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If you are here from Tiktok, nice;
if you are here directly, this is the relevant link:
Now, let us begin,
Let's say we have an amount of money, say🍎 (if I use letters as variables, it could be messy and exhausting to identify between text and variable)
then the table should be something like this:
┌───┬─────────┐
│spent│ left-over │
├───┼─────────┤
│ 🥐 │ 🍎-🥐 │
│ 🌮 │ 🍎-🥐-🌮 │
│ 🥦 │ 🍎-🥐-🌮-🥦│
│ 🍋 │ 0$ │
└───┴─────────┘
notice that we used up all the money, therefore 🍋 must be equal to 🍎-🥐-🌮-🥦 in order to let Row4Column2 become $0
so let's sum up column 1
🥐+🌮+🥦+🍋
= 🥐+🌮+🥦+(🍎-🥐-🌮-🥦)
= 🍎
and then let's sum up column 2
(🍎-🥐)+(🍎-🥐-🌮)+(🍎-🥐-🌮-🥦)+0$
= 🍎-🥐+🍎-🥐-🌮+🍎-🥐-🌮-🥦
= 3🍎 - 3🥐 - 2🌮 - 🥦
as you can see Column1 and Column 2 don't match up
it is just a coincidence that when 🍎=50$,🥐=20$,🌮=15$,🥦=9$,🍋=6$,
"amazingly", column2 = column1 + 1$
It is possible that there are other combinations such that Column2 is one more than Column1.
since we know the criteria is column2 = column1 + 1$, then
column2 = column1 + 1$
3🍎 - 3🥐 - 2🌮 - 🥦 = 🍎 + 1$
2🍎 - 3🥐 - 2🌮 - 🥦 = 1$
2🍎 = 1$ + 3🥐 + 2🌮 + 🥦
🍎 = (1$ + 3🥐 + 2🌮 + 🥦) ÷ 2
as you can see we can simply assign a value to 🥐,🌮and🥦, we can then work out the value of 🍎.
If you want 🍎 to be a whole number (not a fraction or a decimal),
we have to make sure 1$ + 3🥐 + 2🌮 + 🥦 is divisible by 2
2🌮 is already even, so 🌮 can be any positive integer.
if (🥐 is even)
{
then 3🥐is even and
1$ + 🥦 must be even,
therefore 🥦 must be odd
}
else if (🥦 is even)
{then 3🥐 + 1$ must be even,
therefore 3🥐 must be odd, and 🥐 must be odd
(because if 🥐 is even, 3🥐 would have been even and it will break the rule)
}
here are the 2 methods that we can assign the value to 🥐,🌮and🥦 such that 🍎is a whole number and Column1 + 1$ = Column2 :
🌮: any, 🥐: even, 🥦: odd OR
🌮: any, 🥐: odd, 🥦: even
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