7
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*) SQUARE makes 4 SQUARES.
**) CUBE makes 2 SQUARES.

Add, and almost a square I am.

What am I?

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  • 1
    $\begingroup$ it seems like that this puzzle comes from elsewhere. please provide the source. unattributed work may get downvoted or even closed. happy puzzling ;) $\endgroup$ – Omega Krypton Nov 20 '19 at 11:51
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    $\begingroup$ No this is honestly my own creation. I might have been influenced by naming it "what am I". But that's it. $\endgroup$ – Prim3numbah Nov 20 '19 at 11:57
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    $\begingroup$ Ok then, thanks for clarifying! $\endgroup$ – Omega Krypton Nov 20 '19 at 12:01
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You are 24 because ii) 2^3=2^2+2^2 and i) 4^2=2^2+2^2+2^2+2^2 16+8=24 which is -1 of 25 and so nearly 5^2

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    $\begingroup$ But 4^3=64 while 5^2 (25) + 7^2 (49) = 74, so the two are not equal...? Also, how would that part relate to the overall answer being 145? $\endgroup$ – Stiv Nov 20 '19 at 18:25
  • $\begingroup$ Oops, my mental math skills suck $\endgroup$ – Deepthinker101 Nov 20 '19 at 18:34
  • $\begingroup$ Changed it I think my math is cool now $\endgroup$ – Deepthinker101 Nov 20 '19 at 18:39
3
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Based on a hunch I had in a comment on another answer to this question, I wonder whether its solution is that:

* = 8 and ) = 0, i.e. the numbers which share a key with these symbols on the top row of a standard keyboard. (There is, however, one proviso to this...)

*) SQUARE makes 4 SQUARES.

Substituting the digits for the symbols, *) is 80. 80 squared is 6400, whose digits hide 4 square numbers: 64, 4, 400 and 6400 itself (the squares of 8, 2, 20 and 80, respectively).

**) CUBE makes 2 SQUARES.

Performing the same substitutions, **) is 880. 880 cubed is 681472000. However, here comes the proviso... If the OP has miscounted the '2 SQUARES' this number's digits actually hide THREE square numbers: 81, 1 and 4 (the squares of 9, 1 and 2, respectively).

Add, and almost a square I am.

Adding our two numbers, 80 and 880, together we get 960 - this is one short of 961, the square of 31. Hence their sum is 'almost a square'.

Obviously, this answer relies on an error by the OP, but the rest of the solution fits so well it makes me wonder if this may indeed have been the intended answer... Happy to be corrected if my inference is wrong!

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  • $\begingroup$ :D Forgot about this one, and you should too because this one was so bad I even forgot the answer(created it that exact day and remember it was a very bad riddle and haven't thought about since then)! But nice job with the explanations :) $\endgroup$ – Prim3numbah Jan 10 at 17:54
  • $\begingroup$ *) and **) was just 1 and 2 $\endgroup$ – Prim3numbah Jan 10 at 17:57
  • $\begingroup$ @Prim3numbah Well, I very nearly found another way to make it work then! $\endgroup$ – Stiv Jan 10 at 18:23
0
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You are

2

Your square

is 4, $4*1^2$

Your cube

is 8, $2*2^2$

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  • $\begingroup$ I feel like there must be something else to it. This same trick works if you replace 2 with any 2n^2. Also, the last line doesn't seem to fit. $\endgroup$ – hexomino Nov 20 '19 at 12:48
  • $\begingroup$ I'm thinking that '*' and ')' are placeholders for specific digits. However, the obvious choice (8 and 0 - their keyboard shift-equivalents) don't work out. A scan of all 4-digit squares for numbers where each digit is a square number also leads nowhere... $\endgroup$ – Stiv Nov 20 '19 at 14:50
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4^2+4^2+4^2+1^2=7^2

[(sqrt3)/2]^2+(1/2)^2=1^3

49+1=50 which is almost a square

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