Quandaries and Queries
My question is about two puzzles called the "22" puzzle and "1089" puzzle.
the following steps are needed:
1. Choose 3 digits from 1-9
Another one related to this is t:
it doesn't matter what numbers are used, the results are alwasy the
same. eg 22 or 1089
The six possibilities in step 2 are generated by placing one digit in the ten's place and another in the one's place - to get all six possibilities, each digit will show up in the ten's place a total of 2 times and in the one's place a total of 2 times. Thus, when we sum the numbers generated in step two, each digit will occur 10+10+1+1 = 22 times. So, this sum is actually 22 times the sum of the three digits. Thus, afterperforming step 3, we are left with 22.
check and confirm that this "works" for 99x4, 99x5, 99x6, 99x7, 99x8 and 99x9
So, all we must do is show that step 1 and 2 generate one of the multiples of 99 checked above. To see this, consider what happens when you reverse the digits of a three digit number:
When we subtract, because of (a), the digits in the tens place cancel; because of (b), the hundreds digit (of the original number) gets reduced by one copy of itself, so we have 99 times that digit; similarly, the digit that started in the ones place, because of (c), there is a negative(since subtract in opposite order) multiple of 99 of this digit. Thus, the subtraction produces 99 copies of the larger digit minus 99 copies of the smaller digit, which is 99 times the differrence of the two digits. But the diiference between the two digits must be greater than or equal to 2 (since condition in step 1, and since we subtracted the smaller one from the larger of these) and less than 10 (since they are both digits). Hence, the result of step 2 is one of the multiples of 99 we checked above.