MAGIC SQUARE: Calculate A-B*C
[3769] MAGIC SQUARE: Calculate A-B*C - The aim is to place the some numbers from the list (3, 4, 5, 7, 8, 9, 55, 56, 57, 58, 85) into the empty squares and squares marked with A, B an C. Sum of each row and column should be equal. All the numbers of the magic square must be different. Find values for A, B, and C. Solution is A-B*C. - #brainteasers #math #magicsquare - Correct Answers: 27 - The first user who solved this task is Eugenio G. F. de Kereki
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MAGIC SQUARE: Calculate A-B*C

The aim is to place the some numbers from the list (3, 4, 5, 7, 8, 9, 55, 56, 57, 58, 85) into the empty squares and squares marked with A, B an C. Sum of each row and column should be equal. All the numbers of the magic square must be different. Find values for A, B, and C. Solution is A-B*C.
Correct answers: 27
The first user who solved this task is Eugenio G. F. de Kereki.
#brainteasers #math #magicsquare
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Gerd Faltings

Born 28 Jul 1954.German mathematician who wasawardedthe 1986 Fields Medal (the highest honour that a young mathematician can receive) primarily for his proof of the Mordell Conjecture which he achieved using methods of arithmetic algebraic geometry. He has also been closely linked with the work leading to the final proof of Fermat's Last Theorem by Andrew Wiles. In 1983 Faltings proved that for every n > 2 there are at most a finite number of coprime integers x, y, z with xn + yn = zn. This was a major step but a proof that the finite number was 0 in all cases did not seem likely to follow by extending Falting's arguments. However, Faltings was the natural person that Wiles turned to when he wanted an opinion on the correctness of his repair of his proof of Fermat's Last Theorem in 1994.
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