Rom

\(72 = 2^3 3^2 \\ \text{This gets us, given the constraints on row length,}\\ 6\times 12,~8\times 9,~9\times 8,~12\times 6\\ \text{so 4 different row lengths}\)

\(\text{The odd numbers will be relatively prime to }2^{100} \\ \text{so }\phi(2^{100}) = \dfrac 1 2 2^{100} = 2^{99}\)

\(ab^4 = 12,~a^5b^5=7776 \\ \text{we want to get rid of }b \text{ by dividing a power of the first equation by a power of the 2nd}\\ \dfrac{(ab^4)^5}{(a^5b^5)^4}=\dfrac{(12)^5}{(7776)^4} \\ \dfrac{a^5b^{20}}{a^{20}b^{20}} = \dfrac{1}{a^{15}} = \dfrac{(2^23)^5}{(2^53^5)^4}=\dfrac{2^{10}3^5}{2^{20}3^{20}}=\dfrac{1}{2^{10}3^{15}}\\ a = (2^{10}3^{15})^{1/15} = 2^{2/3}3\)

Solving by substitution means that you solve for one variable in terms of the other from one equation, 

and then substitute that into the 2nd equation to actually solve for values.

\(y = -18x - 11 \\ \text{Now we substitute this into the 2nd equation}\\ -18x-11 = -22x+15 \\ \text{Now we have an equation in one variable and we solve it as usual}\\ 4x=26 \\ x=\dfrac{26}{4} = \dfrac{13}{2}\\ y=-18\left(\dfrac {13}{2}\right) - 11 = -128\)

\(\text{we need the lines to have the same slope and intercept}\\ 3x+4y=7 \Rightarrow y = -\dfrac 3 4 x + \dfrac 7 4\\ 6x+4y=c-4y \Rightarrow y= -\dfrac 3 4 x + \dfrac c 8\\ \dfrac c 8 = \dfrac 7 4 \\ c = 14 \)

do you mean f(x) = 3a² + 2bx + x

or do you mean f(x) = 3a² + 2bx + x² ?

\(\text{Oh, my bad it's }-25 - 2i \\ 2(x+i y) + 3 (x - i y ) = -25 - 2i \\ 5x - i y = -25 - 2i \\ x = -5, ~y=2 \\ z= -5 +2 i \)

\(2z+3z^* = -25 - 2i \\ \text{let }z=x+i y \\ 2(x+i y) + 3(x - i y) = 25 -2i \\ 5x - i y = 25-2i\\ x=5, y=2 \\ z = 5 + 2i \\\)

2 digit multiples of 19 are (19, 38, 57, 76, 95)

2 digit multiples of 31 are (31, 62, 93)

1 -> 19

19 -> 193 or 195

193 -> 1931 or 1938

195 -> 1957

1931 -> 19319

1938 -> X

1957 -> 19576

19319 -> 193193 or 193195

19576 -> 195762

193193 -> 1931931 or 1931938

193195 -> 1931957

195762 -> X

1931931 -> 19319319

1931938 -> X

1931957 -> 19319576

continuing this way for a bit we see the following pattern

strings of (4+3k) digits will have the last digit in the set (1,7,8)

strings of (5+3k) digits will have the last digit in the set (6, 9)

strings of (6+3k) digits will have the last digit in the set (2, 3, 5)

2002 = 4 + 3(666)

so the possible last digits in the string will be 1, 7, or 8

8 is the largest of these