Find the least value of $m+n$, where $1\le m

Question

I am no so sure how to do this without a calculator...

My method is like this.

So, $k=1978^n-1978^m$ is divisible by $1000$ which means $k$ is divisible by $8$ and $125$.

Since $8$ does not divide into $1978=2\times7\times 127$, $8$ divides $1978^m$ implies that $m\ge 3$.

Then obviously $125$ does not divide into $1978$ as well, so I need to figure out what $r$ is such that $1978^r\equiv 0\mod 125$. I have no idea how to find $r$ without using a calculator...

Could anyone help please?

I believe you meant to search for a power $s$ such that $1978^s\equiv 1 \pmod {125}$ Since $\varphi(5^3)=100$ we know that $s=100$ works but a priori it is possible that a smaller power will work as well, so you must rule out the divisors of $100$. — lulu, Dec 21 '20 at 10:44

score 3 · Accepted Answer · answered Dec 21 '20 at 11:19

You are on the right track but you've made a small error. Let $a = 1978$. Then as you noted, $1000 \mid a^n - a^m$, which leads to $a^m (a^{n-m} - 1) = 1000k$ for some integer $k$. Since $a^r - 1$ is odd for all positive integer $r$, we must have $8 \mid a^m$ so $m \ge 3$, and $125 \mid a^{n-m} - 1$. So instead of having a power of $a$ being congruent to $0$ modulo $125$, it leaves a remainder of $1$. From Euler's theorem, for $\gcd(a, b) = 1$, we have $$a^{\varphi(b)} \equiv 1 \pmod b$$ where $\varphi(b)$ is the Euler totient function that counts the number of positive integers between $1$ and $b$ that are relatively prime to $b$. One property of $\varphi$ is that if $b$ is the power of a prime, say $b = p^k$, then $$\varphi(p^k) = p^k - p^{k-1}.$$

Since $\gcd(1978,125) = 1$ and $125 = 5^3$, it follows that $\varphi(125) = 5^3 - 5^2 = 100$ and $$1978^{100} \equiv 1 \pmod {125}.$$ Thus $(m,n) = (3, 103)$ satisfies the given conditions. All that is left is to show that such an $n$ is minimal; i.e., there is no smaller exponent $r^* < 100$ for which $a^{r^*} \equiv 1$. This is not difficult to show and I leave as an exercise for the reader.

@heropup how could you say, $1978(i.e. 103)$ is a generator in $U(125)$, is there any easier way to say it? — A learner, Aug 05 '21 at 15:42

Sarvesh Ravichandran Iyer · Answer 2 · 2020-12-21T13:51:02.270

If $n>m$ are such that $1978^n - 1978^m$ is a multiple of $1000$, then we factorize as $1978^{m}(1978^{n-m} - 1)$ is a multiple of $1000$.

This is equivalent to $1978^{m}(1978^{n-m}-1)$ being a multiple of $8$ and $125$.

In the first case, note that $1978^{n-m} - 1$ is odd because $n \neq m$, hence coprime to $8$. $1978$ as a multiple of $2$ has multiplicity $1$ . So it follows that $m \geq 3$ is necessary and sufficient for divisibility by $8$ to hold.

In the second case, $1978^m$ is coprime to $125$ (by prime factorization) so we need the smallest value of $T = n-m$ such that $1978^T \equiv 1 \pmod{125}$.

Note that $1978 \equiv 103 \pmod{125}$. So we need to find $T$ such that $103^T \equiv 1 \pmod{125}$. Now, since $\phi(125) = 100$ it follows that $T$ is a divisor of $100$.

Start with the basic : if $103^T \equiv 1 \pmod{125}$ then in particular $103^T \equiv 1 \pmod{5}$, which comes to $3^T \equiv 1 \pmod{5}$, which is true only if $T$ is even. Thus, all odd values of $T$ are ruled out.

Next, we go modulo $25$ to see that $3^T \equiv 1 \pmod{25}$ must be true. We see that $\phi(25) =20$, so $T$ must be a factor of $20$. It is easy (because $3$ is small) to see that no number smaller than $20$ works. Thus, $T=20$ or $T = 100$.

Unfortunately, checking that $103^{20} \not \equiv 1 \pmod{125}$ is to be checked by hand. One can use repeated squaring to make this easier. (Also refer to the answer linked by J.W.Tanner below).

Thus, $T=100$, and $n = 3+100 = 103$ and $m=3,n=103$ form the smallest set of values with $m+n = 106$.

for alternative ways to figure $1978^{20}\pmod{125}$, see this question — J. W. Tanner, Dec 21 '20 at 13:33

score 0 · Answer 3 · answered Dec 21 '20 at 10:48

Not a 'real' answer, but it was too big for a comment. I know that you're looking for a solution without using a calculator or PC but maybe this gives some insight.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[1 <= m < n && 
     Last@IntegerDigits[1978^n] == Last@IntegerDigits[1978^m] && 
     Part[IntegerDigits[1978^n], 1 + Floor[Log10[1978^n]] - 1] == 
      Part[IntegerDigits[1978^m], 1 + Floor[Log10[1978^m]] - 1] && 
     Part[IntegerDigits[1978^n], 1 + Floor[Log10[1978^n]] - 2] == 
      Part[IntegerDigits[1978^m], 1 + Floor[Log10[1978^m]] - 2]], {m, 
    n, m + n}, Nothing], {n, 1, 200}, {m, 1, 200}] //. {} -> Nothing

