Key Concepts: Combinatorial Construction and Counting, Principle of Least Favorability

Hello, young mathematicians! Today, we’re going to solve a fun math problem. Imagine you have a string of numbers, and you’re trying to find the smallest number of consecutive positive integers where at least one of them has a digit sum that is a multiple of 8. Sounds exciting, right? Let’s dive in!

First, let’s understand a basic concept. If we have any 8 consecutive positive integers, and there’s no carry-over between the digits (like going from 9999 to 10000), then at least one of these numbers will have a digit sum that is a multiple of 8.

Now, the trickiest situation is when there is a carry-over. Let’s imagine a situation where we have a string of 7 consecutive numbers ending in 999…9, and another string of 7 consecutive numbers starting from 1000…0. In this case, none of these 14 numbers have a digit sum that is a multiple of 8.

But if we add one more number to this string, making it 15 numbers long, we can guarantee that at least one number will have a digit sum that is a multiple of 8. So, the smallest number of consecutive positive integers we need is 15.

Let’s see this with an example. Let’s say we have 7 consecutive numbers ending with 9999999, and another 7 consecutive numbers starting from 10000000.

For the numbers ending with 9999999, the digit sum of each number is 9 times the number of 9’s in the number. For example, the digit sum of 9999999 is 9*7 = 63. If we want 7 consecutive numbers where the digit sum is not a multiple of 8, we need the digit sum to be 1 more than a multiple of 8. So, we need 9x (where x is the number of 9’s) to be 1 more than a multiple of 8. This gives us x = 7.

Knowledge Points: Combinatorial Construction and Counting, Principle of Least Favorable Outcome

Hello young mathematicians! Today, we are going to embark on a fun journey into the world of consecutive integers. Our mission is to solve a fascinating math problem: “What is the smallest number of consecutive positive integers, such that there is always at least one number whose digits add up to a multiple of 8?”

Before we dive in, let’s understand a key concept. We know that if we have any 8 consecutive positive integers, and if the sum of their digits is consecutive (meaning there is no carry-over), then there is definitely one number that is a multiple of 8.

But what happens when there is a carry-over? The most unfavorable situation is when we have a carry-over, for example, from 9999… to 10000…

In this case, following the principle of least favorable outcome, we can have at most 7 consecutive positive integers on the left side of the carry-over, and each of these numbers’ digits add up to a number that is not a multiple of 8. Similarly, we can have at most 7 numbers on the right side of the carry-over, and each of these numbers’ digits also add up to a number that is not a multiple of 8.

So, we can have a sequence of 14 consecutive positive integers, none of which have a digit sum that is a multiple of 8. But if we add one more number to this sequence, we can guarantee that there is at least one number whose digits add up to a multiple of 8. Therefore, the smallest number of consecutive positive integers we need is 15.

Now, let’s try to find a specific example. Suppose we have $x$ number of 9’s on the left side of the carry-over, and $x$ number of 0’s on the right side. Let’s try to find the smallest value of $x$ that will give us this sequence.

Looking at the left side, suppose the last number is a number made up of $x$ number of 9’s: 999…9. The sum of the digits of this number is $9x$. If we want to have 7 numbers on the left side that do not satisfy the condition, then $9x$ modulo 8 should be 7, i.e., $9x \bmod 8 = 7$.

How do we calculate $x$? We know that $8x \bmod 8 = 0$, therefore $9x \bmod 8 = (x + 8x) \bmod 8 = x \bmod 8 = 7$, so the smallest value for $x$ is 7. That is, $9 * 7 \bmod 8 = 7$.

So, the first 7 numbers are:

Number 9999993 9999994 9999995 9999996 9999997 9999998 9999999
Digit Sum 57 58 59 60 61 62 63
Modulo 8 1 2 3 4 5 6 7

The next 7 numbers are:

Number 10000000 10000001 10000002 10000003 10000004 10000005 10000006
Digit Sum 1 2 3 4 5 6 7
Modulo 8 1 2 3 4 5 6 7

The digit sum of all these 14 numbers is not a multiple of 8. But if we add one more number to make it a sequence of 15 numbers:

  1. If we add it to the front, the number is 9999992, and the digit sum is 56, which is a multiple of 8.
  2. If we add it to the back, the number is 10000007, and the digit sum is 8, which is also a multiple of 8.

So, the magic number is 15! Isn’t it fascinating how numbers work? Keep exploring, young mathematicians!