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Find the last three digits of $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$$

I don't know how to continue my work: $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}=\sqrt{2^{2(2016)}+2\cdot (3\cdot 2)^{2016}+3^{2(2016)}}=\sqrt{2^{4032}+2\cdot 3^{2016}\cdot 2^{2016}+3^{4032}}=\sqrt{2^{4032}+2^{2017}\cdot 3^{2016}+3^{4032}}=?$$

Jyrki Lahtonen
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n2494460
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4 Answers4

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HINT: This can be factorised like this $$(2^{2016})^2+2\cdot2^{2016}\cdot3^{2016}+({3^{2016}})^{2}=(2^{2016}+3^{2016})^2$$ So the question is to find last three digits of $$2^{2016}+3^{2016}$$

LM2357
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The presence of the squares $4$ and $9$ here is a big hint to help you find that the expression $E := \sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}} = \sqrt{(2^{2016}+3^{2016})^2} = 2^{2016}+3^{2016}$.

Then finding hte last three digits of this expression is a matter of finding the remainders $2^{2016}\bmod 1000$ and $3^{2016}\bmod 1000$ and then adding.

The Carmichael function $\lambda$ gives us the maximum cyclic order of exponentiation and here $\lambda(1000) = {\rm lcm}(\lambda(2^3), \lambda(5^3)) = {\rm lcm}(2, 100) = 100 $. Any smaller cycle will divide this value. We could also use Euler's totient $\phi(1000)=400$ to similar effect. So this gives us $a^{2016} \equiv a^{16} \bmod 1000$ (noting $16\ge 3$, the greatest prime exponent in $1000$) and we can calculate this easily enough for both cases:

$ 2^{16}\equiv 1024\cdot 64 \equiv 24.64 \equiv 1536 \equiv 536 \bmod 1000$
$ 3^{16} \equiv 81^4 \equiv 6561^2 \equiv 561^2 \equiv 314721 \equiv 721 \bmod 1000$

And then we can add to get $E\equiv 257 \bmod 1000$.

Joffan
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    You do not have to calculate $561^2$. Render $561=500+61$, square the sun with the binomial theorem, and only the last terms $61^2$ contributes mod $1000$. – Oscar Lanzi Mar 21 '17 at 15:37
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    @OscarLanzi good point. I got lazy and just started hitting calculator buttons. – Joffan Mar 21 '17 at 15:55
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If you follow the hint from @user35508, lets make a congruence. $$2^{2016}+3^{2016} \text{ mód } 1000$$ Thanks to the Euler's totient function, we know that $a^{\phi(b)}\equiv 1 \text{ mod } b$ if mcd(a,b)=1. The factorization of $1000 = (5*2)^3$ so its trivial that $mcd(2,3,1000)=1$. $$\phi(1000)=\phi((5*2)^3)=\phi(5^3)\cdot\phi(2^3)=(5^3-5^2)\cdot(2^3-2^2)=400$$ Finally, lets divide $2016/400 = 5\cdot 400 + 16$ and apply it to the first formula. $$2^{5*400+16}+3^{5*400+16}\text{ mod } 1000 = (2^5)^{400}\cdot{2^{16}}+(3^5)^{400}\cdot{3^{16}}\text{ mod } 1000 = 2^{16}+3^{16}\text{ mod } 1000$$ The remainder is $43112257\text{ mod } 1000 = 257$ that is also the three last digits.

LukasHaring
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  • Wait. $2$ and $1000$ are coprime ? Turns out your answer works because the exponent $16$ is larger than the highest power shared between $1000$ and $2$ (which is $1$), but I don't get your reasoning... – Evariste Mar 21 '17 at 15:38
  • @Evariste mcd(2,3,1000) = 1 – LukasHaring Mar 21 '17 at 17:37
  • Right, but it does not justify the simplification of the power of $2$ since $\gcd(2,1000)\neq 1$. Following your reasoning, $\gcd(3,5,10)=1$ so the last digit of $3^4+5^4$ would be $1+1=2$ (since $\varphi(10)=4)$ but it's $6$ in reality. – Evariste Mar 21 '17 at 18:12
  • Euler's th. is for two variables a and b and $a^{\phi b} = 1 \mod b$. It doesn't follow for three variabes $a^{\phi b} + c^{\phi b} \equiv 1 + 1 \mod b$ unless both mcd(a,b) and mcd(c,b) are 1. As mcd(a,b) is 2 this is not valid. You can do $2^{2013} \equiv x \mod 125$ so $2^{2016} \equiv 8x \mod 1000$ though. – fleablood Mar 21 '17 at 18:38
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Let $x = 2^{2016}$ and $y = 3^{2016}$

So this is $\sqrt{x^2 + 2*xy + y^2} = x+y = 2^{2016} + 3^{2016}$ and we need to find the last three digits of that.

Eulers formula tells us that $3^{\phi(1000)} \equiv 1 \mod 1000$ so the totient function of $\phi(1000) = \phi(2^3*5^3) = \phi(2^3)\phi(5^3) = (2-1)2^2(5-1)5^2 = =400$ so $3^{400}$ will have the last three digits of $001$.

$3^{2016} = (3^{400})^5*3^{16} \equiv 3^{16} \mod 1000$ and

$3^2 = 10 - 1$ so

$3^{16} = (10 -1)^8 = \sum 10^i{8 \choose i}(-1)^{8-i} \equiv 1 - 8*10 + 28*100 \mod 1000 \equiv 800 - 80 + 1 \equiv 721$. So last three digits of $3^{2016}$ are $721$.

We can't use Euler's thereom to find $2^{2016} \mod 1000$ as $2$ and $1000$ are not relatively prime.

So .... well, $1000 = 2^3*125$ so we can solve $2^{2013} \equiv x \mod 125$.

As $2^{\phi (125)} = 2^{100} \equiv 1 \mod 125$ so $2^{2013}\equiv 2^{13}\mod 125$ and $2^{13} = 1024*8 \equiv 8*24 = 8(25-1) =192 \equiv 67 \mod 125$

So $2^{2013} = k*125 + 67$ so $2^{2016} = k*1000 + 8*67 \equiv 536 \mod 1000$

So $2^{2016} + 3^{2016} \equiv 536 + 721 = 1257\equiv 257 \mod 1000$

So the last three digits would be $257$

fleablood
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