Fido Dido's secret
LAST week i challenged readers to make sense of a website recommended by educator and critic Dr. Isagani Cruz: www.digicc.com. Fido, the icon of 7-Up, gives this puzzle. First, write down a 3- or 4-digit number (not all digits should be alike). Jumble all the digits to make another number. Then subtract the smaller number from the larger one. Encircle one of the digits in your answer, except zero. Jumble up all the numbers which are left, and Fido can read your mind by guessing the digit you circled. What is Fido's secret?
Thank you, readers, for your overwhelming response. To date I have received more than 50 replies, and most of them are on the right track. However, some solutions are clearer and more detailed than others. The best explanation comes from creative writer and math aficionado Jose Edmundo Reyes [rilkepoet at yahoo dot com]. Look at how carefully he marshals his proof:
Making it simple
For the sake of simplicity, let's start with a 4-digit number ABCD. Mathematically, that can be represented by 1000A+100B+10C+D. When we jumble the digits, there are 24 possible 4-digit numbers (multiply 4, 3, 2 and 1), because we can only use those four digits from my original number.
Let's say we wind up with BCDA. Mathematically that's 1000B+100C+10D+A. If we subtract the lesser of the two from the greater, we end up with the absolute value of the following quantity: [1000(A-B)+100(B-C)+10(C-D)+(D-A)]. I claim that this quantity is divisible by 9. Why?
We know that a number is divisible by 9 if the sum of its digits is divisible by 9. We know that at least one of the quantities (A-B), (B-C), (C-D) and (D-A) is negative; in other words, not all can be positive. If A is greater than B, and B is greater than C, and C is greater than D, then D must be less than A, for instance. Let's say for the moment that this is the case. Then because we "borrow" from the tens digit, the quantity above is equal to the absolute value of [1000(A-B)+100(B-C)+10(C-D-1)+(10+D-A)].
Assuming that all the quantities in parentheses are positive, then they are all the digits of the resulting number. Adding the digits we get (A-B)+(B-C)+(C-D-1)+(10+D-A). Because of associative and commutative properties, this simplifies to 9, which is of course divisible by 9.
Let's say (C-D-1) is negative. Then because we borrow from the hundreds digit, the quantity is equal to [1000(A-B)+100(B-C-1)+10(10+ C-D-1)+(10+D-A)]. Again, assuming that all the quantities in parentheses are positive, then they are all the digits of the resulting number. Adding the digits we get (A-B)+(B-C-1)+(10+C-D-1)+(10+D-A). This simplifies to 18, which is divisible by 9.
If (B-C-1) is still negative, we simply repeat the process by borrowing from the thousands digit. This will result in a sum of 27, which is divisible by 9. (We only "borrow" as many times as necessary. No matter how many times we borrow, we'll wind up with a sum of digits equal to nine.)
No matter what the order of original digits A,B,C,D is, they will all cancel out using the above process, because each digit will always be canceled by its inverse. We'll wind up with a sum of resulting digits 0,9,18, or 27, meaning that the difference is divisible by 9.
The missing digit
That the difference is divisible by 9 is important because all Fido has to do is work backward from the digits that we provide him to find the missing digit, which is the one we encircled. If, for instance, we picked 1234 and jumbled it into 4321, the difference is 3087. Let's say we circled 8 (notice that Fido is sneaky by not allowing us to circle a 0) and give him 307. The only possible digit between 1 and 9 that will result in a sum of digits equal to a multiple of 9 is 8.
Metrobank outstanding math educator Allan Canonigo of Ormoc City [allan_mc2000 at yahoo dot com] also gives a similar, albeit shorter, proof. Ivan Quentin [try1cmor at yahoo dot com] explains using modulus (or remainders after dividing by 9). These three readers will each get an autographed copy of my latest book In Love with Science: Outstanding Filipino Scientists Tell Their Stories (Anvil, 2006).
The following readers also gave an essentially correct, but not complete, solution: Janice Cham [chamjanice at yahoo dot com], Rock Finn Gonzaga [rock_finn at yahoo dot com], Katherine Marie Tanyu [km_tanyu at yahoo dot com], Eribert Padilla [esp888 at yahoo dot com], Connie Rose Gabor [dammit_ 0072000 at yahoo dot com], Fatima Sia [fatimsia at yahoo dot com], Emet Tenerife [emet.tenerife at gmail dot com], William Anakin [anakinwilliam at yahoo dot com dot ph], Jun Cabrera [brojun29 at yahoo dot com], Bob [fzbear at gmail dot com] and Rey [rze at smprime dot com]. Good thinking!
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