Solution: Seven-Up Puzzle
You’re asked to pick a 4-digit number … let’s call it N1. Let a, b, c, and d
be the 4 digits of the number N. So …
N1 = 1000a + 100b + 10c + d
Then you’re asked to scramble the digits of N1 … let’s call the resulting
number N2 . Suppose the new order of the digits is c, d, a, and b …
N2 = 1000c + 100d + 10a + b
We’re asked to subtract the lesser of the two numbers from the other.
Suppose N1 is greater than N2 …
N1 – N2 = (1000a + 100b + 10c + d) – (1000c + 100d + 10a + b)
= 1000a – 1000c + 100b – 100d + 10c – 10a + d – b
= 990a + 99b – 990c – 99d
= 9(110)a + 9(11)b – 9(110)c – 9(11)d
= 9(110a + 11b – 110c – 11d)
To simplify things, let’s let K be the integer value in this last set of
parentheses …
K = (110a + 11b – 110c – 11d)
Then we’re left with …
N1 – N2 = 9K
So we’ve just proven that the difference between N1 and N2 is evenly
divisible by 9. In fact, we can do a similar proof for every possible ordering
of the 4 digits of N1. (See the “Side Note” below.) In other words, the
difference between ANY arbitrary 4-digit number, and another 4-digit
number formed by using the same 4 digits, MUST be evenly divisible by 9.
Now let’s examine the digits that make up the number (N1 – N2) … let’s call
them w, x, y, and z, in that order. So …
N1 – N2 = 1000w + 100x + 10y + z
Let’s let S be sum of the digits that make up the number (N1 – N2) …
S=w+x+y+z
If we subtract the sum of the digits, S, from the number (N1 – N2) itself, we
get …
(N1 – N2) – S = (1000w + 100x + 10y + z) – (w + x + y + z)
Substituting 9K for (N1 – N2) and doing the algebra …
9K – S = 1000w – w + 100x – x + 10y – y + z – z
= 999w + 99x +9y
Rearranging terms …
S = 9K – 999w – 99x – 9y
= 9(K – 111w – 11x – y)
So we have just also proven that S, the sum of the digits of the number (N1 –
N2), MUST be evenly divisible by 9.
Conclusion: The trick asks you to provide all but one of the digits of the
number (N1 – N2). The digit that you hold back must always be the
number you have to add to the sum of the 3 digits that you do provide,
in order to add up to the next even multiple of 9.
Side Note:
It doesn’t really matter in which order we actually scramble the digits of the
number N1 in order to arrive at the second number, N2 … because the
coefficients of the variables of a, b, c, and d in both N1 and N2 are all powers
of 10. For example, we know that N1 looks like the following …
N1 = 103 a + 102 b + 101 c - 100 d
Similarly, regardless of the order in which we scramble the digits, N2 must
be of the form …
N2 = 10i a + 10j b + 10k c – 10t d
When we subtract the lesser of these from the other, consider the resulting
coefficient of just the digit a …
103 a - 10i a = 10i a (103-i - 1)
But ANY power of 10, less 1, will ALWAYS be evenly divisible by 9 … so
the coefficient of a in the number (N1 – N2) must be evenly divisible by 9.
In the same fashion, we can show that the coefficients of b, c, and d in the
number (N1 – N2) must also be evenly divisible by 9. So ALL coefficients
of (N1 – N2) must be evenly divisible by 9 … which proves that the number
(N1 – N2) must itself also be evenly divisible by 9.
Again … we’ve just shown that it doesn’t really matter in which order we
actually scramble the digits of the number N1 in order to arrive at the second
number, N2 … the above proof holds true, and the puzzle will always be able
to return the number you are holding back.