2014/061) Solve for x if p(x) is sum of digits of x and x +p(x) + p(p(x))) = 1993

x mod 9 = P(x) mod 9 = P(P(x)) mod 9 = say k

so x + P(x) + P(P(x)) = 3 k mod 9

but x + P(x) + P(P(x)) mod 9 = 4 mod 9

as 3 is a factor of 9 there is no y such that 3y = 4 mod 9

so no solution

2014/060) using properties of proportions solve ( x^3 + 3x )/341 = (3x^2 + 1)/91

we have

(x^3 + 3x)/341 = (3x^2 + 1)/91

which is same as
(x^3 + 3x)/(3x^2 + 1)= 341/91

using componendo dividendo we have
(x^3+3x+3x^2+1)/(x^3+3x-3x^2-1) = (341+91)/(341-91)

or (x+1)^3/(x-1)^3 = (216)/125

hence (x+1)/(x-1) = 6/5

using componendo dividendo we have

(2x/2) = (11/1) or x= 11

2014/059) find real solutions

a- b  + c - d = 0 .... (1)
ab = cd  ...(2)
a^2 - b^2 + c^2 - d^2 = - 24 ...(3)
a^2 + b^2 + c^2 + d^2 = 50 ...(4)

from (1)
a- b = d- c... (5)

square above and using (2)

(a-b)^2 + 4ab = (c-d)^2 + 4cd

or (a+b)^2 = (c+d)^2

hence a + b = c + d ... (6)
or a+ b = -c - d ..(7)
from (5) and (6) a = d and b= c but it is not possible as LHS of (3) is zero which is contradiction
from (5) and (7) a = -c and b = - d
so we get from (3) and (4)
a^2 - b^2 = - 12
a^2 + b^2 = 25

add above to get 2 a^2 = 13, subtract to get 2b^2 = 37

this gives 4 set of solutions

(a,b,c,d) = ((13/2)^(1/2), (37/2)^(1/2), -(13/2)^(1/2),-  (37/2)^(1/2))
or  ((13/2)^(1/2),- (37/2)^(1/2), -(13/2)^(1/2),  (37/2)^(1/2))
or (- (13/2)^(1/2), (37/2)^(1/2), (13/2)^(1/2),-  (37/2)^(1/2))
or (-(13/2)^(1/2),-  (37/2)^(1/2), (13/2)^(1/2), (37/2)^(1/2))

2014/058) Find three positive numbers whose sum is 27 and such that the sum of their squares is as small as possible.

we have (x+y+z)^2 = (x^2+y^2+z^2) + 2(xy+yz+zx)  ... (1)

further 2(x^2+y^2+z^2-xy-xz-yz)= (x-y)^2 + (y-z)^2+(z-x)^2

so 2(xy + yz+xz) = 2(x^2+y^2+z^2) - (x-y)^2 - (y-z)^2 - (z-x)^2

from (1) and above
(x+y+z)^2 = 3((x^2+y^2+z^2) - (x-y)^2 - (y-z)^2 - (z-x)^2
as x+y +z = 27 we have

27+ (x-y)^2 + (y-z)^2 + (x-z)2  = 3(x^2 +y^2 +z^2)

so x^2+y^2 + z^2 is lowest when x = y = z that is x=y=z = 9

2014/057) given p+q+r+s = 63 find the maximum of pq+qr+rs

pq + qr + rs = pq + qr + rs + sp - sp = (p+r)(q+s) - sp

now we need to maximize (p+r)(q+s) and minimise sp as s and p are in different expressions

 p+r and q + s should be as close as possible as p+r + q + s = 63

so p+ q = 32 and r+ s = 31 or vice versa and p = s = 1

so p = 1 , q = 31, r = 30, s = 1 or p =1, q = 30, r = 31, s = 1
 
you can find some discussion at http://mathhelpboards.com/challenge-questions-puzzles-28/find-maximum-sum-11061.html#post51454

