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How many perfect squares can be formed using 20 1's, 20 2's and 20 3's. This is a recent exam question, which I had no clue how to solve?

There is some kind of trick here, since time allotted to solve it was just 4 minutes.

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All $60$ digits? None. –  Ivan Loh May 14 '13 at 2:19
    
That's what I thought too, Ivan, but apparently from looking at the answers the question means "formed by addition using ...". –  Doug McClean May 14 '13 at 13:35
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3 Answers 3

up vote 19 down vote accepted

Hint: The resulting number is divisible by $3$ but not by $9$, since the digit sum is $120$.

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Thanks! Looks too simple now. But still I couldn't have figured it out in the exam. –  Uma kant May 14 '13 at 5:00
    
Well, maybe you could now. It looks as if the criterion for divisibility by $3$, by $9$ is part of the curriculum. Ideas tend to repeat themselves on exams. –  André Nicolas May 14 '13 at 5:07
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How about this: $$\begin {array} &1&2&3&1&2&3&1&2&3&1&2&3&1&2&3&1\\ 2&&&&&&&&&&&&&&&2\\ 3&&&&&&&&&&&&&&&3\\ 1&&&&&&&&&&&&&&&1\\ 2&&&&&&&&&&&&&&&2\\ 3&&&&&&&&&&&&&&&3\\ 1&&&&&&&&&&&&&&&1\\ 2&&&&&&&&&&&&&&&2\\ 3&&&&&&&&&&&&&&&3\\ 1&&&&&&&&&&&&&&&1\\ 2&&&&&&&&&&&&&&&2\\ 3&&&&&&&&&&&&&&&3\\ 1&&&&&&&&&&&&&&&1\\ 2&&&&&&&&&&&&&&&2\\ 3&&&&&&&&&&&&&&&3\\ 1&2&3&1&2&3&1&2&3&1&2&3&1&2&3&1\\ \end {array}$$
If done in a properly spaced font, it looks like a perfect square to me. The sum along each side is the same

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\begin{array}{ll}
    20 *1 &= 20 \\
    20 *2 &= 40 \\
    20 *3 &= 60 \\
    ---------------
 \text{total} &= 120
 \end{array}

Now, Perfect square(s)... $1+4+9+16+25+36+49 = 140$ ( I can remove $2$ number with total of $20$ ($4+16$))

that will lead to

$1+9+25+36+49$ (i.e. $5$ numbers)

Answer: $5$ perfect squares can be formed..

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