Divisibility of 100 digits number

Question

prove that the $100$ digits number $1.....11$ is divisible by $101$

To show that the $100$-digit number consisting of $1$'s only, i.e. $11...1$ (with $100$ digits), is divisible by $101$, we can use the fact that if a number is divisible by $101$, then its alternating digit sum (the sum of every other digit, starting with the first digit) must be divisible by $101$.

Let's consider the alternating digit sum of the given number: $1 - 1 + 1 - 1 + ... + 1 - 1$

We can see that this alternating sum consists of $50$ pairs of $+1$ and $-1$. Thus, it simplifies to:

$(1 - 1) + (1 - 1) + ... + (1 - 1) = 0 + 0 + ... + 0 = 0$

Since the alternating digit sum is equal to $0$, which is divisible by $101$, we can conclude that the $100$-digit number consisting of $1$'s only is indeed divisible by $101$.

Welcome to Mathematics Stack Exchange. What's your question? — J. W. Tanner, May 05 '23 at 14:55
"if a number is divisible by $101$, then its alternating digit sum (the sum of every other digit, starting with the first digit) must be divisible by $101$" is false—it fails for $101$ itself for example. Also, your calculation doesn't calculate the sum you described anyway. In fact what you have proved is that the number is divisible by $11$. — Greg Martin, May 05 '23 at 15:19
The block $1111$ is divisible by $101$. You have $25$ such blocks. — Lozenges, May 05 '23 at 15:46
$10^2\equiv-1\bmod101$, so $10^{100}\equiv1\bmod101$, and your number is $\dfrac{10^{100}-1}9$ — J. W. Tanner, May 05 '23 at 16:27
Factor Theorem $\Rightarrow 100!+!1\mid 100^{50}!-1 = 9n,,$ so $101\mid n,$ by $,(101,9)=1$ and Euclid's Lemma. Or, equivalently, use mod arithmetic as in prior comment (which is one of the linked proofs of the Factor Theorem). — Bill Dubuque, May 07 '23 at 04:43
For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. — Bill Dubuque, May 07 '23 at 04:47
@BillDubuque: did you mean $100$ where you typed $50$ in the above comment? — J. W. Tanner, May 07 '23 at 04:56

score 0 · Answer 1 · edited May 05 '23 at 16:16

Number is $\underbrace{111\dots111}_{100}$ and we know that $101$$\underbrace{00\dots000}_{97}$ is divisible by $101$.

Then subtract $101$$\underbrace{00\dots000}_{97}$ from $\underbrace{111\dots111}_{100}$

we get $101$$\underbrace{111\dots111}_{96}$

Subtract $101$$\underbrace{000\dots000}_{96}$(which is divisible by $101$) from $101$$\underbrace{111\dots111}_{96}$

we get $\underbrace{111\dots111}_{96}$ .And if it is divisible by $101$ so is $\underbrace{111\dots111}_{100}$

we subtract $101$$\underbrace{000\dots000}_{93}$ (divisible by $101$) from $\underbrace{111\dots111}_{96}$

we get $101$$\underbrace{111\dots111}_{92}$

then subtract $101$$\underbrace{000\dots000}_{92}$ from $101$$\underbrace{111\dots111}_{92}$

we get $\underbrace{111\dots111}_{92}$. And if it is divisible by $101$ so is $\underbrace{111\dots111}_{100}$

Then just repeat the same procedure until we get $\underbrace{1\dots1}_{4}$ which is divisible by 101. And it means that $\underbrace{111\dots111}_{100}$ is divisible by $101$

In other words we know that $1111$ is divisible by $101$ then so is $\underbrace{11111111}_{8}=1111\cdot10000+1111$ , so is $\underbrace{111\dots1111}_{12}=1111\cdot100000000+1111\cdot10000+1111$ and so on.Which means every $\underbrace{111\dots1111}_{4n}$ number is divisible by $101$. So is $\underbrace{111\dots111}_{100}=\underbrace{111\dots111}_{4\cdot25}$

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. — Bill Dubuque, May 07 '23 at 04:45

Divisibility of 100 digits number

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