2014/038) Find any/all 6 digit numbers that are squares given the condition that the last 3 digits is 1 more than the first 3 digits.

Let the 3 digit number be x
Now we have x writing to the right and add 1 we get the value
1001x + 1 = y^2
So 1001 x = y^2 -1 = (y-1)(y+1)
Or 7 .11.13 x = (y-1) (y+ 1)
Let x factor into ab
So 7.11.13 ab = (y-1)(y+1)
This gives rise to 6 cases
1) 7a = y- 1, 11.13 b = y+1
2) 7a = y+ 1, 11.13 b = y-1
3) 11a = y- 1,7.13 b = y+1
4) 11a = y+ 1, 7.13 b = y-1
5) 13a = y- 1, 7.13 b = y+1
6) 13a = y+ 1, 7.13 b = y-1
Using Chinese remainder theorem and modular arithmetic we can solve these and get solution and as 7 * 11 * 13 > 1000 we shall get not more than one solution in each set
I shall solve one and let others solve the rest
Say 7 a = y-1 and 11.13b = y+1
Let y = z + 1
7a = z and 143b = z + 2 => 3b = 2 taking mod 7
b =3 or z = 143 * 3 – 2=>y = 428
this gives a = 61 or x = 183
so we get 183184 = 428^2
similarly we get 5 more 3 digit for x solutions
024025 = 155^2
075076= 274^2
183184= 428^2
328329=573^2
528529= 727^2
715716= 846^2
The 1st 2 solutions to be removed as they are 5 digit numbers. So rest 4 are solutions

2014/037) show that in a set of any 5 consecutive numbers there is at least one number that is co-prime to all the rest 4 (for example (2,3,4,5,6- 5 is co-prime to 2,3,4,6)

because we have 5 consecutive number the largest difference is 4. so if there is a common factor between 2 numbers of the 5 it has to be <=4. So a common prime factor has to be 2 or 3.

now in a set of 5 consecutive numbers one of the numbers has to be of the form 6n + 1 or 6n - 1 which is neither divisible by 2 nor 3. so it is co-prime to rest of the 4.

2014/036) Demonstrate that prime numbers (except for the "2"), can only be expressed as the sum of two consecutive natural numbers.

let the number of numbers be n and 1st number a+1

then sum of numbers= an + n(n+1)/2

it is integer

 if n is odd (n+1)/2 is integer so the expression  is divisible by n

if n is even an and n(n+1)/2 is divisible by n/2

so if n > 2 and odd it is not prime as divisible by n

if n > 2 and even it is divisible by n/2(which is >= 2) so not prime

2014/035) x+y+z=2π, prove that cos^2 x+cos^2 y+cos^2 z+2 cos x cos y cos z=1

cos²x+cos²y+cos²z+2cos x cos  y cos z

=        cos²x+cos²y+cos²(π−z)+2 cos x cos y cos z
= cos²x+cos²y+cos²(x+y)+2 cos x cos y cos z
= cos² x+cos² y+(cos x cos y−sin x sin y)^2+2cos x cos y cos z
=cos² x+cos² y+cos^2 x cos^2 y+sin^2 x sin^2 y−2cos x cos y sin x sin y+2 cos x cos y cos z
= cos² x+cos² y+cos^2 x cos^2 y+( 1 - cos ^2 x)( 1- cos ^ 2  y)−2cos x cos y sin x sin y+2 cos x cos y cos z
= cos² x+cos² y+cos^2 x cos^2 y+( 1 - cos ^2 x  - cos ^ 2  y + cos ^2 x cos^2y - 2 cos x cos y sin x sin y+2 cos x cos y cos z
=   2 cos^2 x cos^2 y+  1  - 2 cos x cos y sin x sin y+2 cos x cos y cos z
=  1+2cos xcos y(cos xcos y–sin xsin y)+2 cos x cos y cos z
= 1+2 cos x cos y cos(x+y)+2 cos x cos ycos z
= 1–2 cos x cos y cos (π–(x+y))+2 cos  x cos y cos z
= 1–2 cos  x cos y cos z + 2 cos  x cos y cos z
= 1

