABC : CD = EF - * + FDF + EG = AHD ----------------- BB + FGA = AIJRef: Bernhard Berchtold, www.mathematik.ch/puzzle/, Puzzle 9
A - B = C * D : E = F = G + H = IThe Solution
VV RE HR AH VE RL RV HE VRThis is a magic square with magic constant
HEH
.
If you order the letters according there values you get the hidden word.
HG GH UF LI U IF I EU GG UH NU UG IH FF EI FH HF NI RU IG RI LU FG HH GFThis is a magic square with magic constant
RHH
.
The 25 numbers form an arithmetic progression.
If you order the letters according there values you get the hidden word.
The problem - What is the value of D ?
BALLET 45 POLKA 59 CELLO 43 QUARTET 50 CONCERT 74 SAXOPHONE 134 FLUTE 30 SCALE 51 FUGUE 50 SOLO 37 GLEE 66 SONG 61 JAZZ 58 SOPRANO 82 LYRE 47 THEME 72 OBOE 53 VIOLIN 100 OPERA 65 WALTZ 34A problem form 'Tough Puzzles'. Solution
161 + 134 + 145 = 503Ref: Geoffrey Mott-Smith, Mathematical Puzzles, 1954, Problem 32
R O M E - S U M = R U S ERef: Stephen Barr, Mathematical Brain Benders 2nd Miscellany of Puzzles, 1969, Problem 22
What SENSE does it make, if nine HENS give seven EGGS?Author: J.A.H. Hunter
W E ) G E T ( I T W E --- C U T x x x =====Author: J.A.H. Hunter
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 10Replace the 'x' by '+', '-', '*', '/'. No operator occurrs more than twice. Evaluate left to right.
Home | FSP Mathematisierung | Fakultät für Mathematik | Universität Bielefeld
Last Update: 2005-12-24