bonanova 84 Posted December 12, 2008 Report Share Posted December 12, 2008 Solve for the number t and the digit represented by a. [3(230 + t)]^{2} = 492,a04. Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 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 Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 Solve for the number t and the digit represented by a. [3(230 + t)]^{2} = 492,a04. t = 4 a = 8 Quote Link to post Share on other sites

0 bonanova 84 Posted December 12, 2008 Author Report Share Posted December 12, 2008 a is 8 by process of elimination to find a number that was a perfect square, and therefore t=4 Is there a simpler path to a? Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 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. Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 Is there a simpler path to a? are there any restrictions on t? it seems that t can be parametrized by a. Quote Link to post Share on other sites

0 bonanova 84 Posted December 12, 2008 Author Report Share Posted December 12, 2008 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? Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 Is there a simpler path to a? Sure. Figure out t first! Let x^{2} = 492,a04. x must end in 02, for x^{2} to end in -04. x^{2} is roughly half a million, or (sqrt(1/2)*1000)^{2}, or about 700^{2}. So x is 702. The rest follows. Quote Link to post Share on other sites

0 bonanova 84 Posted December 12, 2008 Author Report Share Posted December 12, 2008 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. Quote Link to post Share on other sites

0 bonanova 84 Posted December 12, 2008 Author Report Share Posted December 12, 2008 Sure. Figure out t first! Let x^{2} = 492,a04. x must end in 02, for x^{2} to end in -04. x^{2} is roughly half a million, or (sqrt(1/2)*1000)^{2}, or about 700^{2}. 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. Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 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. Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 Yeah, that's cool. I like the other way [the one about the 3] but to each his own. Right. Now I see... Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 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. Quote Link to post Share on other sites

0 Guest Posted December 12, 2008 Report Share Posted December 12, 2008 Yes. Once a is found, you know [can find] t. The elegant path to a is being discussed. 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. Quote Link to post Share on other sites

0 Prime 15 Posted December 13, 2008 Report Share Posted December 13, 2008 While there is just one solution for a (a=8), there are two possible solutions for t. t=4 t=-464 Quote Link to post Share on other sites

0 Guest Posted December 13, 2008 Report Share Posted December 13, 2008 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 Quote Link to post Share on other sites

## Question

## bonanova 84

Solve for the number

and the digit represented byt.a[3(230 +

)]t^{2}= 492,04.a## Link to post

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