5
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The following are tricks with numbers that you must explain how they work.

Trick #1

  • Think of a number between 1 and 9
  • Multiply by 9
  • Add up all of the digits of this number
  • Divide by 3
  • Subtract 2

Your answer? 1 of course!

Trick #2

  • Think of any positive number!
  • Double it.
  • Add 2.
  • Half it.
  • Add 3.
  • Subtract your original number.

Your answer is obvious, it's 4

Trick #3

  • Pick any positive 3 digit number in the universe.
  • Multiply by 7.
  • Multiply by 11.
  • Multiply by 13.

Your answer is your original number twice :O

Trick #4

  • Think of a number between 1 and 9.
  • Double it.
  • Add 5.
  • Multiply by 5.
  • Add another digit between 1 and 9 to it.
  • Subtract 25.

The first digit of your new number was your original number. The second is your second number.

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  • 15
    $\begingroup$ Trick #1 keeps giving me infinity because I keep thinking of pi. $\endgroup$ – candied_orange Oct 28 '14 at 5:12
10
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  1. Digits of any number divisible by 9 always add up to 9.
  2. It's actually always 4. Steps 1 and 2 bring the total to n+1, then you add 3 and subtract n, leaving you with 4.
  3. The numbers are actually 7, 11, and 13, because multiplying these is 1001. A three digit number times 1001 is the same as taking the number with 3 0's after, then add the number.
  4. Steps 1 through 3 give us 2n+5, then 4, 5, and 6 give us 10n + 1m.
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  • $\begingroup$ I'll quickly fix #3. I meant 11 after I took out my calculator, but I must have forgotten to fix. $\endgroup$ – warspyking Oct 28 '14 at 0:33
  • $\begingroup$ Digits of any number divisible by 9 always add up to 9 that's not true unless you say you should add up digits of any number that results infinitely till you reach a single digit, which is not specified in the question - it just says "Add up all of the digits of this number" so for 9999189999 the addition gives me 81 so my result for step 1 would be 81/3 = 27 -2 = 25 $\endgroup$ – skv Oct 28 '14 at 9:21
  • $\begingroup$ @skv It's not specified in the question, but it's still the right answer. $\endgroup$ – mdc32 Oct 28 '14 at 12:05
  • $\begingroup$ @mdc32 yes indeed :) it was a comment on the clarity of the question more than your answer $\endgroup$ – skv Oct 28 '14 at 12:17
  • $\begingroup$ It might be noteworthy to mention that for #3, 7, 11, and 13 are prime numbers above the point where you can easily check a number to see if it's a factor. (it's easy to check if a number is a multiple of 2, 3, or 5) so it's slightly less obvious that 1001 will turn out to be their multiple. Maybe I'm reading too much into it, but that might help to explain the design of that particular puzzle. $\endgroup$ – Kingrames Sep 11 '15 at 12:10
12
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Trick 1

  • Think of a number between 1 and 9
  • Multiply by 9
  • Add up all of the digits of this number
  • Divide by 3
  • Subtract 2

$$\begin{align} x_1 & \in\left[1,9\right] \\ x_2 & = 9 \times x_1 \implies x_2 \in \left\{9,18,27,36,45,54,63,72,81\right\} \\ x_3 & \in \left\{9,1+8,2+7,3+6,4+5,5+4,6+3,7+2,8+1\right\} \implies x_3 = 9 \\ x_4 & = \tfrac{9}{3} = 3 \\ x_5 & = 3-2 = 1 \end{align}$$

Your answer? 1 of course!


Trick 2

  • Think of any positive number!
  • Double it.
  • Add 2.
  • Half it.
  • Add 3.
  • Subtract your original number.

$$\begin{align} x_1 & \in \mathbb Z_{\ge 0} \\ x_2 & = 2x_1 \\ x_3 & = 2x_1 + 2 \\ x_4 & = x_1 + 1 \\ x_5 & = x_1 + 4 \\ x_6 & = 4 \end{align}$$

Your answer is obvious, it's 4


Trick 3

  • Pick any positive 3 digit number in the universe.
  • Multiply by 7.
  • Multiply by 11.
  • Multiply by 13.

$$\begin{align} x_1 & = 100a + 10b + c\quad|\, a \in\left[1,9\right]\,\, b,c \in\left[0,9\right] \\ x_2 & = 700a + 70b + 7c \\ x_3 & = 7700a + 770b + 77c \\ x_4 & = 100100a + 10010b + 1001c \\ x_4 & = 100000a + 10000b + 1000c + 100a + 10b + c\\ \end{align}$$

Your answer is your original number twice


Trick 4

  • Think of a number between 1 and 9.
  • Double it.
  • Add 5.
  • Multiply by 5.
  • Add another digit between 1 and 9 to it.
  • Subtract 25.

$$\begin{align} x_1 & = m\quad|\, m \in \left[1,9\right] \\ x_2 & = 2m \\ x_3 & = 2m + 5 \\ x_4 & = 10m + 25 \\ x_5 & = 10m + 25 + n\quad|\, n \in \left[1,9\right] \\ x_6 & = 10m + n\quad|\, m,n \in\left[1,9\right] \\ \end{align}$$

The first digit of your new number was your original number. The second is your second number.

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