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let be a number with k-cipher $ number= \sum_{n=0}^{k}a_{n}10^{n} $

this number satisfies

$$ \sum_{n=0}^{k}a_{n}10^{n} = 2\prod_{n=0}^{k}a_{n} $$

the number is equal to double the product of its ciphers

for k=2 is easy the number is 36 however for fixed positive 'k' can this only be tested by brute force ?? , i mean w must test all the EVEN numbers with k ciphers

Rasmus
  • 18,404
Jose Garcia
  • 8,506

1 Answers1

6

Hint:$$a_k10^{k}\leq\sum_{n=0}^{k}a_{n}10^{n} = 2\prod_{n=0}^{k}a_{n}\leq2a_k\cdot 9^{k},$$ hence $10^k\leq2\cdot9^k,k\leq 6$. Even you do a brute force , it will not take much time.

lsr314
  • 15,806