lll result reduces show that, in this case, hi to the standard Pythagorean t 8.

Question

lll result reduces show that, in this case, hi to the standard Pythagorean t 8. This problem comes from Al-Kh Algebra. You should nwarizmi S solve it. A woman dies and leaves her daugh ter, her mother, and her husband. She bequeaths to some person as much as the share of her mother and to another person as much as one-ninth of her entire cap- ital. Find the share of each person [Note: It is known, from Arab legal principles of the time, that the mother's share would be 2/13 and the husband's share 3/13.] 9. Abu Kamil (850 C.E.-930 C.E.) wrote a commentary on izmi's Algebra. In it, he contributed a number of ingenious algebra problems. Solve the following one: The number 50 is divided by a certain other number. If the divisor is increased by 3, then the quotient decreases by 33/4. What is the divisor? 10. The method of "Casting out Elevens" is mathematically learned in the text for casting out nines). Casting out elevens is performed on a positive integer by (i) adding up the digits in the odd positions, (ii) adding up the digits in the even positions, and (iii) subtracting the second sum from the first. Explain why this method

Explanation / Answer

10. Every decimal number n can be written as n0 + 10n1 + 100n2 + 1000n3 + .......

Thus n mod 11 = (n0 + 10n1 + 100n2 + 1000n3 + .......) mod 11

=> n mod 11 = n0 mod 11 + 10n1 mod 11 + 100n2 mod 11 + 1000n3 mod 11 + .....

=> n mod 11 = n0 mod 11 + 11n1 mod 11 - n1 mod 11 + 99n2 mod 11 + n2 mod 11 + 1001n3 mod 11 - n3mod 11 + .....

=> n mod 11 = n0 mod 11 - n1 mod 11 + n2 mod 11 - n3mod 11 + .....

=> n mod 11 = [ (n0 + n2 + n4 + .......) - (n1 + n3 + n5 + ........) ] mod 11

Thus casting out elevens works.

12. For n = 2,

h = 3.2n - 1 t = 3.2n-1 - 1 s = 9.22n-1 - 1

=> h = 3*4 - 1 = 11 t = 3*2 - 1 = 5 s = 9*8 - 1 = 71

Since h, t and s are all prime, 2n * h * t and 2n * s are both amicable

=> 22 * 11 * 5 and 22 * 71 are amicable

=> 220 and 284 are amicable as given in the problem statement.

For n = 4,

h = 3.2n - 1 t = 3.2n-1 - 1 s = 9.22n-1 - 1

=> h = 3*16 - 1 = 47 t = 3*8 - 1 = 23 s = 9*128 - 1 = 1151

Since h, t and s are all prime, 2n * h * t and 2n * s are both amicable

=> 24 * 47 * 23 and 24 * 1151 are amicable

=> 16 * 47 * 23 and 16 *1151 are amicable

=> 17296 and 18416 are amicable.

Proper divisors of 17296 = 1, 2, 4, 8, 16, 23, 46, 47, 92, 94, 184, 188, 368, 376, 752, 1081, 2162, 4324, 8648

Sum = 1 + 2 + 4 + 8 + 16 + 23 + 46 + 47 + 92 + 94 + 184 + 188 + 368 + 376 + 752 + 1081 + 2162 + 4324 + 8648 = 18416

Proper divisors of 18416 = 1, 2, 4, 8, 16, 1151, 2302, 4604, 9208

Sum = 1 + 2 + 4 + 8 + 16 + 1151 + 2302 + 4604 + 9208 = 17296

This proves the result.

That nobody knows whether there are infinitely many amicable numbers or not implies it is not trivial to decompose large integers into prime factors.

1. (a) Let the equation of the circle be x2 + ax + y2 + by = c

The circle passes through (1,2), (2,3) and (4,9)

Substituting in the equation

=> 1 + a + 4 + 2b = c, 4 + 2a + 9 + 3b = c, 16 + 4a + 81 + 9b = c

=> a + 2b + 5 = c, 2a + 3b + 13 = c, 4a + 9b + 97 = c

Subtracting the first two equations,

=> a + b + 8 = 0 (1)

Subtracting the second and third equations,

=> 2a + 6b + 84 = 0

=> a + 3b + 42 = 0 (2)

Subtracting (1) from (2)

=> 2b + 34 = 0

=> b = -17

Substituting in (1)

=> a - 17 + 8 = 0

=> a = 9

=> c = a + 2b + 5 = 9 - 34 + 5 = -20

Thus the equation of the circle is x2 + 9x + y2 - 17y = -20

(b) P = (1,0) Q = (0,1) R = (1,1)

Slope of PQ = (0-1)/(1-0) = -1/1 = -1

Mid point of PQ = (1/2,1/2)

Thus the perpendicular bisector passes through (1/2,1/2) and has slope 1

=> (y-1/2)/(x-1/2) = 1

=> y - 1/2 = x - 1/2

=> y - x = 0 (1)

Slope of QR = (1-0)/(1-1) = infinity

Mid point of PQ = (1,1/2)

Thus the perpendicular bisector passes through (1,1/2) and has slope 0

=> (y-1/2)/(x-1) = 0

=> y - 1/2 = 0

=> y = 1/2 (2)

Substituting (2) in (1)

=> x = 1/2

Thus the point of intersection or center of the circle = (1/2,1/2)

Radius = ((1/2-1)2 + (1/2-0)2)

= (1/4 + 1/4)

= (1/2)

= 1 / 2

Thus the equation is (x-1/2)2 + (y-1/2)2 = 1/2

5. Consider the very simple parabola (y-k)2 = 4a(x-h)

Let (1,2) and (2,3) be two points on the parabola

=> (2-k)2 = 4a(1-h) and

(3-k)2 = 4a(2-h)

Subtracting,

5 - 2k = 4a (1)

Since there are two variables k and a, we cannot arrive at an immediate solution.

However if there was one more point say (3,4)

(4-k)2 = 4a(3-h)

Subtracting (2-k)2 = 4a(1-h)

=> 12 - 4k = 8a

=> 3 - 2k = 2a (2)

Subtracting (2) from (1)

=> 2 = 2a

=> a = 1

Substituting in (2)

=> 3 - 2k = 2

=> k = 1/2

=> (3-1/2)2 = 4 (2-h)

=> 25/4 = 8 - 4h

=> 4h = 8 - 25/4 = 7/4

=> h = 7/16

The equation is (y-1/2)2 = 4 (h-7/16)

Thus it is possible to determine the equation with three points.

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