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708

Mmm 

I do not know the best way to do this but I will give it a go

13^1031 mod 599

13^1=13          = 13 mod 599

13^2=169        =  169 mod 599

13^3 = 2197     = 400 mod 599

13^4=28561    = 408 mod 588

13^5                = 512 mod 599

13^6                = 67 mod 599

13^7                = 272 mod 599

13^1031= 13^(6*171)*13^5

13^1031= ((67mod599)^171)*512mod599

13^1031= ((67mod599)^5)^34*(67mod599)*512mod599

13^1031= (72mod599)^34*(67*512mod599)

13^1031= (72mod599)^34*(161mod599)

13^1031= ((72mod599)^4)^8*(72mod599)^2*(161mod599)

13^1031= (320mod599)^8*(392mod599)*(161mod599)

13^1031= (320mod599)^3*(392mod599)^3*(320mod599)^2*(392mod599)*(161mod599)

13^1031= (304mod599)*(304mod599)*(570mod599)*(392*161mod599)

13^1031= ((304*304)mod599)   * (570mod599)*(217mod599)

13^1031= (170)mod599   *  (570*217)mod599

13^1031= (170)mod599   *  (296)mod599

13^1031= (170*296)mod599

13^1031   =     4   mod 599      

the answer is 4

Verification:

SORRY    WRONG ANSWER:

A hectoliter is 100 liters     so the bathtub has  2 x 100 = 200 liters

200 - 25 = 175 liters remaining

5*2^(-5) = 0.15625

Ambiguous syntax     Here are two possibilities:

2 / (3/18) =  2 x 18/3 = 36/3 = 12

(2/3) / 18 =  2/3  /  18/1 =  2/3 x 1/18 = 2/54 = 1/27

why

post best meme

ln(3x) = 5

3x = e^(5)

x = e^(5)/3 =

e^5/3 = 49.4710530341922011

A hectoliter is 100 liters     so the bathtub has   25 x 100 = 2500 liters

2500 - 25 = 2475 liters remaining

You will have to have common denoinator (30) to add fractions

3 1/3 = 10/3          10/3 = 100/30

1 7/10 = 17/10    17/10 =   51/30

100/30 + 51/30 = 151/30 = 5 1/30

What is the remainder of: 13^1031 mod 599 =?

\(\begin{array}{|rcll|} \hline && 13^{1031} \pmod{599} \\ &\equiv & 13^{2\cdot 515+1} \pmod{599} \\ &\equiv & (13^{2})^{515}\cdot 13 \pmod{599} \quad & | \quad 13^2 \pmod{599} = 169 \\ &\equiv & 169^{515}\cdot 13 \pmod{599} \\ &\equiv & 169^{2\cdot 257+1}\cdot 13 \pmod{599} \\ &\equiv & (169^{2})^{257}\cdot 169 \cdot 13 \pmod{599} \quad & | \quad 169^2 \pmod{599} = 408 \\ &\equiv & 408^{257}\cdot 169 \cdot 13 \pmod{599} \\ &\equiv & 408^{2\cdot 128+1}\cdot 169 \cdot 13 \pmod{599} \\ &\equiv & (408^{2})^{128}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \quad & | \quad 408^2 \pmod{599} = 541 \\ &\equiv & 541^{128}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \\ &\equiv & (541^{2})^{64}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \quad & | \quad 541^2 \pmod{599} = 369 \\ &\equiv & 369^{64}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \\ &\equiv & (369^{2})^{32}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \quad & | \quad 369^2 \pmod{599} = 188 \\ &\equiv & 188^{32}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \\ &\equiv & (188^{2})^{16}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \quad & | \quad 188^2 \pmod{599} = 3 \\ &\equiv & 3^{16}\cdot 408 \cdot 169 \cdot 13 \pmod{599} \quad & | \quad 3^{16} = 43046721 \pmod{599} = 185 \\ &\equiv & 185\cdot 408 \cdot 169 \cdot 13 \pmod{599} \\ &\equiv & 165829560 \pmod{599} \\ &\equiv & 4 \pmod{599} \\ &\equiv &\mathbf{ 4 }\\ \hline \end{array}\)

Looking for an equation that looks like   y= mx + b     and you are given m = slope = -1/5

y = -1/5x + b     Now substitute in you point (given) to calculate 'b'

7=-1/5(-3) + b

7 = 3/5 + b

b= 6 2/5

y = -1/5 x + 6 2/5     Done!

What is the remainder of: 13^1031 mod 599 =? Thanks for help.

I =?

P=2000

R= .072

T= 2

?=(2000)(.072)(2)

2000x.072=144

144x2=288

I = 288

431431/9 = 47.8888888888888889

431431/9 = 47.8888888888888889

0.00001

Solve for x over the real numbers:
32^(3 x) = 1/8

Take the logarithm base 32 of both sides:
3 x = -3/5

Divide both sides by 3:
Answer: |x = -1/5

if 32^3x=1/8, what is x

\(\begin{array}{|rcll|} \hline 32^{3x} &=& \frac18 \\ (2^5)^{3x} &=& 2^{-3} \\ 2^{15x} &=& 2^{-3} \\ 15x &=& -3 \\ x &=& -\frac{3}{15} \\ x &=& -\frac{1}{5} \\ \mathbf{x} &\mathbf{=}& \mathbf{-0.2} \\ \hline \end{array}\)

Random math IV

1.

