Coefficient in a product

Consider the exponents in base $3$ . In each of the factors, the exponents are in the form of $10...0$ or $20...0$ . Since there are no $2$ factors that have the same number of zeroes at the end of each exponent of $x$ , the exponent of each term of $x$ after the expansion would have a different base $3$ representation than any other term. Hence there is only one term that has $x^{2013}$ in it even after we expand but do not combine like terms.

Now we identify the coefficient of that term.

$2013=2202120$ . The only coefficients in the factors that are not $1$ s are the ones with an even exponent ). We multiply those coefficients to get $(6)(5)(3)(1)=90$ . Hence the coefficient of $x^{2013}$ is $\boxed{90}$ .

We write out the product: $\begin{aligned}&(1+x^3+x^6)(1+x^9+2x^{18})(1+x^{27}+3x^{54})\\ &(1+x^{81}+4x^{162})(1+x^{243}+5x^{486})(1+x^{729}+6x^{1458})\end{aligned}$ Now, consider the final term of the product. To get powers that add to 2013, we must take $6x^{1458}$ . Otherwise, we get a maximum of 729+486+162+54+18+6=1455. We now add 2013-1458=555 to the exponent. $\begin{aligned}&(1+x^3+x^6)(1+x^9+2x^{18})(1+x^{27}+3x^{54})\\ &(1+x^{81}+4x^{162})(1+x^{243}+5x^{486})(6x^{1458})\end{aligned}$ Consider the fifth term. If we do not use $5x^{486}$ , we can only get a total exponent of 243+162+54+18+6=483. Therefore, we must use $5x^{486}$ , and must add 555-486=69 to the exponent. $\begin{aligned}&(1+x^3+x^6)(1+x^9+2x^{18})(1+x^{27}+3x^{54})\\ &(1+x^{81}+4x^{162})(5x^{486})(6x^{1458})\end{aligned}$ In the fourth term in the product, we must clearly use the 1. $\begin{aligned}&(1+x^3+x^6)(1+x^9+2x^{18})(1+x^{27}+3x^{54})\\ &(1)(5x^{486})(6x^{1458})\end{aligned}$ In the third term of the product, if we don't use $3x^{54}$ , we can only achieve an exponent of 27+18+6=51. Therefore, we must use $3x^{54}$ , and must add 69-54=15 to the exponent. $(1+x^3+x^6)(1+x^9+2x^{18})(3x^{54})(1)(5x^{486})(6x^{1458})$ Now, it is clear we must use $x^6$ and $x^9$ . Therefore, there is only one way to get a power of 2013 on $x$ . $(x^6)(x^9)(3x^{54})(1)(5x^{486})(6x^{1458})$ The answer is $3\cdot 5\cdot 6=90$ .

Firstly, we write $2013 = 2 \cdot 3^6 + 2 \cdot 3^5 + 0 \cdot 3^4 + 2 \cdot 3^3 + 1 \cdot 3^2 + 2 \cdot 3^1$ . Note that this representation is unique, and corresponds to multiplying $6 \cdot x^{2 \cdot 3^6}$ , $5 \cdot x^{2 \cdot 3^5}$ , $1$ , ..., $x^{2 \cdot 3^1}$ . Clearly, the coefficient of $x^{2013}$ is $6 \cdot 5 \cdot 1 \cdot 3 \cdot 1 \cdot 1 = 90$ .

$\displaystyle\coprod_{i=1}^{6}(1+x^{3^i}+ix^{2\cdot 3^i})$

$=\displaystyle\coprod_{i=1}^{6}(x^{0\cdot 3^i}+x^{1\cdot 3^i}+ix^{2\cdot 3^i}),$

$=\displaystyle\sum x^{k_1\cdot 3^1+k_2\cdot 3^2+k_3\cdot 3^3+k_4\cdot 3^4+k_5\cdot 3^5+k_6\cdot 3^6}$

at which $\left \{0;1;2 \right \} \ni k_i$ is a coefficient of $3^i$ in the exponents of terms in the sum $(x^{0\cdot 3^i}+x^{1\cdot 3^i}+ix^{2\cdot 3^i}).$

Assume $x^{k_1\cdot 3^1+k_2\cdot 3^2+k_3\cdot 3^3+k_4\cdot 3^4+k_5\cdot 3^5+k_6\cdot 3^6}=x^{2013}$

$\Leftrightarrow k_1\cdot3^0+k_2\cdot 3^1+k_3\cdot 3^2+k_4\cdot 3^3+k_5\cdot 3^4+k_6\cdot 3^5=671$

