catmg

 (a-bi)^2 = 8 - 6i + 13 - 4i

a^2 - b^2 - 2abi = 21 - 10i

-2abi = -10i

ab = 5

a^2 - b^2 = 21

Can you take it from here?

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LCM (2, 3, 5, 7, 19) = 3990

3990, 3990*2

It stops there since 3990*3 is not a 4 digit number. 

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g(2x) = 2(g(x))

(2x + 3)/4 = 2((x + 3)/4)

(2x + 3)/4 = (x + 3)/2

2(2x + 3) = 4(x+3)

4x + 6 = 4x + 12

6 = 12???

I don't think this is possible.

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Nice, I had the same idea but I thought that 0 was a very suspicious answer. 

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sin(7x) = cos (3x − π/14) + 2

cos(7x - pi/2) = cos(3x − π/14) + 2

Note that cos(a) has to return a number from -1, to 1. 

So the only way this equation is possible is if cos(3x − π/14) = -1 and cos(7x - pi/2) = 1. 

cos(3x − π/14) = -1

cos(7x - pi/2) = 1

Can you take it from here?

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Your question is incomplete. 

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8 choices for the first 4 digit. 

4 choices for the last digit. (1, 3, 5, 7)

Any idea how you could approach it from here?

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12 with line = 12/99

321 with line = 321/999

(12/99)/(321/999)

Can you take it from here?

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I will give this my best shot. 

110 = 2*5*11

I'm going to assume that n only has factors of 5 and 11. 

n^3 = 5^a * 11^b

x = number of factors of 2 in 110n^3

y = number of factors of 5 in 110n^3

z = number of factors of 11 in 110n^3

(x+1)(y+1)(z+1) = number of factors in 110n^3

(2)(2+a)(2+b) = 110

a = 3, b = 9 or a = 9, b = 3

It doesn't really matter which one you choose. 

n^3 = 5^9 * 11^3

n = 5^3 * 11

n^4 = 5^12 * 11^4

81*n^4 =  3^4 * 5^12 * 11^4

5*13*5 = 325

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Oh oops.

Thanks for pointing it out.

Tsk tsk, can't believe I misread that. 

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