Try this fun little game.

http://www.digicc.com/fido/

How is the 7up guy able to do this?

The difference value that you compute in the game is such that the sum of its digits, except for the 9's, is nine or zero.

For example, suppose the difference value is 4320. You can't circle 0 so you circle one of the others. Let it be 3. You enter 420 (duuuuuuude!) or some other combination of 4, 2, and 0. Shitface just counts up 4+2+0 = 6 and knows that the other one has to be a 3 for everything to add up to 9. Suppose the difference is 9000. You cannot circle 0 so you circle the 9 and report '000'. Homeslice knows that 9 is the only value that it could be.

Why is the sum of the digits in the difference, except for the 9's, always nine? The difference values are always multiples of nine, and the digits in multiples of nine, excluding the 9's, always sum to nine or zero.

Proof

definitions:

abs(i-j) = (i-j) if i is greater than j, and = -(i-j) if i is less than j

max{a,b} = greater of a and b

min{a,b} = lesser of a and b

a(b) = a*b = a multiplied by b

[i,j] = max{10(i-1)+(j-1), 10(j-1) + (i-1)} - min{10(i-1)+(j-1), 10(j-1) + (i-1)}

_ab_ = 10(a) + b

Note that [(i+1),(j+1)] is the difference value computed when the switching around of the number _ij_ is done so that you just write it backwards. For example,

[5,2] = 41 - 14 = 27

[7,1] = 60 - 06 = 54

[4,5] = -(45 - 54) = 9

Part 1.

Proof that the difference values are all multiples of 9:

Start with two digit numbers.

Claim: [i,j] = 9*abs(i-j)

[i,j] can be reduced. Start from the definition above. It can be reduced to

[i,j] = max{10i+j, 10j+i} - 11 - (min{10i+j, 10j+i} - 11)

= max{10i+j, 10j+i} - min{10i+j, 10j+i}

[1,1] = max{0,0} - min{0,0} = 0 = 9*abs(1-1)

Assume [i,j] = 9*abs(i-j)

[(i+1),j] = max{10(i+1)+j, 10j+(i+1)} - min{10(i+1)+j, 10j+(i+1)}

= max{10i+j+10, 10j + i +1} - min{10i+j+10, 10j + i +1}

case 1: 10i+j+10 > 10j+i+1

[(i+1),j] = 10i + j + 10 - (10j + i +1) = 9i -9j + 9 = 9(i-j+1)

case 2: 10i+j+10 < 10 =" -(9i" style="font-weight: bold;">Note: This only works for two digit numbers. I am tiring of this shit, so I am not going to show how it extends to work for three or four digit numbers. With three and four digit numbers you have to show how it works with all permutations of the digits, not just when they get reversed in order. I could totally fucking do it though, you can bet your sweet ass on that.

Part 2.

Handwavey proof that the digits in all multiples of nine, excluding the 9's, add up to nine or zero:

Suppose y=9(x) holds. The last digit in y=9(x+1) is always one less than it is in y(x) and the next to last digit is always one more than it was in y(x), with the exception of when the last digit of y(x) is 0. In that case, only the last digit of y(x+1) changes, and it increases by 9. When all digits from the leading nonzero digit to the last digit are 9's, the next multiple has an 8 in the last place, a one in the zero place one ahead of where the leading 9 had been, and zeros in each place between the two. Note that this will always add to 9, since the only nonzero places will have an 8 and a 1.

Let x=0. y(0)=0, which holds.

y(x+1) =9 because y(x)=0.

Now let x=1, which was just shown to hold.

The only digit here is 9, so the last place becomes an 8 and the lead digit where the last zero had been becomes a 1 and everything inbetween becomes a 0. In this case there is nothing between them, so it is just 18. Of course 1+8=9, so this holds.

Now let x=2.

y(x+1) will have a decrease of 1 in the last place and an increase of one in the next place.

8 becomes 7 and 1 becomes 2.

y(x+1) = 27. 2+7 = 9, so this again holds.

When the last place in y(x) is not a zero, the sum of the digits from y(x) to y(x+1) remains the same, since one is being added to the second to last digit and one is being subtracted from the last digit.

Note: I may fix part 2 to be non-shitty. I likely will not explore finishing part 1.

http://www.digicc.com/fido/

How is the 7up guy able to do this?

