Peter picked an arbitrary positive integer, multiplied it by 5, multiplied the result by 5, then multiplied the result by 5 again and so on. Is it true that from some moment all the numbers that Peter obtains contain 5 in their decimal representation?
Problem
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Tags: number theory
kznil96
20.06.2022 16:29
Yes, given any positive integer, suppose the largest power of 2 which divides it is m, and the largest power of 5 which divides it is n, where both m and n are non-negative integers. Upon multiplying the positive integer with 5 for at least m+1 times, we will have an odd integer which is divisible by 5, followed by n trailing zeros. Therefore, there is at least one 5 in the decimal representation after multiplying our initial number by 5 for m+1 times.
Lasitha_Jayasinghe
20.06.2022 17:13
Let $N=2^a\cdot 5^b\cdot x$, even if it is not so at the beginning, when we keep on multiplying by 5, the power of 5 in the prime factorization will at some point exceed the power of 2, then $N=10^k \cdot x \cdot 5^l$ for some $k,l$ for all the number received from that point onwards.
Thus, since all odd multiple of 5 end in 5, the number will be of the form, $$N=D_1D_2\cdots D_y50\cdots0$$with a set of trailing zeros. Thus, the number will always contain a 5 in its decimal representation from that point onwards.
sanyalarnab
21.06.2022 08:29
The construction : If Peter picks a number $A$ then we will say that $A.5^k$ satisfies the second condition. It's quite trivial to see that it holds true
Matherer9654
30.07.2022 11:34
If the number chosen is even then it will still have a 5 in the decimal representation eventually, because it will be multiplied by 5 many times