Michael's base 6 remainder

Notice that $1043_6=6^3+4\times 6^2+3=3^5$ Therefore we desire to find the remainder when $104221432510_6=6^{11}+4\times 6^9+2\times 6^8+2\times 6^7+1\times 6^6+4\times 6^5+3\times 6^4+2\times 6^3+5\times 6^2+1\times 6$ is divided by $3^5$ .
Notice that $6^i\equiv 0\pmod{3^j}$ for $i>j$ therefore when reducing the above expression mod $3^5$ the first terms cancel and we arrive at $104221432510_6\equiv 3\times 6^4+2\times 6^3+5\times 6^2+1\times 6\equiv 0+432+210+1\equiv 132\pmod{3^5}$

The trick to this problem is to restate the question in terms of long division.

$104221432510$ base 6 is equivalent to:

$6^{11} + 4 * 6^{9} + 2 * 6^{8} + 2 * 6^{7} + 6^{6} + 4 * 6^{5} + 3 * 6^{4} + 2 * 6^{3} + 5 * 6^{2} + 6$

Which if we let $x = 6$ is the same thing as:

$x^{11} + 4x^{9} + 2x^{8} + 2x^{7} +x^{6} + 4x^{5} + 3x^{4} + 2x^{3} + 5x^{2} + x$

Using the same argument, $1043$ base 6 equals:

$x^{3} + 4x + 3$

Now we divide:

$\frac {x^{11} + 4x^{9} + 2x^{8} + 2x^{7} +x^{6} + 4x^{5} + 3x^{4} + 2x^{3} + 5x^{2} + x}{x^{3} + 4x + 3}$

Which, through long division, equals: $x^{8} - x^{5} +2x^{4} +5x^{3} - x^{2} - 23x - 9$ with a remainder of:

$\frac {100x^{2} + 106x +27}{x^{3} + 4x +3}$

Plugging $6$ back in for $x$ , we get

$\frac {3600 + 636 + 27} {216 + 24 +3} = \frac {4263}{243} = 17$ with a remainder of $\boxed{132}$

1043 *1 = 1043

1043 * 2 = 2130

1043*3 = 3213

1043 * 4 = 4300

1043 * 5 = 5343

now that we have the table of 1043 in base 6 just do long division and find the remainder....we will get 340 as the remainder...convert it to decimal

$104221432510_6 \rightarrow 407109210$

$1043_6 \rightarrow 243$

$407109210mod(243)=132$

First, we change the numbers into base $10$ .

$104221432510_{6}$ becomes $6^{11}+4.6^{9}+2.6^{8}+2.6^{7}+6^{6}+4.6^{5}+3.6^{4}+2.6^{3}+5.6^{2}+6$

Well, since $1043_{6}$ is $243$ , which is $3^{5}$ , then all terms that is divisible by $3^{5}$ when divided by $243$ produces a positive integer k , which does nothing with the remainder. So, we can eliminate those terms.

We're left with just

$\frac{2.6^3+5.6^2+6}{243}$

So, the remainder is $\boxed{132}$ .

Sorry if anything isn't where it should be since this is the first time I write my solution here. Thanks!

Convert both in base 10

104221432510 (6)==407109210 (10) and 1043(6)==243 (10)

Answer:: 407109210 % 243 == 132

The main issue of the problem is the conversion from base 6 to base 10.

Basically, the conversion can be done as so:

$104221432510 = 1$ x $6^{11} + 0$ x $6^{10} + 4$ x $6^9 + 2$ x $6^8 + 2$ x $6^7 +$

$1$ x $6^6 + 4$ x $6^5 + 3$ x $6^4 + 2$ x $6^3 + 5$ x $6^2 + 1$ x $6^1 + 0$ x $6^0$

As you can see from above, each of the digit that is initially expressed as x $10^x$ , where $x$ is the placing of the digit, like hundredth => 2, is now converted to $6^x$

The final value of $104221432510$ base $6$ to base $10$ is $407109210$

Similarly, $1043$ base $6$ to base $10$ is $243$

Using basic long division, $\frac{407109210}{243} = 1675346\frac{132}{243}$

Thus, $132$ is the remainder.

As the conversion to base 10 is initially done before the division, the remainder itself would already be in base $10$ .

Hence, the answer is $\boxed{132}$