"I can't tell you how I did it, but I'll leave you with this final thought...Asshole. Well, 7up boy or whatever the fuck your name is, I'm hear to Tell the World!

Invest wisely...Spend on Yourself."

The difference value that you compute in the game is such that the sum of its digits, except for the 9's, is nine or zero.

For example, suppose the difference value is 4320. You can't circle 0 so you circle one of the others. Let it be 3. You enter 420 (duuuuuuude!) or some other combination of 4, 2, and 0. Shitface just counts up 4+2+0 = 6 and knows that the other one has to be a 3 for everything to add up to 9. Suppose the difference is 9000. You cannot circle 0 so you circle the 9 and report '000'. Homeslice knows that 9 is the only value that it could be.

Why is the sum of the digits in the difference, except for the 9's, always nine? The difference values are always multiples of nine, and the digits in multiples of nine, excluding the 9's, always sum to nine or zero.

Proof

definitions:

abs(i-j) = (i-j) if i is greater than j, and = -(i-j) if i is less than j

max{a,b} = greater of a and b

min{a,b} = lesser of a and b

a(b) = a*b = a multiplied by b

[i,j] = max{10(i-1)+(j-1), 10(j-1) + (i-1)} - min{10(i-1)+(j-1), 10(j-1) + (i-1)}

_ab_ = 10(a) + b

Note that [(i+1),(j+1)] is the difference value computed when the switching around of the number _ij_ is done so that you just write it backwards. For example,

[5,2] = 41 - 14 = 27

[7,1] = 60 - 06 = 54

[4,5] = -(45 - 54) = 9

Part 1.

Proof that the difference values are all multiples of 9:

Start with two digit numbers.

Claim: [i,j] = 9*abs(i-j)

[i,j] can be reduced. Start from the definition above. It can be reduced to

[i,j] = max{10i+j, 10j+i} - 11 - (min{10i+j, 10j+i} - 11)

= max{10i+j, 10j+i} - min{10i+j, 10j+i}

[1,1] = max{0,0} - min{0,0} = 0 = 9*abs(1-1)

Assume [i,j] = 9*abs(i-j)

[(i+1),j] = max{10(i+1)+j, 10j+(i+1)} - min{10(i+1)+j, 10j+(i+1)}

= max{10i+j+10, 10j + i +1} - min{10i+j+10, 10j + i +1}

case 1: 10i+j+10 > 10j+i+1

[(i+1),j] = 10i + j + 10 - (10j + i +1) = 9i -9j + 9 = 9(i-j+1)

case 2: 10i+j+10 < 10 =" -(9i" style="font-weight: bold;">Note: This only works for two digit numbers. I am tiring of this shit, so I am not going to show how it extends to work for three or four digit numbers. With three and four digit numbers you have to show how it works with all permutations of the digits, not just when they get reversed in order. I could totally fucking do it though, you can bet your sweet ass on that.

Part 2.

Handwavey proof that the digits in all multiples of nine, excluding the 9's, add up to nine or zero:

Suppose y=9(x) holds. The last digit in y=9(x+1) is always one less than it is in y(x) and the next to last digit is always one more than it was in y(x), with the exception of when the last digit of y(x) is 0. In that case, only the last digit of y(x+1) changes, and it increases by 9. When all digits from the leading nonzero digit to the last digit are 9's, the next multiple has an 8 in the last place, a one in the zero place one ahead of where the leading 9 had been, and zeros in each place between the two. Note that this will always add to 9, since the only nonzero places will have an 8 and a 1.

Let x=0. y(0)=0, which holds.

y(x+1) =9 because y(x)=0.

Now let x=1, which was just shown to hold.

The only digit here is 9, so the last place becomes an 8 and the lead digit where the last zero had been becomes a 1 and everything inbetween becomes a 0. In this case there is nothing between them, so it is just 18. Of course 1+8=9, so this holds.

Now let x=2.

y(x+1) will have a decrease of 1 in the last place and an increase of one in the next place.

8 becomes 7 and 1 becomes 2.

y(x+1) = 27. 2+7 = 9, so this again holds.

When the last place in y(x) is not a zero, the sum of the digits from y(x) to y(x+1) remains the same, since one is being added to the second to last digit and one is being subtracted from the last digit.

Note: I may fix part 2 to be non-shitty. I likely will not explore finishing part 1.

Labels: awful chief, math

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