Will concatenation always works?

Let three consecutive numbers be n , n+1 , n+2 .

We know that a number is divisible by 3 if the sum of its digits is divisible by three .

n+n+1+n+2 = 3n+3 = 3(n+1). [ For example, 333435 here n=33 , n+1=34, n+2=35 . n+n+1+n+2=33+34+35=102 =3x34=3(n+1) ]

Here I am not adding the digits of the number rather I am adding the consecutive numbers that form the number, either way it can shown that it is divisible by 3.

This implies that the concatenation of any three consecutive numbers is always divisible by 3.

The three numbers, as they are consecutive, will contain:

• A number which has remainder 0 on division by 3
• A number which has remainder 1 on division by 3
• A number which has remainder 2 on division by 3

This means that the digit sums of the three consecutive numbers will have:

• A remainder of 0 on division by 3
• A remainder of 1 on division by 3
• A remainder of 2 on division by 3

This means that the digit sum of the concatenated number will have a remainder of $0 + 1 + 2 = 3 = 0$ on division by 3, meaning that it will be a multiple of 3

Divisibility by $3$ is true if the sum of the digits of a number add up to $3$ or a multiple thereof. Let's define the first of the consecutive numbers as $a$ . Notice that $a$ can also be written as $a = a_0 + 10a_1 + 100a_2 + 1000a_3 \cdots$ and consequently $a + 1 = 1 + a_0 + 10a_1 + 100a_2 + 1000a_3 \cdots$ $a + 2 = 2 + a_0 + 10a_1 + 100a_2 + 1000a_3 \cdots$ Notice that $a_0, a_1, a_2, a_3, \cdots$ are the digits of $a$ , thus we will only concern ourselves with those with the following expressions. Adding all three right hand sides of the previous equations (and disregarding the powers of $10$ ) results in $3 + 3a_0 + 3a_1 + 3a_2 + 3a_3 \cdots$ $3(1 + a_0 + a_1 + a_2 + a_3 \cdots)$ Clearly, adding the digits of $a, a+1,$ and $a+2$ results in a number with its digits also adding up to $3$ or a multiple thereof. Thus, a number formed by the concatenation of $3$ consecutive integers is always divisible by $3$ .

Such a number can be written as $(n-1)\cdot 10^a + n\cdot 10^b + (n+1).$ Modulo 3, this is equivalent to $(n-1)\cdot 1^a + n\cdot 1^b + (n+1) = (n-1)+n+(n+1) = 3n \equiv 0.$ Thus the number is a multiple of 3.

Let us take the middle number as 'n' then the three consecutive numbers are (n-1) , n, (n+1) respectively.

Now, the sum of the three numbers= (n-1) + n +(n+1)=3n.

It is clear sum of three consecutive numbers always multiple of 3. This implies concatenation of any three consecutive numbers is always multiple of 3.

Therefore, the numbers Dr. Frankenthree creates in this way always multiples of 3

Let the three consecutive integers be $a$ , $a+1$ and $a+2$ , with $a$ having $n$ digits. Then the concatenated number $N = a\times 10^{2(n-1)}+(a+1)\times 10^{n-1} + (a+2)$ . Then, we have:

$\begin{aligned} N & \equiv a\times 10^{2(n-1)}+(a+1)\times 10^{n-1} +(a+2) \text{ (mod 3)} \\ & = a +(a+1) +(a+2) \text{ (mod 3)} \\ & = 3a + 3 \text{ (mod 3)} \\ & = 0 \text{ (mod 3)} \end{aligned}$

Yes , it is true that all numbers obtained in such a way are always divisible by 3.

All the numbers can be expressed as

n(n+1)(n+2)

where n is the root number, and the brackets DON'T signify multiplication.

How do we know it is divisible by three?

Simply sum the digits.

Summing

n+n+n+1+2 gives 3n + 3. Which is clearly divisible by 3.

This will also hold for ANY three consecutive rational numbers written side to side. Thank you.

Consider that with three consecutive numbers, you can substract one from the highest number and add it to the lowest number without changing the sum total. This results in three identical numbers, which obviously always can be divided by three.

大家好！these numbers is regular because they follow a rule.For example:789 = 888 - 99,and 141516 = 151515 - 99999.It's obvious that 888,99,151515 and 99999 can be divided by 3,so the result of their subtraction can be divided by 3.I'm Chinese so my English is not good,some of the words I even used translator.I might make some grammatical errors,sorry.

Method 1:

When we observe carefully we come to know that for any three numbers there will be at least one number which will be a multiple of 3 . Therefore, Dr. Frankenstein can always create such numbers .