Running the code gives:

Out[1]={{{3, 103, 106}}, {{4, 104, 108}}, {{5, 105, 110}}, {{6, 106, 
   112}}, {{7, 107, 114}}, {{8, 108, 116}}, {{9, 109, 118}}, {{10, 
   110, 120}}, {{11, 111, 122}}, {{12, 112, 124}}, {{13, 113, 
   126}}, {{14, 114, 128}}, {{15, 115, 130}}, {{16, 116, 132}}, {{17, 
   117, 134}}, {{18, 118, 136}}, {{19, 119, 138}}, {{20, 120, 
   140}}, {{21, 121, 142}}, {{22, 122, 144}}, {{23, 123, 146}}, {{24, 
   124, 148}}, {{25, 125, 150}}, {{26, 126, 152}}, {{27, 127, 
   154}}, {{28, 128, 156}}, {{29, 129, 158}}, {{30, 130, 160}}, {{31, 
   131, 162}}, {{32, 132, 164}}, {{33, 133, 166}}, {{34, 134, 
   168}}, {{35, 135, 170}}, {{36, 136, 172}}, {{37, 137, 174}}, {{38, 
   138, 176}}, {{39, 139, 178}}, {{40, 140, 180}}, {{41, 141, 
   182}}, {{42, 142, 184}}, {{43, 143, 186}}, {{44, 144, 188}}, {{45, 
   145, 190}}, {{46, 146, 192}}, {{47, 147, 194}}, {{48, 148, 
   196}}, {{49, 149, 198}}, {{50, 150, 200}}, {{51, 151, 202}}, {{52, 
   152, 204}}, {{53, 153, 206}}, {{54, 154, 208}}, {{55, 155, 
   210}}, {{56, 156, 212}}, {{57, 157, 214}}, {{58, 158, 216}}, {{59, 
   159, 218}}, {{60, 160, 220}}, {{61, 161, 222}}, {{62, 162, 
   224}}, {{63, 163, 226}}, {{64, 164, 228}}, {{65, 165, 230}}, {{66, 
   166, 232}}, {{67, 167, 234}}, {{68, 168, 236}}, {{69, 169, 
   238}}, {{70, 170, 240}}, {{71, 171, 242}}, {{72, 172, 244}}, {{73, 
   173, 246}}, {{74, 174, 248}}, {{75, 175, 250}}, {{76, 176, 
   252}}, {{77, 177, 254}}, {{78, 178, 256}}, {{79, 179, 258}}, {{80, 
   180, 260}}, {{81, 181, 262}}, {{82, 182, 264}}, {{83, 183, 
   266}}, {{84, 184, 268}}, {{85, 185, 270}}, {{86, 186, 272}}, {{87, 
   187, 274}}, {{88, 188, 276}}, {{89, 189, 278}}, {{90, 190, 
   280}}, {{91, 191, 282}}, {{92, 192, 284}}, {{93, 193, 286}}, {{94, 
   194, 288}}, {{95, 195, 290}}, {{96, 196, 292}}, {{97, 197, 
   294}}, {{98, 198, 296}}, {{99, 199, 298}}, {{100, 200, 300}}}

So, we can see that the lowest number it found is $\text{m}=3$ and $\text{n}=103$ such that $\text{m}+\text{n}=106$. This is true because:

In[2]:=1978^(3)
Out[2]=7738893352
In[3]:=1978^(103)
Out[3]=3245694113774954778728642579697330790826430286814377973924914348349172
1776912127907693002626855998102667117472974423953886955433471331120533
2387497964129735391536809964330889620953818913346685089195593781962547
1616597048762535229892714999678772716316673743259286605339715498711285
462611923097914966231058794697896139243678438050112682852352

Bill Dubuque · Answer 4 · 2020-12-21T20:45:53.450

As mentioned, it boils down to computing the order $\,K\,$ of $\,a = 1978\equiv -22\pmod{\!5^3}$.

Note $\!\bmod 5^3\!:\ 1\equiv a^K\Rightarrow\bmod 5\!:\ 1\equiv a^K\equiv 3^K\Rightarrow 4\mid K,\,$ so $\, K = 4k$

and $\bmod 5^2\!:\ a^4 \equiv 3^4\not\equiv 1,\,$ therefore we infer that $\,\color{#c00}{\nu_5(a^4\!-1) = 1}$
so applying LTE = Lifting The Exponent as here

$$\begin{align} &5^{\large 3}\!\mid a^{\large 4k}-1\\[.2em] \iff\ &3 \le \nu_5(a^{\large 4k}-1) = \color{#c00}{\nu_5(a^{\large 4}-1)}+\nu_5(k)\\[.2em] \iff\ & 2\le \nu_5(k)\\[.2em] \iff\ &5^{\large 2}\!\mid k\end{align}\qquad$$

so the least such $\,k\,$ is $\,5^2,\,$ so $\,a\,$ has order $\,K = 4k = 4\cdot 5^2=100\pmod{\!5^3}$.

Find the least value of $m+n$, where $1\le m

4 Answers4

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