A probability problem by César Castro
Probability
Level
3
In an infinite succession of decimal digits 1999851 ..., each digit, starting from the fifth, is the digit of the units of the sum of the four that precede it.
Do the four digits 1, 8, 9, 8 appear sequentially in succession?
No
Yes
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1 solution
La respuesta es no. Basta demostrar que si la sucesión se divide en bloques de 5 dígitos consecutivos cada uno, entonces en todos los bloques los 4 dígitos iniciales son impares y el quinto es par. Esto es verdadero para el primer bloque 19998. Supongamos que la afirmación se verifica para el k-ésimo bloque abcde, es decir, sean a, b, c, d dígitos impares mientras que el dígito e es par. Si el (k+1)-ésimo bloque es a'b'c'd'e', entonces a' es el último dígito de b+c+d+e, que es una suma de tres sumandos impares y uno par. De modo que a' es impar, y ahora el mismo argumento muestra sucesivamente que b', c' y d' también son impares. Luego a'+b'+c'+d' es par, y en consecuencia su último dígito, e', es impar. Esto demuestra, por inducción, la afirmación hecha al comienzo. En consecuencia no puede haber cuatro dígitos consecutivos iguales a 1, 8, 9, 8.
The answer is no. It suffices to show that if the sequence is divided into blocks of 5 consecutive digits each, then in all the blocks the initial 4 digits are odd and the fifth is even. This is true for the first block 19998. Suppose the statement is verified for the kth block abcde, that is, they are a, b, c, d odd digits while the digit e is even. If the (k + 1) -th block is a'b'c'd'e ', then a' is the last digit of b + c + d + e, which is a sum of three odd addends and one even. So a 'is odd, and now the same argument shows successively that b', c 'and d' are also odd. Then a '+ b' + c '+ d' is even, and consequently its last digit, e ', is odd. This shows, by induction, the statement made at the beginning. Consequently there can not be four consecutive digits equal to 1, 8, 9, 8.