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Solve for the number t and the digit represented by a.

[3(230 + t)]2 = 492,a04.

a is 8 by process of elimination to find a number that was a perfect square, and therefore t=4

:D

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a is 8 by process of elimination to find a number that was a perfect square, and therefore t=4
:D

Is there a simpler path to a?

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Is there a simpler path to a?

For a natural number N, the digits of N^2 add up to either 1, 4, 7, or 9. - all squares except 7, which is prime. Hmmmm

I just guessed 8 at the first pass, so I went with it.

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Is there a simpler path to a?

are there any restrictions on t? it seems that t can be parametrized by a.

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For a natural number N, the digits of N^2 add up to either 1, 4, 7, or 9. - all squares except 7, which is prime. Hmmmm

I just guessed 8 at the first pass, so I went with it.

the "3" inside the square brackets?

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Is there a simpler path to a?

Sure. Figure out t first!

Let x2 = 492,a04.

x must end in 02, for x2 to end in -04.

x2 is roughly half a million, or (sqrt(1/2)*1000)2, or about 7002. So x is 702. The rest follows.

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are there any restrictions on t? it seems that t can be parametrized by a.

Yes.

Once a is found, you know [can find] t.

The elegant path to a is being discussed. B))

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Sure. Figure out t first!

Let x2 = 492,a04.

x must end in 02, for x2 to end in -04.

x2 is roughly half a million, or (sqrt(1/2)*1000)2, or about 7002. So x is 702. The rest follows.

Yeah, that's cool.

I like the other way [the one about the 3] but to each his own. :)

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the "3" inside the square brackets?

The number on the right must be divisible by 9, and hence 3. This means the digits add up to a number divisible by 3 - which mean they must add up to 9.

This does happen, although with 2 iterations, first they sum to 27, then to 9.

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Yeah, that's cool.

I like the other way [the one about the 3] but to each his own. :)

Right. Now I see...

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The number on the right must be divisible by 9, and hence 3. This means the digits add up to a number divisible by 3 - which mean they must add up to 9.

This does happen, although with 2 iterations, first they sum to 27, then to 9.

Multiples of 9 also have digits adding up to a multiple of 9, so you can stop when you get 27.

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Yes.

Once a is found, you know [can find] t.

The elegant path to a is being discussed. B))

well, what i was getting was

if t is allowed to be complex, then the quadratic formula produces an equation t(a) depending on a, where a is a base 10 digit. so there will be several solutions for t (possibly 18 unique ones). for instance, t(1) and its conjugate, etc. in fact, by this (admittedly unelegant but elementary) approach, one obtains at least two unique solutions to the problem.

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While there is just one solution for a (a=8), there are two possible solutions for t.

t=4

t=-464

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Solve for the number t and the digit represented by a.

[3(230 + t)]2 = 492,a04.

[3(230 + t)]2 = 492,a04 mean

9x(230 + t)2 = 492,a04.

that mean 492,a04 is able to be divided by 9, and hence

4+9+2+a+0+4 = multiplier of 9

19+a = multiplier of 9 (which the nearest is 27)

a = 8!

so 492,804 / 9 = 54756

(230 + t)2 = 54756

(230 + t) = 234

t = 4

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