Hi Marcelle,

Puzzle "22"

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.

Algebraically:

Let the digits chosen be x, y and z
Then the 6 possibilities generated by step 2 are

10x+y, 10x+z, 10y+x,10y+z, 10z+x, 10z+y

So the sum is

(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)
=10x+10x+x+x + 10y+10y+y+y + 10z+10z+z+z
=22x + 22y + 22z
=22(x+y+z)

Performing step 3 is 22(x+y+z)/(x+y+z) = 22

Puzzle "1089"
99x2=198
198+891=1089

99x3=297
297+792=1089

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:

(a) the tens digit is unchanged
(b) the digit in the ones place goes into the hundreds place
(c) the digit in the hundreds place goes into the ones place

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.

Algebraically:

Let the digits chosen in step 1 be x, y and z, where x is hundreds digit, y is the tens and z is the ones digit. Since we want to subtract the smaller from the larger, choose x to be larger than z. Then, the 3 digit number chosen is 100x+10y+z Reversing the digits produces 100z+10y+x Step 2 produces

(100x+10y+z) - (100z+10y+x)
=100x + 10y + z - 100z - 10y - x
=100x - x +10y - 10y + z - 100z
=99x - 99z
=99(x-z)

By the conditions in step 1, x and z differ by at least 2, so x-z is greater than equal to 2. Also, x and z are digits so x-z is less than or equal to 9. Thus, 99(x-z) is one of 99x2, 99x3, ... 99x9

Paul

Go to Math Central