second smallers integer?
2011-11-01
What's the second smallest integer with a units digit of 5, which is the square of an integer, and the cube of an integer?
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Let's call the number we are looking for `n`.
we know it's the square of an integer, so:
`n = "int"_a^2`
and we know it's the cube of an integer, so:
`n = "int"_b^3`
since any integer can be prime-factorized, let's consider the prime factorization of `"int"_1` and `"int"_2`:
`"int"_a = "prime"_1^("int"_(a1)) * "prime"_2 ^("int"_(a2)) ...`
`"int"_b = "prime"_1^("int"_(b1)) * "prime"_2 ^("int"_(b2)) ...`
`n = ("int"_a)^2 = ("prime"_1^("int"_(a1)) * "prime"_2^("int"_(a2))...)^2 = "prime"_1^(2*"int"_(a1)) * "prime"_2^(2*"int"_(a2))...`
and
`n = "int"_b^3 = ("prime"_1^("int"_(b1)) * "prime"_2^("int"_(b2))...)^3 = "prime"_1^(3*"int"_(b1)) * "prime"_2^(3*"int"_(b2))...`
so `"prime"_1^(2*"int"_(a1))*"prime"_2^(2*"int"_(a2))... = "prime"_1^(3*"int"_(b1))*"prime"_2^(3*"int"_(b2))...`
and so we can say:
`"prime"_x^(2*"int"_(ax)) = "prime"_x^(3*"int"_(bx))` for all x.
and therefore: `2*"int"_ax = 3*"int"_bx` this is the number of `"prime"_x` in the prime factorization of `n`.
since it has a factor of `2` and `3`, then the number of `"prime"_x` in `n` must be a multiple of 6.
to summarize, for any prime factor of `n`,there are a multiple of 6 of them in `n`.
we also know that the units digit of `n` is `5`, which means it has a factor of `5`, so:
`n = 5 * "int"_3`
so `n` has at least one `5` in it's prime factorization, and consequently it must have at least 6.
so `5^6` is the least integer that fits the description, but we are looking for the 2nd smallest.
well `5^12` would fit the description, and it's not the smallest, but maybe it's not the 2nd smallest,
`5^6 * 2^6` is worth considering, but `5^6 * 2^6 = 10^6`, which is one million, which doesn't have a `5` as the units digit. This does not fit the description in the problem.
`5^6 * 3^6` fits the description and is the second smallest integer that does.
`n = 5^6 * 3^6 = 11390625`