We have the alternating 3-digit sum for 7, 11 and 13:
To find $210600900 \bmod{} 7$, we can proceed by: $(900-600+210) \bmod{} 7$.
The idea behind is
$$210600900 =210\times1000^{2}+600\times1000+900$$ $$1 =1$$ $$1000 =\left(1001\right)-1$$ $$1000^{2} =1001M+1$$ where $M$ is some integer, hence $1000^2$ is the sum of $1$ and a multiple of $1001$.
Since $7 \times 11 \times 13 = 1001 = 1000 + 1 = 10^3 + 1$, any multiple of $1001$ is automatically that of $7$, $11$ and $13$.
Therefore, $$210600900 =210\times\left(1001M-1\right)+600\times\left(1001-1\right)+900\times1$$ $$\therefore 210600900 \equiv 210-600+900 \pmod{7}$$
The above operation can be written compactly:
$$ 1000^n = \left(-1\right)^n \pmod{7} $$ $$ 210600900 = 210 \times (-1)^2 + 600 \times (-1) + 900 \pmod{7} $$ $$ 210600900 = 210 - 600 + 900 \pmod{7} $$
Note: the clock element $-1$ is the same as $6$.
Is $1023$ a multiple of $11$?
One can check as follows:
$$1023 = 10 + 23 = 33 \equiv 0 \pmod{11} $$
since we have $100 = 99 + 1 \equiv 1 \pmod{11}$.
The above is called the 2-digit sum of 1023, where we group digits two by two and compute the total sum.
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.Mod
or Integers
.Definition: Let $a$ be an element of $Z_n$ (i.e. all operations are under modulo $n$).
The additive order of 3 in $Z_11$ is 4 by checking.
The multiplicative order of 1000 in $Z_7$ is 2 as shown above. Alternatively, we have $1000 = 6 \pmod{7}$ and hence $$1000^2 = 36 = 1 \pmod{7}$$
The additive order is computed by element.order()
:
Mod(3,11).order()
The multicative order is computed by element.multiplicative_order()
:
Mod(1000,7).multiplicative_order()