# Answer to Question #25605 in Algebra for raghav

Question #25605

show that a number and its cube always leave the same remainder when divided by 6.

Expert's answer

Any integer N can be represented in the form

N = 6M + R, R<6,

where M is a quotient and R is a remainder. Let's consider N³:

N³ = (6M + R)³ = (6M)³ + 3(6M)²R + 3*6M*R² + R³.

The first three terms of the sum are divisible by 6. Let's check what remainder does R³ leave when divided by 6:

R = 1: 1³ (mod 6) = 1 = R;

R = 2: 2³ (mod 6) = 8 (mod 6) = 2 = R;

R = 3: 3³ (mod 6) = 27 (mod 6) = 3 = R;

R = 4: 4³ (mod 6) = 64 (mod 6) = 4 = R;

R = 5: 5³ (mod 6) = 125 (mod 6) = 5 = R.

Therefore, a number and its cube always leave the same remainder when divided by 6.

N = 6M + R, R<6,

where M is a quotient and R is a remainder. Let's consider N³:

N³ = (6M + R)³ = (6M)³ + 3(6M)²R + 3*6M*R² + R³.

The first three terms of the sum are divisible by 6. Let's check what remainder does R³ leave when divided by 6:

R = 1: 1³ (mod 6) = 1 = R;

R = 2: 2³ (mod 6) = 8 (mod 6) = 2 = R;

R = 3: 3³ (mod 6) = 27 (mod 6) = 3 = R;

R = 4: 4³ (mod 6) = 64 (mod 6) = 4 = R;

R = 5: 5³ (mod 6) = 125 (mod 6) = 5 = R.

Therefore, a number and its cube always leave the same remainder when divided by 6.

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