Divisibility Rule for 6-Digit Consecutive Numbers

We need to determine which number always divides a 6-digit number formed using consecutive natural numbers.

Understanding Consecutive Digits

Let the 6 consecutive natural digits start from 'n'. The digits will be n, n+1, n+2, n+3, n+4, and n+5. For example, if n=1, the digits are 1, 2, 3, 4, 5, 6, forming the number 123456.

Applying the Divisibility Rule for 3

A number is divisible by 3 if the sum of its digits is divisible by 3. Let's find the sum (S) of these consecutive digits:

$ S = n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) $

Combine like terms:

$ S = 6n + (1+2+3+4+5) $

$ S = 6n + 15 $

Factor out 3:

$ S = 3(2n + 5) $

Since the sum of the digits ($S$) is always a multiple of 3, any 6-digit number formed by consecutive natural digits is guaranteed to be divisible by 3.

Checking Other Divisors

The divisibility by 2, 4, or 5 depends on the specific last digit(s) of the number, which varies. For instance:

• The number 123456 is divisible by 2, 4, and 3, but not 5.
• The number 234567 is divisible by 3, but not 2, 4, or 5.

Therefore, only divisibility by 3 is constant for all such numbers.

Conclusion

A 6-digit number formed by consecutive natural numbers is always divisible by 3.