Method 2 :

Let the three numbers be $\color{#3D99F6}a , \color{#20A900}(a - 1) , \color{#D61F06}(a - 2)$ . Now we will add the numbers and if it is divisible by 3 the number is also divisible by three .

a + a - 1 + a - 2 = 3a - 3 = 3(a - 1)

So, now it is clear that the number will always be a multiple the 3 .

If Dr. Frankenthree ALWAYS multiplies by three as shown in the problem, then all of products will be a multiples of three. It’s as simple as that!

Recall that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.

Let S(n) == n mod 3. So S(n) = 0 iff n is divisible by 3 iff the sum of its digits is divisible by 3. More generally, S(n) is the remainder when n is divided by 3.

Given an arbitrary integer n we wish to prove that the concatenation N=n|(n+1)|(n+2) is divisible by 3.

Using the 'sum of digits' formulation, it is clear that S(N) = S(n) + S(n+1) + S(n+2). Also it's clear that {S(n),S(n+1),S(n+2)} = {0,1,2} : if S(n) = 0, S(n+1) = 1, and S(n+2) = 2, and so forth. Each value appears exactly one. Thus S(N) = S(n) + S(n+1) + S(n+2) = 0 + 1 + 2 == 0 mod 3. Thus N is divisible by 3.

If we add all the intergars of a number up and its product is divisible by three, we know the number is divisible by three.

For this problem, because all ranges come in a group of three consecutive digits, we just have to break down the range into 3 singular groups of 3 of numbers. This would look like turning the number 123 into the categories of 1,2, and 3. Then we see that a pattern occurs where all numbers will fit into one of two possibilities: one number divisible by three paired with either two other odd numbers (such as 567) or paired with one odd and one even number (such as 678).

Since all odd numbers can be factored down into a 1 (a prime number multiplied by 1 and itself) and all even numbers can be factored down to a 2, the pair of one even, one odd, and one divisible by three will clearly add up to a number divisible by three since this version factored down will always look like this: 2+1+3=6 and 6/3=2 and is divisible.

Now for the second option where one digit is divisible by three and the other two are odd numbers, we can see that consecutively, all odd numbers one digit away from one another have a difference of three (see 567: 7-5=3). This means any number with option two will factor down to 3+3=6 and then 6/3=2 to show that this option will always be divisible by three as well.

Therefore, any three consecutive numbers will contain one number divisible by 3 and two other numbers that will add up to 3, thus creating a product that is always divisible by 3.

Three consecutive integers always have exactly one that is divisible by 3, this that number can be discarded altogether from the whole process, as the sum of its digits is surely divisible by three. The concatenation of the other two can be written as $10^{n}\times x + y$ where x and y are the numbers, n being the length of the number x (concatenating 46 and 47 means $10^2 \times 46 + 47$ ). But $10^{n}\times x + y = 9..9 \times x + x +y$ , 9..9 having n digits, all nines. Since that multiplication will always end up as divisible by 3, we can discard it as well, and what we are left with is x + y. If the original number we discarded was before or after these numbers (like 45, 46, 47, where we discarded 45), then x-1 is divisible by 3, y=x+1, so $x+y=(x-1+1)+(x-1+2)=2\times(x-1)+3$ which is divisible by three. If the original number we discarded was between these numbers (like 44,45, 46, where we discarded 45), then x+1 is divisible by 3, y=x+2, so $x+y=2x+2=2(x+1)$ , again, divisible by three. Q.E.D.

(x)+(x+1)+(x+2) = 3x+3 =3(x+1) hence divisible by 3

Given any $3$ consecutive numbers, there will always be exactly one with $mod$ $3 = 1$ , one with $mod$ $3 = 2$ , and one with $mod$ $3 = 0$ . Adding up all the mod's for the three numbers will result in something with $mod$ $3 = 0$ . The divisibility rule for three is that the sum of all the digits is also a multiple of $3$ . This also means that adding up the digits of the numbers doesn't change the module for it in $mod$ $3$

If the three numbers that you choose are N, N+1, N+2 We can make them all equal by saying N+1, N+1, N+1 By doing this, you are adding the same number 3 times. The sum is a multiple of three.

Consider three consecutive positive integers, one smaller, one slightly bigger and one that's the biggest. We'll call them n-1, n and n+1. If you wanna see that a number is divisible by 3, all you gotta do is sum up the integers of each digit (Ex 678 >>>>> 6+7+8=21=7X3).

So as a general rule (n-1)+n+(n+1)=n+n+n=3n.

See? All you need to know your times tables and you'll see how it all ADDS UP.

[N.B. You can also think of it this way. When ever you pick three consecutive positive numbers, there will always already be one divisible by 3 and the other two will be divisible by three when summed together.]

