Find the least value of $m+n$, where $1\le m<n$ such that the last three digits of $1978^n$ are the same as $1978^m$. 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?
If $n&gt;m$ 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 $m+n = 106$.
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[&quot;Global`*&quot;];
ParallelTable[
  If[TrueQ[1 &lt;= m &lt; n &amp;&amp; 
     Last@IntegerDigits[1978^n] == Last@IntegerDigits[1978^m] &amp;&amp; 
     Part[IntegerDigits[1978^n], 1 + Floor[Log10[1978^n]] - 1] == 
      Part[IntegerDigits[1978^m], 1 + Floor[Log10[1978^m]] - 1] &amp;&amp; 
     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}] //. {} -&gt; 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