Prove it Too 2!
Let be the set of the digits of . Find the smallest natural number such that is a proper subset of .
Clarification
:
A set
is a
proper subset
of another set
if all elements of
is in
, but
.
The answer is 235.
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1 solution
If n is a 2-digit number, n^2 must be composed of only 1 digit as S(n^2) is the proper subset to S(n), so it can't have 2 members as S(n). However, those beautiful numbers, such as 111 or 1111, can not be a perfect square, so n must be at least a 3-digit number.
By multiplying the same number, we know what the last digit of n^2 would be from looking at the last digit of n itself. That is, for example, if n has the last digit of 1, 1 times 1 equals 1, so n^2 will have 1 has the same last digit. Other digits of choice would be 5 and 6, where n^2 will end with 5 and 6 respectively.
Starting from n over 100-199, we should compose a number with 3 different digits to maximize the elements in S(n). Therefore, we can't use 1 as the last digit as it will repeat with 1 on the first place, and after trying 1X5 and 1X6 numbers, no S(n^2) is a proper subset to S(n).
Moving on to 200-299, we resume the same methods with 1 as a free choice now, and we will soon realize that S(235) = {2, 3, 5} and S(235^2) = S(55225) = {2, 5}. So S(n^2) is a proper subset to S(n), and the smallest natural n is, therefore, 235.