\(\begin{array}{|rcll|} \hline \cos(\frac{19}{12}\pi) &=& \cos(285^{\circ}) \\ &=& \cos(285^{\circ}-360^{\circ}) \\ &=& \cos(-75^{\circ}) \\ &=& \cos(15^{\circ}-90^{\circ}) \\ &=& \cos(-(90^{\circ}-15^{\circ})) \\ &=& \cos( 90^{\circ}-15^{\circ} ) \\ &=& \mathbf{\sin( 15^{\circ} )} \\\\ \mathbf{\cos(\frac{19}{12}\pi) } &\mathbf{=}& \mathbf{\sin( 15^{\circ} )}\\ \hline \end{array} \)

2.

\(\begin{array}{|lrcll|} \hline (1) & \cos(45^{\circ}-15^{\circ})= \cos(30^{\circ}) &=& \cos(45^{\circ})\cdot\cos(15^{\circ})+\sin(45^{\circ})\cdot \sin(15^{\circ})\\ (2) & \cos(45^{\circ}+15^{\circ})= \cos(60^{\circ}) &=& \cos(45^{\circ})\cdot\cos(15^{\circ})-\sin(45^{\circ})\cdot \sin(15^{\circ})\\ \\ \hline (1)-(2): & \cos(30^{\circ})-\cos(60^{\circ}) &=& 2\cdot\sin(45^{\circ})\cdot \sin(15^{\circ})\\\\ & \cos(30^{\circ}) &=& \frac{\sqrt{3}}{2} \\ & \cos(60^{\circ}) &=& \frac{1}{2} \\ & \sin(45^{\circ}) &=& \frac{\sqrt{2}}{2} \\ \\ & \frac{\sqrt{3}}{2}-\frac{1}{2} &=& 2\cdot \frac{\sqrt{2}}{2}\cdot \sin(15^{\circ})\\ & \frac{\sqrt{3}-1}{2} &=& \sqrt{2}\cdot \sin(15^{\circ})\\ & \frac{\sqrt{3}-1}{2\cdot \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} &=& \sin(15^{\circ})\\ & \frac{\sqrt{2}\sqrt{3}-\sqrt{2}}{4} &=& \sin(15^{\circ})\\ & \frac{\sqrt{6}-\sqrt{2}}{4} &=& \sin(15^{\circ})\\\\ & \mathbf{\cos(\frac{19}{12}\pi) = \sin( 15^{\circ} ) }&\mathbf{=}& \mathbf{\frac{\sqrt{6}-\sqrt{2}}{4}} \\ \hline \end{array} \)

EP: What you have calculated is "simple" annual rate:

FV=PV[1 + R]^N

102 =74[1 + R]^2

R =17.40% =0.1740 - annual compound rate. Well, now they have both "simple" and "compound" rate!.

17

Suppose that the 'heights' of the rectangles in the top row are A, those of the second row B and those of the third row C, and suppose that the widths of the rectangles in the  first column are D those in the second column E and those in the third column F, then we can write down the set of nine equations,

(first column)

(1) A + D = 5 , (2) B + D = s/2 ,  (3) C + D = 8,

(second column)

(4) A + E = t/2 , (5) B + E = 5 ,  (6) C + E = q/2,

(third column)

(7) A + F = 8 ,  (8) B + F = r/2 ,  (9) C + F = p/2.

Eq(3) - Eq(1) : C - A = 3,

Eq( 9) - Eq(7): C - A = p/2 - 8, 

so p/2 - 8 = 3 , p = 22.

Eq(6) - Eq(4): C - A = q/2 - t/2 = 10 - t/2 (since q = 20) = 3, so t = 14 etc, etc.

Tiggsy

18      Here's a geometrical method.

Begin by finding the length of BE.

Let BE = 4L, then BA = 5L, but the figure is a rhombus, so 4L + 2 = 5L, so L = 2 in which case, BE = 8.

Now, take a line through E perpendicular to AB and extend it to the other side of AB to a point E* so that E and E* are equidistant from the line AB.

CE* then cuts AB at the required point P, (for minimum CP + PE).

To see that this is the case, see that PE = PE* so that CP + PE = CP + PE* = CE*, and that if P moves either way along AB, then CP + PE will become bigger.

(Alternatively the line perpendicular to AB could be taken through C, extended to C*. C*E (equal in length to CE*) then gets you the same point P as earlier).

The length of CE*  can be found by some simple trig, tanB in the smaller triangle at B followed by the cosine rule in the triangle CEE*.

I got the answer as 2sqrt(745)/5, approx 10.9179.

Tiggsy