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four digit numbers
- Thread starter lynnehowe
- Start date
Let's do it for a two digit number. The same principle holds for a four digit number, although a little more complicated.
94 - 49 = 45 which is a multiple of 9.
9(10)+4 - 4(10)+9 is same problem written in standard form.
Let a two digit number be represented by ab.
ab - ba = ?
Standard form gives us...
(10a+b) - (10b+a) =
10a+b - 10b - a =
10a - 10b -a + b =
10(a - b) -(a-b) =
9(a-b) which is a multiple of 9.
94 - 49 = 45 which is a multiple of 9.
9(10)+4 - 4(10)+9 is same problem written in standard form.
Let a two digit number be represented by ab.
ab - ba = ?
Standard form gives us...
(10a+b) - (10b+a) =
10a+b - 10b - a =
10a - 10b -a + b =
10(a - b) -(a-b) =
9(a-b) which is a multiple of 9.
Hello, lynnehowe!
\(\displaystyle \text{Suppose the number looks like this: }\:abcd\)
\(\displaystyle \text{This represents the quantity: }\:N_1 \:=\:1000a + 100b + 10c + d\)
\(\displaystyle \text{There are: }4! = 24\text{ possible orderings of the digits.}\)
\(\displaystyle \text{Let's take one of them: }\;dabc\)
\(\displaystyle \text{This represents: }\;N_2 \;=\;1000d + 100a + 10b + c\)
\(\displaystyle \text{Their difference is: }\;d \;=\;N_1-N_2\)
. . . . . . . . . . . . . . . \(\displaystyle =\;(1000a + 100b + 10c + d) - (1000d + 100a + 10b + c)\)
. . . . . . . . . . . . . . . \(\displaystyle =\; 1000a + 100b + 10c + d - 1000d - 100a - 10b - c\)
. . . . . . . . . . . . . . . \(\displaystyle =\;900a + 90b + 9c - 999d\)
. . . . . . . . . . . . . . . \(\displaystyle =\; 9(100a + 10b + c + 111d) \quad \hdots\;\;\text{a multiple of 9}\)
If you take a whole number containing 4 digits (eg 4321) and take away a second number containing the same 4 digits
but in a different order (eg 1234), why does the answer always end up being a multiple of 9 regardless of the 4 digits used?
\(\displaystyle \text{Suppose the number looks like this: }\:abcd\)
\(\displaystyle \text{This represents the quantity: }\:N_1 \:=\:1000a + 100b + 10c + d\)
\(\displaystyle \text{There are: }4! = 24\text{ possible orderings of the digits.}\)
\(\displaystyle \text{Let's take one of them: }\;dabc\)
\(\displaystyle \text{This represents: }\;N_2 \;=\;1000d + 100a + 10b + c\)
\(\displaystyle \text{Their difference is: }\;d \;=\;N_1-N_2\)
. . . . . . . . . . . . . . . \(\displaystyle =\;(1000a + 100b + 10c + d) - (1000d + 100a + 10b + c)\)
. . . . . . . . . . . . . . . \(\displaystyle =\; 1000a + 100b + 10c + d - 1000d - 100a - 10b - c\)
. . . . . . . . . . . . . . . \(\displaystyle =\;900a + 90b + 9c - 999d\)
. . . . . . . . . . . . . . . \(\displaystyle =\; 9(100a + 10b + c + 111d) \quad \hdots\;\;\text{a multiple of 9}\)