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Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are $$ 24,2534,253534,25353534, \ldots $$ What is the least value of $\mathrm{n}$ for which $\mathrm{a}_{\mathrm{n}}$ is divisible by $99 ?$

So, In this post,the present answer there get to the result that $a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$

But after that what i did is to multiply both sides by $9$ in the congruence

$35n - 11 \equiv 0 \pmod{99}$

$315n \equiv 0 \pmod{99}$

$18n \equiv 0 \pmod{99}$

dividing by $18$ ,we get $n \equiv 0 \pmod{11}$ but I did not get the other part of the congruence which was arrived in that answer by this process ?

thankyou

Ishan
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  • It is not true that $n\equiv 0 \pmod {11}\implies 35n\equiv 11\pmod {99}$. Just try $n=0$. – lulu Jan 09 '21 at 14:06
  • @lulu ok,so divison rule in congruences is not iff right ? – Ishan Jan 09 '21 at 14:08
  • Your "multiplication by $9$" introduced false solutions. There was no point to that anyway...just use the euclidean algorithm (or whatever) to find that $35\times 17\equiv 1\pmod {99}$. – lulu Jan 09 '21 at 14:09
  • @lulu but we can always multiply both sides of the congruences ? I just want to see where is the mistake. – Ishan Jan 09 '21 at 14:10
  • Since $\gcd(9,99)=9>1$ multiplying by $9$ introduced false solutions. for instance, if you started with $m\equiv 22\pmod {99}$ and multiplies by $9$ you'd get $9m\equiv 0\pmod {99}$ but this does not permit you to conclude that all $m\equiv 0\pmod {11}$ are solutions to the original (they aren't). – lulu Jan 09 '21 at 14:15
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    If you stick to factors that are prime to $99$, everything works. Multiplying by $17$ gives us $n\equiv 11\times 17\equiv 88 \pmod {99}$. – lulu Jan 09 '21 at 14:17
  • ok got it, thanks – Ishan Jan 09 '21 at 14:18

5 Answers5

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It's an easy one-liner using a fractional form of mDL = mod Distributive Law, viz.

Notice $\,\ \dfrac{\color{#c00}{11}}{35}\bmod \color{#c00}{11}(9)\,=\, \color{#c00}{11}(\color{#0a0}8)\,$ by $\color{#0a0}{\bmod 9\!:\ \dfrac{1}{35}\equiv \dfrac{1}{-1}\equiv 8},\ $ via

Theorem $\ \ \dfrac{\color{#c00}ab}d\bmod \color{#c00}ac\, =\, \color{#c00}a\left(\color{#0a0}{\dfrac{b}d\bmod c}\right)\ \ $ if $\ \ (d,ac) = 1$

Proof $\, $ Bezout $\Rightarrow$ exists $\, d' \equiv d^{-1}\pmod{\!ac}.\,$ Factoring out $\,\color{#c00}a\,$ by mDL

$$\color{#c00}abd'\bmod \color{#c00}ac\, =\ \color{#c00}a(bd'\bmod c)\qquad\qquad\qquad$$

and $\,dd' \equiv 1\pmod{\!ac}\Rightarrow dd' \equiv 1\pmod{\!c},\,$ so $\,d'\bmod c = d^{-1}\bmod c$

Bill Dubuque
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From $a_n= 24+(n - 1)35 \equiv 0 \pmod{99} $ you can see that $n-1$ is a multiple of $3$. Let it be $3m$, then dividing throughout by 3, $$0\equiv8 +35k \equiv 8+2k \pmod{33} $$ and therefore $$k \equiv -4 \pmod{33}$$

Then take $k=29$ and $n=88$.

N.B. Another way to look at your 'multiplying by 9' idea is that you can, of course, do this but you must multiply the base by the factor as well! In general this is unlikely to be a useful procedure. In your case you would change an equation mod $99$ into one mod $891$.

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You need to solve $35n \equiv 11 \pmod{99}$. This can be done via a variation of the (Extended) Euclidean Algorithm.

$$ \begin{array}{ll} 99n \equiv 0 \pmod{99} & (1)\\ 35n \equiv 11 \pmod{99} & (2)\\ -6n \equiv -33 \pmod{99} & (3) = (1)-3\times(2)\\ -n \equiv -187 \equiv -88 \pmod{99} & (4) = (2) + 6\times(3) \end{array} $$ Therefore $n\equiv 88 \pmod{99}$.

Neat Math
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    Yes, the above equational form of the fractional extended Euclidean algorithm is generally very efficient, but here it is a bit simpler to use the mod Distributive Law as I explain in my answer, – Bill Dubuque Jan 09 '21 at 20:20
  • @BillDubuque Thanks. I myself solved it with inverse reciprocity: $\frac{11}{35} \equiv \frac{11-99 \left( \frac{11}{99} \pmod{35} \right) }{35} \equiv \frac{11-99\cdot 4}{35} \equiv \frac{11(1-36)}{35} \equiv -11 \equiv 88 \pmod{99}$ with $\frac{11}{99} \equiv \frac 19 \equiv \frac{4}{36} \equiv 4 \pmod{35}$ and noticed I could factor out $11$ in the numerator. Great that I learned another trick. – Neat Math Jan 09 '21 at 22:18
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    Yes, I.R. can also leverage the common factor $9$ in the numerator and modulus. – Bill Dubuque Jan 09 '21 at 22:36
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Write the recursion $\bmod 99$ $$a_n = a_{n-1} + 35\bmod 99;\;a_1=24$$ we get the general solution: $$a_n=35 n-11$$ and $35n-11\equiv 0\bmod 99$ if $$35n\equiv 11\bmod 99\tag{1}$$

Consider the inverse of $35\bmod 99$, that is $17$.

$17\cdot 35\equiv 1 \bmod 99$

Multiply both sides of $(1)$ by $17$

$$n\equiv 187\bmod 99\to n\equiv 88\bmod 99$$

Raffaele
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$35n\equiv11\bmod99$ could also be solved with the Chinese remainder theorem.

It's equivalent to $-n\equiv2\bmod9$ and $n\equiv0\bmod11$.

Can you find a multiple of $11$ between $0$ and $99$ that is $2$ less than a multiple of $9$?

J. W. Tanner
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  • The claim that "It's equivalent to ..." requires justification at this level (it's not usually explicitly proved in textbooks). In particular the nontrivial direction $(\Leftarrow)$ often proves troubling for beginners. The necessity of proving this direction is often overlooked and can lead to problems such as extraneous solutions due to use of unidirectional (vs. bidirectional) inferences. – Bill Dubuque Jan 10 '21 at 08:21