Because the way a number is shown in a counting system is unique,

and $\left\{ \begin{array}{l l} k_1\cdot3^0+k_2\cdot 3^1+k_3\cdot 3^2+k_4\cdot 3^3+k_5\cdot 3^4+k_6\cdot 3^5_{(10)}=\overline{k_6k_5k_4k_3k_2k_1}_{(3)}\\ 671_{(10)}=\overline{220212_{(3)}} \end{array} \right.$ ,

so that $k_1=2,k_2=1,k_3=2,k_4=0,k_5=2,k_6=2.$

Infer the coefficient of $x^{2013}$ is $1\cdot1\cdot3\cdot1\cdot5\cdot6=90.$

To observe things easier, we rewrite the expression as $(1+x^3+x^6)(1+x^9+2x^{18})(1+x^{27}+3x^{54})(1+x^{81}+4x^{162})(1+x^{243}+5x^{486})(1+x^{729}+6x^{1458})$

In the expansion, we select one term from each of the $6$ brackets and multiply them together.

We will argue that the only way to get $x^{2013}$ is to multiply the terms in $x^6,x^9,x^{54},1, x^{486}$ and $x^{1458}$ together (one from each of the brackets). We do this by selecting the terms in the brackets from back to front.

Indeed, $x^{1458}$ must be chosen, or else the maximum power of $x$ we can get is only $6+18+54+162+486+729 = 1455$ , not enough to reach $2013$ .

Once $x^{1458}$ is already chosen, the sum of the powers of the remaining terms must be $2013-1458=555$ . Arguing similarly, the term $x^{486}$ must be chosen, or else we would only get a maximum power of $6+18+54+162+243 = 483$ .

Thus, the total power of the rest of the terms is $555-486 = 69$ . The only possible term with power less than $69$ from the fourth bracket is $1$ . Using the similar reasoning, the term in $x^{54}$ will be chosen from the third bracket. Till this point, the remaining powers we need is $2013 - 1458 - 486 - 54 = 15$ , and the only way to get the power of $15$ from the first two brackets is to multiply $x^6$ and $x^9$ .

Thus the term in $x^{2013}$ is $(x^6)(x^9)(3x^{54})(1)(5x^{486})(6x^{1458}) = 90x^{2013}$ . The coefficient is $\fbox{90}$

what i've done:

$\prod_{i=1}^{6}3^i = \frac{3^7 - 1}{3 - 1} = 1093$ and $\prod_{i=1}^{6}2*3^i = 3^7 - 1= 2186$

we've $\prod_{i=1}^{6}3^i = \frac{3^7 - 1}{3 - 1} < 2013$ so we choose $2*3^6 = 1458$ , and we subtract it from $2013$ , gives us $2013 - 2*3^6 = 555$

then $\prod_{i=1}^{5}3^i = \frac{3^6 - 1}{3 - 1} < 555$ we choose $2*3^5 = 486$ , and we subtract it from $555$ , gives us $555 - 2*3^5 = 69$

we've $3^4 = 81 > 69$ so we ignore implies we ignore also $2*3^4$

then $\prod_{i=1}^{3}3^i = \frac{3^4 - 1}{3 - 1} < 69$ we choose $2*3^3 = 54$ , we subtract it from $69$ , gives us $69 - 2*3^3 = 15$

finally we've $2*3^2 > 15$ so we choose $3^2$ and we subtract it from $15$ to get $6$ and $2*3^1 = 6$ .

$kx^{2013} = 6x^{2*3^6} * 5x^{2*3^5} * 3x^{2*3^3} * x^{3^2} * 1x^{2*3^1}$ so

$k = 6 * 5 * 3 * 1 = 90$

Each factor can be written as $x^{0.3^i} + x^{1.3^i} + ix^{2.3^i}$ . Then, each term at the expansion is determined uniquely by the 6 factors, depending on their representation in base 3. In other words, for $x^k$ , looking at the representation of $k$ in base 3, we can infer how $x^k$ was formed, and there's only one way of doing so. Of course since $k$ has to be a multiple of 3, then the last digit is zero, and we ignore it. For any digit corresponding to $3^i$ , if that digit is 0, then it must have come from the "1" term. If it's 1, then it must have come from the $x^{3^i}$ term, and if it's 2, it must have come from $i.x^{2.3^i}$ term, for that particular factor.

Note that each term contributes a coefficient of 1, except the $i$ if it comes from $i.x^{2.3^i}$ , so we need to only care for "2" in base 3 representation.

Next, we find 2013 in base 3, and that is 2202120. The twos belong to $i = 1,3,5,6$ so the coefficient is $1.3.5.6$