Note that \(81239744=81238123+1621=8123(10001)+1621\). Thus the least residue is \(1621\). (Long division also works.)

We work modulo \(10\) to find: \[\begin{align*} &12580 \times (4236 + 1284) + (1201+412)^3\\ \equiv &0 \times (6 + 4) + (1+2)^3\\ \equiv &(3)^3 \equiv 27 \equiv 7 \pmod{10} \end{align*}\] Since the original number is positive, its last digit must be \(7\).

Dividing by \(2\), we obtain the congruence \[5x \equiv 3 \pmod{17}.\] Adding \(17\) to the right hand side we have \[5x \equiv 20 \pmod{17}\] and we can cancel \(5\) since it is relatively prime to \(17\). Thus \(x \equiv 4 \pmod{17}\) is the unique solution modulo \(17\). However, the original problem was modulo \(34\), thus the solutions are \(4\) and \(4+17=21\).

First we solve \(4x \equiv 1 \pmod{3}\) to obtain \(x=1+3y\) for an integer \(y\). Now we plug this into the second equation: \[\begin{align} 2x &\equiv 1 &\pmod{5}\\ 2(1+3y) &\equiv 1 &\pmod{5}\\ 6y &\equiv -1 &\pmod{5}\\ y &\equiv -1 &\pmod{5} \end{align}\] Thus \(y=-1+5z\) for some integer \(z\). Thus \(x=1+3(-1+5z)=15z-2\) Now we plug this into the third equation: \[\begin{align} 3x &\equiv 1 &\pmod{11}\\ 3(15z-2) &\equiv 1 &\pmod{11}\\ 45z-6 &\equiv 1 &\pmod{11}\\ z &\equiv 7 &\pmod{11} \end{align}\] Thus \(z=7+11k\) for some integer \(k\). Thus \[\begin{align*} x&=15z-2\\ &=15(7+11k)-2\\ &=103+165k \end{align*}\] and we conclude that the integer solutions are those of the form \(x=103+165k\) for an integer \(k\).

By Fermat’s Theorem, we have \(70^{10} \equiv 1 \pmod{11}\) since \(11\) is prime. Thus \(70^{60} \equiv (70^{10})^6 \equiv 1^6 \equiv 1 \pmod{11}.\) We conclude the remainder is \(1\).

Alternative approach: We can explicitly work modulo \(11\) without reference to Fermat’s Theorem. For example \(70 \equiv 4 \pmod{11}\). Thus \[\begin{align*} & 70^{60} \equiv 4^{60} \\ \equiv & (4^2)^{30}\equiv 5^{30}\\ \equiv & (5^2)^{15} \equiv 3^{15}\\ \equiv & (3^5)^3 \equiv 1^{3}\\ \equiv & 1 \pmod{11} \end{align*}\]

Let \(f(n)=(n+1)(n+2)(n+3)\). Computing \(f(n)\) for each least residue modulo \(6\) we find: \[\begin{matrix} n & 0 & 1 & 2 & 3 & 4 & 5 \\ f(n) & 6 & 24 & 60 & 120 & 210 & 336. \end{matrix}\] All of these are divisible by \(6\). Thus \(f(n) \equiv 0 \pmod{6}\) for all integers \(n\). We conclude that \((n+1)(n+2)(n+3)\) is divisible by \(6\) for all integers \(n\).

Quicker approach: Working modulo \(2\) and \(3\) right from the beginning, it is very quick to check that \(f(n) \equiv 0 \pmod{2}\) and \(f(n) \equiv 0 \pmod{3}\) for all integers \(n\). Since \(f(n)\) is divisible by both \(2\) and \(3\), it must be divisible by \(6\).

Longer approach: This will work, but is much more involved than using modular arithmetic! We prove the result by induction. The basis step \(n=1\) follows since \(f(1)=6\) is divisible by \(6\). Now suppose \(f(n)\) is divisible by \(6\) for some positive integer \(n\). A tedious computation shows that \[f(n+1)=f(n)+3(n+3)(n+2).\] If \(n\) is even, then \(n=2k\) for some integer \(k\) and \[(n+3)(n+2)=(2k+3)(2k+2)=2(2k+3)(k+1)\] is even. If \(n\) is odd, then \(n=2k+1\) for some integer \(k\) and \[(n+3)(n+2)=(2k+4)(2k+3)=2(k+1)(2k+3)\] is even. Thus, for any integer \(n\), we see \((n+3)(n+2)\) is even. Thus \(3(n+3)(n+2)\) is divisible by \(6\). By the induction hypothesis, \(f(n)\) is divisible by \(6\). Thus \(f(n+1)\) is divisible by \(6\). The result follows by mathematical induction.

We proceed by mathematical induction. For the basis step \(n=1\), we observe that \[\sum_{k=0}^{0} 2^k = 1 = 2^1-1\] as desired. Now suppose the equation is true for a positive integer \(n\); we will show it holds for \(n+1\). Indeed \[\begin{align*} &\sum_{k=0}^{n} 2^k\\ =& \left( \sum_{k=0}^{n-1} 2^k\right) + 2^n\\ =& (2^n-1) + 2^n\\ =& 2\times 2^n-1\\ =& 2^{n+1}-1 \end{align*}\] as desired. Thus the result holds by mathematical induction.

Let’s compute all the cubes modulo \(7\): \[\begin{matrix} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ x^3 & 0 & 1 & 1 & 6 & 1 & 6 & 6 \end{matrix}\] Since \[x^3+14y^3 \equiv 4 \pmod{7}\] is equivalent to \[x^3 \equiv 4 \pmod{7}\] and no cube is congruent to \(4\) modulo \(7\), we conclude the original equation has no integer solutions.