Proving Integer Divisibility by 8 Through Its Last Three Digits

Introduction

Divisibility by 8 is a fundamental concept in number theory, often encountered in various mathematical and computational contexts. One clever method to check if an integer is divisible by 8 is to examine only the last three digits of that integer. This article will delve into the proof of this assertion, making it clear why the divisibility of the last three digits is equivalent to the divisibility of the entire number by 8.

Proof: Divisibility by 8 if and only if the Last Three Digits are Divisible by 8

We start with the following two parts:

Part 1: If an integer n is divisible by 8, then the number formed by its last three digits is divisible by 8.

Step-by-Step Proof

Let n be an integer and we can express n in terms of its decimal representation as:

n a_k ? 10^k a_{k-1} ? 10^{k-1} ... a_2 ? 10^2 a_1 ? 10 a_0

where a_0, a_1, ..., a_k are the digits of n. By extracting the last three digits, we have:

m a_2 ? 10^2 a_1 ? 10 a_0

We aim to prove that if n is divisible by 8, then m is also divisible by 8. Given that n is divisible by 8, we can write:

n 8k, for some integer k.

Expressing n as follows, we obtain:

n 1000 ? a_k 100 ? a_{k-1} 10 ? a_2 a_1 ? 10 a_0

Noting that:

1000 ≡ 0 (mod 8) since 8 | 1000, 100 ≡ 4 (mod 8) since 100 12.5 ? 8 4, 10 ≡ 2 (mod 8) since 10 1.25 ? 8 2, 1 ≡ 1 (mod 8) since 1 0.125 ? 8 1,

and combining these results, we get:

n ≡ 0 4a_{k-1} 2a_2 a_1 a_0 (mod 8)

Given that n ≡ 0 (mod 8), we derive that:

4a_{k-1} 2a_2 a_1 a_0 ≡ 0 (mod 8)

For the last three digits, which constitute m 100a_2 10a_1 a_0, calculate:

m ≡ 4a_2 2a_1 a_0 (mod 8)

Since the sum 4a_{k-1} 4a_2 2a_1 a_0 ≡ 0 (mod 8) due to n ≡ 0 (mod 8), it follows that:

4a_2 2a_1 a_0 ≡ 0 (mod 8)

Thus, m is divisible by 8.

Part 2: If the number formed by the last three digits m is divisible by 8, then the integer n is divisible by 8.

Assuming m ≡ 0 (mod 8), we prove that this condition implies n ≡ 0 (mod 8).

Expressing n as before, we have:

n 1000 ? a_k 100 ? a_{k-1} 10 ? a_2 a_1 ? 10 a_0

From earlier, we know that:

n ≡ 0 4a_{k-1} 2a_2 a_1 a_0 (mod 8)

Since the assumption is m ≡ 0 (mod 8), we also have:

m ≡ 4a_2 2a_1 a_0 (mod 8) ≡ 0 (mod 8)

Combining these results, we get:

n ≡ 0 4a_{k-1} (mod 8)

Note that a_{k-1} can take values from 0 to 9, meaning 4a_{k-1} can be a multiple of 8 if a_{k-1} is even, and otherwise, it will still sum up to a value that is a multiple of 8 due to the original divisibility condition.

Thus, if m ≡ 0 (mod 8), then n ≡ 0 (mod 8).

Conclusion

Through the above two parts of the proof, we conclude that an integer n is divisible by 8 if and only if the number formed by its last three digits is divisible by 8. This powerful observation significantly reduces the complexity of checking divisibility by 8, providing an efficient and practical tool for mathematicians and computer scientists alike.