Question

Find the remainder of the division. 2001 • 2002 • 2003 + 2004^3 by 7.
Show steps.

Answer 1

`1`

Explanation:

The gist of modular arithmetic in a nutshell: the numbers
`a` and
`b` are considered to be congruent by their modulus
`m` if
`m` is a divisor of their difference.

In mathematics:
`a \equiv_(m) b \Leftrightarrow (a - b) \vdots m`

Exemplifying this:
`6 \equiv_(7) -1` because
`6 - (-1) = 6 + 1 = 7`,
`7 \vdots 7`.

Let us have following equivalences:
`a \equiv_(m) b` and
`c \equiv_(m) d`, then:
`(a - b) \vdots m` and
`(c - d) \vdots m` by definition.

Properties:

1.
`a + c \equiv_(m) b + d \Leftrightarrow ((a + c) - (b + d)) \vdots m \Leftrightarrow (a + c - b - d) \vdots m \Leftrightarrow ((a - b) + (c - d)) \vdots m`.

2.
`a - c \equiv_(m) b - d \Leftrightarrow ((a - c) - (b - d)) \vdots m \Leftrightarrow (a - c - b + d) \vdots m \Leftrightarrow ((a - b) - (c - d)) \vdots m`.

3.
`ac \equiv_(m) bd \Leftrightarrow (ac - bd) \vdots m \Leftrightarrow (ac - bc - bd + bc) \vdots m \Leftrightarrow (c(a - b) + b(c - d)) \vdots m`.

4. What if we have
`a \equiv_(m) b` twice? If we abide by property 3, we can come to the conclusion that
`a^2 \equiv_m b^2`. It is fair enough that there is room for the equivalence
`a^n \equiv_(m) b^n`.

`2001 * 2002 * 2003 + 2004^3 = 2002 * (2001 * 2003) + 2004^3 \equiv_(7) 0 * (2001 * 2003) + 2^3 \equiv_(7) 0 + 8 \equiv_(7) 8 \equiv_(7) 1`

Notice that
`2002 \vdots 7` already. There is no need to consider the rest multipliers because the product is
`0` anyways, we can cut corners in this regard.

We used property 4.

Recapitulating this: our remainder is
`1`.