Can a 6-digit number that contains all the digits 1 to 6 in random order be a perfect square?
Number Theory
Level
2
Can a 6-digit number that contains all the digits 1 to 6 in random order be a perfect square?
yes
no
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1 solution
There are 6! = 720 possible numbers using all the digits from 1 to 6, but none of these can be a perfect square. All of these numbers are divisible by 3 (1+2+3+4+5+6=21, 21 is divisible by 3), but not divisible by 9 (21 is not divisible by 9). For a perfect square there must be 2 equal factors, but there can only be one factor divisible by 3, so two equal factors are not possible.