2014/034) Find all positive integers whose squares end with 444 and show that no square can end with 4444

for the 2nd part 1st
ending with 4444 it is 10000n + 4444= 4( 2500n + 1111)
4 is a perfect square and 2500n + 1111 mod 4 = 3

as no square mod 4 can be 3 so 2500n + 1111 cannot be a perfect square. so no square can end with 4444.

now for the 1st part

we have n say x^2 mod 1000 = 444

1000 = 8 * 125 and 8 and 125 are copimes

so x^^2 mod 8 = 4

and x^2 mod 125 = 444 mod 125 = 69

for the case x^2 mod 8

x has to be even and if x mod 4 = 0 then x^2 mod 8 = 0

so x mod 4 = 2

check (4n+2) ^2 = 16n^2 + 16n + 4 = 4 mod 8

now we need to solve

x^2 mod 125 = 69 and as 125 = 5^3 so we solve

x^2 = 4 mod 5

and x^2 = 19 mod 25

we sta69rt with 4 x^2 = 4 mod 5 and then x^2 = 19 mod 25 and finally x^2 = 69 mod 125

x^2 = 4 mod 5 has solution x= 2 or 3 mod 5

we check for x = 2 mod 5 and then for x = 3 mod 5




First let us check for x = 2 mod 5

x= 2 mod 5

=> x = 5k + 2 mod 25

so x^2 mod 25 = 20k + 4 = 19 mod 25

or 20 k = 15 mod 25

or 4k = 3 mod 5

so k = 2 mod 5

so x = 12 mod 25

so x = 25 m + 12 mod 125

so x^2 = 600 m + 144 mod 125 = 69 mod 125

or 600 m = - 75 mod 125 = 50 mod 125

so 12 m = 1 mod 5 or 2m = 1 mod 5 or m = 3

so x = 87 mod 125

x= 2 mod 4

thsese 2 give x = 462 or – 38 mod 500

Now let us solve for x = 3 mod 5

x= 3 mod 5

=> x = 5k + 3 mod 25

so x^2 mod 25 = 30k + 9 = 19 mod 25

or 30 k = 10 mod 25

or 6k = 2 mod 5

so k = 2 mod 5

so x = 13 mod 25

so x = 25 m + 13 mod 125

so x^2 = 650 m + 169 mod 125 = 69 mod 125

or 650 m = - 100 mod 125 = 25 mod 125

so 25m = 25 mod 125 or m = 1

so x =38 mod 125

x= 2 mod 4

these 2 give x = 38 mod 500

So we get x = +/- 38 mod 500 or (500k +/- 38)

some discussion can be found at http://trickofmind.com/?p=1949&cpage=1#comment-29427

 

2014/033) f(x)=x^4−29x^3+mx^2+nx+k, and f(5)=11, f(11)=17 and f(17)=23, find the sum of m+n+k.

we have f(x) = x+ 6 for x= 5 11 and 17

so f(x)= (x-5)(x-11)(x-17)Q(x) +x + 6
ax f(x) is a 4th oder polynomal so Q(x) = linear say x+ a
coefficient of x3=−29=−5−11−17+a so a = 4

so f(x) = (x-5)(x-11)(x-17)(x+4) + x + 6
put x = 1 to get
1- 29 + m + n + k = (-4) *(-10) *(-16) * 5 + 1 + 6 = - 3200 +7 = - 3193

or m + n + k = -3193 + 28 = - 3165

2014/032) How many real solutions does the following system have. a+ b = 2 ab –c^2 = 1

Let a = 1 – x and b = 1 + x

So ab – c^2 = (1-x)(1+x) – c^2 = 1 – x^2- c^2 = 1

Or x^2 + c^2 = 0 => x = c = 0

So c = 0 and a=b = 1,

So one solution a =1, b= 1, c = 0