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Solve the following morning math problems to solve the Math Message:

$7 \times (60 + 103)$
$219 \times 2 \times 3$
$2 \times 3 \times 10^2$
$\sqrt{40000}$
$23 \times 5 $
$16^2 + 26$
$\frac{1000}{2} - \frac{24}{3}$
$(150-7) \times12$
$33\times50$
$\frac{500}{2} + 70$
$\frac{453}{25} \times 10^{-1} \times 500$

Hint 1:

Your math may not be finished when you answer these first questions.

Hint 2:

Each answer will be divided by a number. Remember what number. (1 is a number)

Hint 3:

Each problem corresponds to a single character.

Hint 4:

You should find 40 twice.

Hint 5:

Aside from the two 40's, your numbers will lie between 114 and 166

Hint 6:

You will only see each divisor once.

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  • 2
    $\begingroup$ 1141 1314 600 200 115 282 492 1716 1650 320 906... are you sure this isn't just your math homework? :P $\endgroup$ – Joe Z. Mar 13 '15 at 18:33
  • 3
    $\begingroup$ Oh how I wish I had had homework like this in any of the last however many years of schooling. $\endgroup$ – Aggie Kidd Mar 13 '15 at 18:35
  • $\begingroup$ Of course. This is fifth-grade stuff. And yet I still don't see a pattern... $\endgroup$ – Joe Z. Mar 13 '15 at 18:35
  • $\begingroup$ It is more in depth than that. You've taken the first step in a series of steps that need to be taken. Now to find the hidden message within. I'll post clues as time goes on if no one gets it. $\endgroup$ – Aggie Kidd Mar 13 '15 at 18:37
  • $\begingroup$ I suspected as much. $\endgroup$ – Joe Z. Mar 13 '15 at 18:39
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The solution is

"Math is fun"

The explanation:

We order the factorizations of the eleven numbers by increasing smaller factor, and we interpret the larger factors as octal numbers. The corresponding Ascii symbols then spell out "math is fun" (with 40oct=space etc.)

  • $115= 1*115$; and 115=M
  • $282= 2*141$; and 141=a
  • $492= 3*164$; and 164=t
  • $600= 4*150$; and 150=h
  • $200= 5* 40$; and 40=space
  • $906= 6*151$; and 151=i
  • $1141= 7*163$; and 163=s
  • $320= 8* 40$; and 40=space
  • $1314= 9*146$; and 146=f
  • $1650=10*165$; and 165=u
  • $1716=11*156$; and 156=n
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  • $\begingroup$ Very nicely done! $\endgroup$ – Aggie Kidd Mar 18 '15 at 17:07
  • 1
    $\begingroup$ I did a lot of the work for this (everything except the rearranging); it would've been nice to be mentioned in your answer! Never mind; +1 anyway. $\endgroup$ – Rand al'Thor Mar 19 '15 at 1:02
  • 1
    $\begingroup$ But computing the factorizations is straightforward. My little nephew can also do that, and computer programs can also do that. $\endgroup$ – Alexis Mar 19 '15 at 15:47
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Partial solution:

As already answered in the comments, the equations solve to the following numbers (prime factors),(all factors up to 50):

$1141 (7,163),(7)$

$1314 (2,3,3,73),(2,3,6,9,18)$

$600 (2,2,2,3,5,5),(2,3,4,5,6,8,10,12,15,20,24,25,30,40,50)$

$200 (2,2,2,5,5),(2,4,5,8,10,20,25,40,50)$

$115 (5,23),(5,23)$

$282 (2,3,47),(2,3,6,47)$

$492 (2,2,3,41),(2,3,4,6,12,41)$

$1716 (2,2,3,11,13),(2,3,4,6,11,12,13,22,26,33,39,44)$

$1650 (2,3,5,5,11),(2,3,5,6,10,11,15,22,25,30,33,50)$

$320 (2,2,2,2,2,2,5),(2,4,5,8,10,16,20,32,40)$

$906 (2,3,151),(2,3,6)$

No number is a factor of all the answers, and 40 is a factor of 3 of the numbers.

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  • $\begingroup$ Don't take the numbers down quite so far. You should divide each number by a single factor to get a pair of numbers that will be used to find the solution. $\endgroup$ – Aggie Kidd Mar 16 '15 at 16:33
  • $\begingroup$ @AggieKidd I am quite stuck. I think I should wait for further hints :P (Just wondering, does ASCII have anything to do with the answer?) $\endgroup$ – March Ho Mar 16 '15 at 16:39
  • $\begingroup$ Perhaps. I can't give away too much. $\endgroup$ – Aggie Kidd Mar 16 '15 at 17:24
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$7∗(60+103) = 1141$

$219∗2∗3 = 1314$

$2∗3∗102 = 600$

$\sqrt{40000} = 200$

$23∗5 = 115$

$162+26 = 282$

$\frac{1000}{2}−\frac{24}{3} = 492$

$(150−7)∗12 = 1716$

$33∗50 = 1650$

$\frac{500}{2}+70 = 320$

$\frac{453}{25}∗10^{−1}∗500 = 916$

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  • 1
    $\begingroup$ +1 for having a go, but Joe Z already got this far in the comments 3 days ago! $\endgroup$ – Rand al'Thor Mar 17 '15 at 10:54
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@JoeZ has already done the first calculations, to get the following results:

$1141, 1314, 600, 200, 115, 282, 492, 1716, 1650, 320, 906$

According to the OP's hints, we should now

divide each of these numbers by another number to get results two of which are 40 and the rest of which are between 114 and 166.

Assuming all numbers involved are integers, the only possibilities are:

  • $1141/7=163$
  • $1314/9=146$
  • $600/4=150$ (or $600/5=120$ or $600/15=40$)
  • $200/5=40$
  • $115/1=115$
  • $282/2=141$
  • $492/4=123$ or $492/3=164$
  • $1716/13=132$ or $1716/12=143$ or $1716/11=156$
  • $1650/10=165$ or $1650/11=150$
  • $320/8=40$ (or $320/2=160$)
  • $906/6=151$

So our new list of numbers is:

$163, 146, 150, 40, 115, 141, 123 \mathrm{ or } 164, 132 \mathrm{ or } 143 \mathrm{ or } 156, 165 \mathrm{ or } 150, 40, 151$

Using the octal ASCII codes from here, this becomes:

s, f, h, [space], M, a, S or t, Z or c or n, u or h, [space], i

... which still doesn't make sense. According to the OP's second hint though, we should remember the numbers we divided by ($7,9,4,5,1,2,4 \mathrm{ or } 3, 13 \mathrm{ or } 12 \mathrm{ or } 11, 10 \mathrm{ or } 11, 8,6$). Maybe Caesar-shifting the letters we have by these numbers of places in the alphabet could be the key? Unfortunately that gives zol, lwd, or ncv for the first three letters (I didn't go any further since none of these make sense), according to whether we add or subtract.

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  • $\begingroup$ You're off on one, but definately almost there! $\endgroup$ – Aggie Kidd Mar 16 '15 at 22:14
  • 1
    $\begingroup$ @AggieKidd Since it was important that I was off on one (I missed 600/4), that one must be the actual thing we're intended to do in that clue! Which means 600/5 and 600/15 are irrelevant. Which means the second 40 must be 320/8, and 320/2 is irrelevant. You gave me lots of info in that comment! ;-) $\endgroup$ – Rand al'Thor Mar 16 '15 at 22:29
  • $\begingroup$ Figured that could stand as the final clue :) $\endgroup$ – Aggie Kidd Mar 16 '15 at 22:51

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