1. Do there exist ten consecutive natural numbers that are successively divisible by the numbers $9,7,5,3,1,1,3,5,7,9$?

(Jaroslav Zhouf)
SOLUTION. Yes, such ten numbers do exist. An example is the following ten consecutive numbers, under which we immediately write their required divisors:

![](https://cdn.mathpix.com/cropped/2024_04_17_b127a3223ea29dd23f61g-01.jpg?height=90&width=1189&top_left_y=803&top_left_x=456)

Indeed, divisibility by 1 is trivial, by 3 and 9 according to the sum of digits, by 5 according to the last digits, and divisibility by 7 follows from the equalities $154=7 \cdot 22$ and $161=154+7$.

COMMENT. The given solution is complete, but we will provide the considerations that led to it.

Let the smallest of the ten sought numbers be denoted by $n$. Similarly as in the solution, let's write down clearly what each number should be divisible by:

![](https://cdn.mathpix.com/cropped/2024_04_17_b127a3223ea29dd23f61g-01.jpg?height=93&width=1446&top_left_y=1250&top_left_x=325)

- If we choose $n$ to be divisible by 9, then the number $n+9$ will also be divisible by 9. At the same time, the numbers $n+3$ and $n+6$ will be divisible by 3.
- If the number $n+1$ is divisible by 7, then the number $n+8$ will also be divisible by 7.
- If the number $n+2$ is divisible by 5, then the number $n+7$ will also be divisible by 5.

Since divisibility by 1 is always satisfied, it suffices to find a number $n$ that meets the three conditions:

(i) $n$ is divisible by 9,

(ii) $n+1$ is divisible by 7,

(iii) $n+2$ is divisible by 5.

First, let's focus on conditions (i) and (ii). From all numbers $9, 18, 27, 36, \ldots$ divisible by 9, we select the smallest one that satisfies (ii). This is 27, because 28 is divisible by 7.

Now, let's consider what all other numbers $n$ that satisfy (i) and (ii) look like. We write them as $27+k$ with an integer $k>0$. To satisfy (i), $k$ must be a multiple of 9 (and this is sufficient). Condition (ii) means that the number $n+1=28+k$ must be divisible by 7, so $k$ must also be a multiple of 7. Due to the coprimality of 7 and 9, this means that $k$ is a multiple of the number $9 \cdot 7=63$. Therefore, all $n$ that satisfy both (i) and (ii) are of the form $n=27+63l$, where $l$ is a non-negative integer. These are the numbers 27, 90, 153, 216, ...

It remains to satisfy condition (iii). From the numbers $27+2, 90+2, 153+2, 216+2, \ldots$, we need to select one that is divisible by 5. The first such number is clearly 155, so the smallest $n$ that satisfies the triplet of conditions (i), (ii), (iii) is the number $n=153$ (which we mentioned in the solution). If we repeated the reasoning from the previous paragraph, we would find that (considering the coprimality of 63 and 5 and the equality $63 \cdot 5=315$) all suitable numbers $n$ are of the form $153+315l$. Among them is, for example, the number with the interesting decimal representation 888888888.

NOTE. The task was only to decide on the existence of a suitable group of ten consecutive natural numbers, so it was not necessary to find such a group of numbers. We will show that their existence is a consequence of the Chinese Remainder Theorem. To do this, we will look at the conditions (i), (ii), and (iii) that we derived in the comment, in such a way that the same number $n$ should give the required remainders when divided by several different numbers. Specifically,

(i) $n$ gives a remainder of 0 when divided by 9,

(ii) $n$ gives a remainder of 6 when divided by 7,

(iii) $n$ gives a remainder of 3 when divided by 5.

Since the numbers 9, 7, and 5 are pairwise coprime, by the Chinese Remainder Theorem, there indeed exists a number $n$ that satisfies all three conditions, as we were supposed to show.

## GUIDING AND ADDITIONAL PROBLEMS:

N1. Petr wrote 7 consecutive natural numbers on the board. Pavel did not see them, but he claims that one of them is divisible by seven. Is he right? Why? [Pavel is right. Let $n$ be the smallest of the seven numbers and $z$ be its remainder when divided by seven. If $z \neq 0$, then $1 \leq 7-z \leq 6$, so the number $n+(7-z)$ is written on the board and is divisible by seven, because it is the sum of the numbers $n-z$ and 7, both of which are divisible by seven.]

N2. I am thinking of a natural number. If I decrease it by 