As a number is a multiple of 3 if the sum of its digits is a multiple of 3, the three numbers n, n+1, and n+2 are concatenated, not added, multiplied, etc. so if the numbers are always multiples of three, $\frac{n+n+1+n+2}{3}$ must be an integer. We can simplify the numerator into 3n+3, after which you factor out the three. as $\frac{3(n+1)}{3}$ will always be an integer for all integer values of n, so yes, this will always work.

My solution considers the sum of digits of Dr. Frankenthree's number. Let $x$ be a positive integer of $n$ digits $(n \geq 1)$ . Here is very many essentially similar solutions which states that $x + (x+1) + (x+2)$ is divisible by 3. That is obvious, but I cannot myself see, how this leads to the result, that Dr. Frankenthree's number would be divisible by 3. I have asked it many times in the comments of the solution, but I don't think I have got satisfying answer. So I think this problem is not proven for example in that solution most upvoted just now, and i'll try it myself also.

Let's assume first that x is not in form

(1) $999 \ldots 9$ (n pcs. 9's) or

(2) $999\ldots 98$ (n-1 pcs. 9's and 8)

That means, x is not number which every digits are 9 (case 1) or every other digits are 9 but the last is 8 (case 2) and then $x+1$ and $x+2$ are also n-digit numbers. Now, Dr. Frankenthree's number is (let's name this A)

$A = 10^{2n}x + 10^n(x+1) + (x+2) = (10^{2n} + 10^n + 1)x + (10^n + 2)$

Sum of digits of number $(10^{2n} + 10^n + 1)$ is $1+1+1=3$ . So, it is divisible by 3 and then $(10^{2n} + 10^n + 1)x$ is divisible by 3. Sum of digits of number $(10^n + 2)$ is $1+2=3$ , so it is also divisible by 3. Then A is sum of two numbers, both of them divisible by 3 and therefore A itself is divisible by 3.

At last, we have to consider cases (1) and (2). Let's examine them separately. In case (1), $x = 999 \ldots 9$ , and sum of digits of Dr. Frankenthree's number is

$n\cdot 9 + 1 + (1+1) = 9n +3 = 3(3n+1)$

Then it is divisible by 3 and the whole Dr. Frankenthree's number is divisible by 3. In case (2), $x = 999 \ldots 98$ , sum of digits of Dr. Frankenthree's number is

$((n-1)\cdot 9 +8) + (n\cdot 9) + 1 = 9n - 9 + 8 + 9n + 1 = 18 n = 3\cdot 6n$

And in the similar way Dr. Frankenthree's number is divisible by 3 in this last case also.

Is my solution understandable? I believe i've got it correct. If you find mistakes or something to clarify, please comment! I'm sorry about my english, it's not my native language.

Scala:

(1 to 2000).sliding(3).forall(_.mkString("").toLong % 3 == 0)

returns true

Such numbers are of the form $100n+10(n+1)+(n+2)=111n+12=3\cdot(37n+4)$ .

Which means they are all a multiple of 3. QED

A number is dividable by 3 if and only if the sum of its digits is dividable by 3.

The sum of the digits of the combined number is the sum over the sum of the digits of its components.

To make the sum over the sums dividable by 3 all the partial sums either need to be dividable by 3 themselves or 1 of them is dividable by 3 while the other are of the form 3n-1and 3m+1.

Now the question becomes, if the sum of the digits of 3 sequencal numbers falls in one of the cases above.

So what does it mean to add 1 to a number for the sum of its digits? Well if the last digit is anything from from 0 to 8 it's just increasing the sum by 1 as well. If the last digit is a 9 the next digit will be increased (adding 1 to the sum) while setting the last digit to 0 (subtracting 9 from the sum).

So adding 1 to a number either adds 1 to the sum of its digits or it adds 1 while subtracting a multiple of 9 (if there are multiple 9s at the end of the original number). The subtraction of 9 doesn't change the remainder of the number when dividing by 3.

So starting at some number n adding 1 and adding 1 again will create 3 numbers with sums of their digits of 3x-1 3y and 3z+1 (maybe in different order).

The sum of those sums is 3x + 3y + 3z +1 -1= 3(x+y+z). Therefore the sum of the digits of the combined number is dividable by 3 and the number is as well.

我们可以假设这个数由三个n位数组成的，分别为x-1、x、x+1，那么这个数L= $10^{2n}*(x+1)+10^n*x+(x+1)$ = $x*(10^{2n}+10^n+1)-(10^{2n}-1)$ 。 其中 $10^{2n}+10^n+1$ 各位数之和等于1+0+0+……+1+0+……+1=3是3的倍数， $10^{2n}-1$ 各位数之和等于9+9+……是3的倍数，那么L就是3的倍数。