2014/056) p(x) = x^3 -6x^2+ 17x, p(m) = 16, p(n) = 20 find m+ n

we have P(x) = x^3 - 6x^2 + 17x
so we have P(x+2) = x^3 + 5x + 18 ( I take x+2 to eliminate the $x^2$ term to see in case we get odd function)

now
P(m) = (m-2)^3 + 5(m-2) + 18 = 16

or (m-2)^3 +5 (m-2) = - 2...(1)

P(n) = (n-2)^3 + 5(n-2) + 18 = 20

or (n-2)^3 + 5(n-2) = 2 ....   (2)

from (1) and (2) as

f(x) = x^3 + 5x

f(m-2) + f(n-2) = 0

so  m- 2 + n -2 = 0

or m+n = 4

2014/055) show that 6^33 > 3^33 + 4^33 + 5^33

we have
6^3 = 3^3 + 4^3 + 5^3

raising both sides to 11th power

6^33 = (3^3+4^3+5^3)^11 > (3^3)^11 + (4^3)^11 + (5^3)^11 or 3^33 + 4^33 + 5^33

2014/054) show that 27x^6+27x^3y^3+8y^6 is composite

We see that

27x^6 + 27x^3 y^3 + 8y^ 6
= 27x^6 - 27 x^3 y^3 + 8y^ 6 + 54 x^3y^3
= (3x^2)^3 + (-3xy)^3 + (2y^2)^3 – 3(3x^2)(-3xy)(2y^2)
Above is
a^3+ b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - cb) where a = 3x^2, b = - 3xy, c = 2y^2

We are not finished yet. Because one term is –ve we need to show that neither a+b+c is 1 nor other term is 1

a+b+ c = 3x(x-y) + 2y^2 >= 2

a^2 + b^2 + c^2 – ab –bc – ca >=   ½(a-b)^2

or >=  1/2(3x^2+ 3xy)^2 so > 2

as both factors are > 1 so this is composite

2014/053) Let P(n) be the sum of the first n terms of the sequence 0,1,1,2,2,3,3,4,4,5,5,6,6,⋯

find a formula for P(n) and prove that P(x+y) - P(x-y) is xy for x and y integers and when x > y
It becomes easier if we take 2 forms one for odd and another for even and knowing that both (x+y) and (x-y) are of same parity that is both are odd or even

Now consider the case that n is even say 2m ( and further x+y and x- y even)

in the sum there is one 0 one m and 2 instances of numbers from 1 to m- 1

so the sum = 2m(m-1)/2 + m = m^2 or (n/2)^2 .. (1)

so p(x+y) - p(x-y) = ((x+y)/2)^2 - ((x-y)/2)^2 = ((x+y)^2-(x-y)^2)/4 = (4xy)/4 = xy

Now consider the case that n is odd say 2m+1 ( and further x+y and x- y both odd )

sum = m^2 + m ( m^2 from previous even calculation and the term m)

= m(m+1) = (n-1)/2 * (n+1)/2 = (n/2)^2 - 1/4 .. (2)

Now consider the case that n is even say 2m ( and further x+y and x- y even)

so p(x+y) - p(x-y) = (((x+y)/2)^2- 1/4) - (((x-y)/2)^2- 1/4) = ((x+y)^2-(x-y)^2)/4 = (4xy)/4 = xy

from (1) and (2)

p(n) = floor((n/2)^2)

　

2014/052) Find the equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

 clearly m is -ve as if m is positive or zero then it shall not cut x and y both in 1st quadrant
so the equation  of line is

y-5 =m(x-3) as it passes through (3,5)

now x intercept when x = 0 is y = 5 - 3m

now y intercept when y  = 0 is x= (3m-5)/m

so area of the triangle in 1st quadrant is xy/2

so we need to minimize xy = - (3m - 5)^2/m = (3p+5)^2/p where p = -m and p >0

(3p + 5)^2/p = (3p^(1/2) + 5p^-(1/2))^2 = (3p^(1/2) -5p^-(1/2))^2 + 60

it is lowest when (3p^(1/2) -5p^-(1/2)) = 0 or p = 5/3

so equation of line is y - 5 = -5/3(x-3) or 3y + 5 x = 30
2)