Magic gopher

The magic gopher is an interactive Adobe Flash game published online by the British Council, presented as an aid in learning English.

Overview

The game involves the 'magic gopher' asking the player to pick a random two-digit number. The player is then asked to sum the digits of the number and subtract them from the original number. The gopher then presents a list of symbols and instructs the player to find the one next to this new number and memorise it. The gopher then proceeds to correctly guess the symbol, supposedly after reading the player's mind.

The trick

The trick to the game is that the gopher assigns the same symbol to every multiple of 9, from 0 through 81 (it is impossible to get a higher number using only two digits). The symbol picked for each game is randomized. No matter what two-digit integer the player chooses, when the subtraction is done, the resulting number will always be a multiple of 9. In fact, no matter which (nonnegative) integer the player chooses, the result will always be a multiple of 9. At least two other sites on the web have similar games using this trick, but without the gopher.^[1]^[2] To conceal the trick, the same symbol is assigned to some non-multiples of 9, but they will never be the number the player is looking for.

The fact that the result will always be a multiple of 9 can be proven using elementary algebra. To prove for n digits, the digits of the number $n$ are indexed with the rightmost digit being assigned position 0 and with the adjacent digits having an index increasing by one each time moving leftwards. For example, for the number '261', the digit '1' is in position 0, '6' is in position 1, and '2' is in position 2.

Proof for 2 digits

Let n be a 2-digit integer. Additionally, let a be the first digit of n and b be the second digit of n. Finally, let c equal the sum of the digits of n, so c = a + b.

An equivalent form for n, by virtue of using a decimal numeral system, is n = 10a + b.

The resulting number, z, is given by z = n − c = (10a + b) − (a + b) = 9a. Hence, z is always a multiple of 9. Q.E.D.

Proof for n digits

Proving that no matter how large $n$ is (and how many digits $n$ has), it is always a multiple of 9 is slightly trickier. The following proof makes use of modular arithmetic:

Let $n$ be an integer with $m$ digits and let $n_{m}$ represent the $m^{t h}$ digit of $n$ .
Thus, $n = 1 0^{m} \cdot n_{m} + 1 0^{m - 1} \cdot n_{m - 1} + \dots + 1 0^{0} \cdot n_{0}$ .
Let $c$ be the sum of the digits of $n$ . So, $c = n_{m} + n_{m - 1} + \dots + n_{0}$ .
Since $1 0^{m} \equiv 1 mod 9, n = n_{m} + n_{m - 1} + \dots + n_{0} = c mod 9$ .
Hence $n - c \equiv 0 mod 9$ so the resulting number $z = n - c$ is a multiple of 9.
Q.E.D.

Alternative proof for n digits

This alternative proof is less mathematically rigorous, relying on some common sense and intuition, but it is still sufficient to demonstrate the same as the above.

Again, let n be an integer with m digits and let n_m represent the m^th digit of n.

Thus,

n = 1 0^{m} \cdot n_{m} + 1 0^{m - 1} \cdot n_{m - 1} + \dots + 1 0^{0} \cdot n_{0} .

Let c be the sum of the digits of n. So

c = n_{m} + n_{m - 1} + \dots + n_{0} .

Now, let

z = n - c = (1 0^{m} \cdot n_{m} + 1 0^{m - 1} \cdot n_{m - 1} + . . + 1 0^{0} \cdot n_{0}) - (n_{m} + n_{m - 1} + \dots + n_{0}) .

This can be written as

z = (1 0^{m} \cdot n_{m} - n_{m}) + (1 0^{m - 1} \cdot n_{m - 1} - n_{m - 1}) + \dots + (1 0^{0} \cdot n_{0} - n_{0}) .

By factoring, we obtain

z = n_{m} (1 0^{m} - 1) + n_{m - 1} (1 0^{m - 1} - 1) + \dots + n_{0} (1 0^{0} - 1) .

Now, $1 0^{m} - 1$ will give a number with $m$ nines, hence, each individual digit of n is being multiplied by a multiple of 9 as the numbers given by $1 0^{m} - 1$ are implicitly multiples of 9. Since the sum of any number of multiples of 9 is always divisible by 9, we conclude that whichever number is picked for n, it will always be a multiple of 9.

If we use the resulting formula for z assuming that n will only be 2 digits in length, we obtain the same formula as with the proof for a 2 digit n:

z = n_{1} (1 0^{1} - 1) + n_{0} (1 0^{0} - 1) = n_{1} \cdot 9 + n_{0} \cdot 0 = 9 n_{1}

where

n_{1} = a .

References

↑ Henri Monjauze, "Multiples of nine", Mathematical Magic Tricks – Numericana, 2008-02-07.
↑ Leigh Lundin, "Puzzles, Part I", SleuthSayers, 2012-04-22.

External links

This article "Magic gopher" is from Wikipedia. The list of its authors can be seen in its historical. Articles copied from Draft Namespace on Wikipedia could be seen on the Draft Namespace of Wikipedia and not main one.

[1] Henri Monjauze, "Multiples of nine", Mathematical Magic Tricks – Numericana, 2008-02-07.

[2] Leigh Lundin, "Puzzles, Part I", SleuthSayers, 2012-04-22.

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