enter image description here Please help me in solving this algebraic puzzle.
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Hint: from the formula $(100T+X)(100T+Y) = (10000T^2+200T(X+Y)+XY)$, you should be able to see that there is only one possible value of $T$. – Connor Harris Mar 21 '17 at 18:53
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$ 100< TWO < 141$ go through these 40 possibilities ... there are only $6$ possibilities that end ** when you square them ... more hints on request. – Donald Splutterwit Mar 21 '17 at 19:00
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@ConnorHarris (100T+10W+O)(100T+10W+O) , right ??? – surya krishna Mar 21 '17 at 19:06
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@suryakrishna 100T+X or 100T+10W+O ... it doesn't matter ... just focus on what happens with the T! – Bram28 Mar 21 '17 at 19:09
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@Bram28 , how come T can take only 1 value. it can take any value from 1-9, right ? – surya krishna Mar 21 '17 at 19:12
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@suryakrishna No, the result of the multiplication TWO * TWO is THREE, so that starts with the T. .. and has only 4 digits after that. There is only one possible value of T for that to happen. Just try a few values to see for yourself, e.g. 389*389 = 151321 ... too many digits ... so you already know T has to be pretty small ... and how can the first digit end up being the same? Again, there is only 1 possibility for that to happen! – Bram28 Mar 21 '17 at 19:18
2 Answers
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$(100T^2 + 10 W + O)^2 = 10,000 T^2 + 2000 TW + 100 (W^2 + 2TO) + 20 (WO) + O^2$
$T^2 = T; T = 1.$
$2W < 10$
$W < 5$ and not $1, W \in \{2,3,4\}.$ And $4$ is quite doubtful as it will be so close to be rounding up.
$O \notin \{6,5,0\}$ or $O^2$ would end with $O$
$O \in\{2,3,4,7,8\}$
If the smallest $O$ can be is $2$, looking back at $W, 4$ no longer works.
We are down to $10$ possible values of $TWO.$ Only one of which ends with repeated digits.
Doug M
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HINT
First figure out the value of T; there is only one possibility just focusing on the T's
After that, there are very few possibilities of W, and also limited possibilities of O (e.g. O can't be 5 or 6 ... and knowing T, that'll rule out more). With that, just try the remaining few options
Bram28
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