Find The Last Three Digits

Since we are asked for the last 3 digits of $a_{2014}$ we can take the modulo $10^3$

$3^{2013}-2^{2013}\equiv_{10^3} 3^{13}-2^{13}\equiv_{10^3} 131$

$3^{131}-2^{131}\equiv_{10^3} 299$

$3^{299}-2^{299}\equiv_{10^3} 979$

$3^{979}-2^{979}\equiv_{10^3} 779$

$3^{779}-2^{779}\equiv_{10^3} 779$

So we get that $a_i\equiv_{10^3}779 \ \forall \ i\geq 5$

Hence $a_{2014} \equiv_{10^3} \boxed{779}$

Lemma $3^{1000n+a}-2^{1000n+a} \equiv 3^a-2^a \pmod{1000}$ , for $n \geq 1$ and $a \geq 3$

Prove Lemma From fermat litle theorem $3^1000 \equiv 1 mod 1000$ . It's mean $3^{1000n+a} \equiv 3^a \pmod{1000}$

$2^1000 \equiv 376 \pmod{1000}$ Suppose that $a=3+x$

$2^{1000n+3+x} \equiv 376 \times 2^{3+x} \equiv 2^{3+x} \equiv 2^a \pmod{1000}$

From apply the lemma:

For next step I use wolframalfa.com for reach the result:

$a_2=3^{2013}-2^{2013} \equiv 3^{13}-2^{13} \equiv131 \pmod{1000}$

$a_3 \equiv 3^{131}-2^{131} \equiv 299 \pmod{1000}$

$a_4 \equiv 3^{299}-2^{299} \equiv 979 \pmod{1000}$

$a_5\equiv 3^{979}-2^{979} \equiv 779 \pmod{1000}$

$a_6 \equiv 3^{779}-2^{799} \equiv 779 \pmod{1000}$

So, $a_{2014} \equiv a_{14} \equiv \fbox{779} \pmod{1000}$

To approach this problem, we must first be able to find the last 3 digits for $3^n$ and $2^n$ .

Consider this method for finding the last 3 digit for $3^n$ , for this case , $3^{2013}$ , $3^{2013}$ = $3(10-1)^{1006}$ . Hence, expanding the binomial expansion, we have :

$3(10-1)^{1006}$ $= 3[(-1)^{1006} + (1006)(-1)^{1005}(10) + \frac{(1006)(1005)}{2}(-1)^{1004}(10)^2]$ (mod $1000$ )

$= 3 [ 1 - 60 + 500 ]$ (mod $1000$ )

$= 1323 = 323$ (mod $1000$ )

For powers of 2, in this case, if you were to write out 100 powers of 2, you would notice that for $2^n$ , the cycle repeats in the period of $100$ , excluding $2^1$ and $2^2$ .

Hence, $2^{2013} = 2^{13} = 192$ (mod $1000$ )

Thus, we now find the value of $a_2$ , which is equals to $323 - 192 = 131$

Repeat the process for finding last 3 digits of $3^n$ using the binomial expansion and finding last 3 digits of $2^n$ by taking a mod $100$ on the power, writing a table of first 100 powers of $2^n$ and refer to. It would be too tedious to type out all 100 powers.

Hence, $3^{131} = 947$ (mod $1000$ ) , $2^{131} = 2^{31} = 648$ (mod $1000$ ) . Thus , $a_3 = 299$ (mod $1000$ ).

Thus, $3^{299} = 667$ (mod $1000$ ) , $2^{299} = 2^{99} = 688$ (mod $1000$ ) . Hence , $a_4 = 979$ (mod $1000$ ).

$3^{979} = 867$ (mod $1000$ ) , $2^{979} = 2^{79} = 779$ (mod $1000$ ) . Hence , $a_5 = 779$ (mod $1000$ )

$3^{779} = 867$ (mod $1000$ ) , $2^{779} = 2^{79} = 779$ (mod $1000$ ) . Hence , $a_6 = 779$ (mod $1000$ )

From this value of $a_6$ , where the powers of 2 and 3 regenerates back the value of its power. Hence, we can deduce that

$a_{n+1} = a{n}$

Thus, $a_{2013} = a{6} = \boxed{779}$