2008/014) Solve the equation 4xABCD=DCBA

A has to be < 3 because 3* 4 = 12 so RHS is a 5 digit number
A cannot be 1 as from RHS A has to be even.
So A has to be 2.
now B can be either 1 or 3 or 5 or 7 but B < 5 because 4*25 = 100 that is 5 digit
So AB = 21 or 23
if AB = 23 DC >= 92do D = 9 which is not possible as 4*8 is 2 ending but 4*9 is notAB = 21
So D = 8
so the number = 4*(2108+10C) = 8032+100C
or 8432+40C = 8012+ 100C
or 60C = 420
so C =7
so number = 2178*4 = 8712

2008/013) 2 is the only prime sum of 2 positive cubes

We know a^3 + b^3 = (a+b)(a^2-ab+b^2)
if a= b then a^3+b^3 = 2 a^3 which not a prime unless it is 2
with out loss of generality we can assume a> b
now a+b >=2
a^2+b^2-ab = a(a-b) + b^2 > b^2 so
a^2 + b^2-ab > 1
as it has 2 factors and both are >2 a^3+b^3 cannot be prime or in other words a prime number > 2 cannot be sum of 2 positive cubes

2008/012) x+y+z=6 , x^2+y^2+z^2=306 ,find maximum and minumum of XYZ

x+y+z = 6
x^2+y^2+ z^2 = 306
find minumum of maximum of xyz
the three unknowns need not be positive.

x+y + z = 6 ..1

x^2+y^2+z^2=306.. 2

square (1)
x^2+y^2 + z^2 + 2xy + 2yz + 2xz = 36

so xy+xz+yz = (36-306)/2 = -135

now let us take

p(t) = (t-x)(t-y)(t-z)
= t^3 - t^2(x+y+z) +t(xy+yz+xz) - xyz

= t^3 - 6t^2 + 135t - k
k = t^3 - 6t^2 - 135t
k is maximum or minimum when t^3 - 6t^2 - 135t
is

now you can differentiate wrt t and equate to zero

3 t^2 - 12 t - 135 = 0

or t^2 - 4t - 45 = 0

(t-9) ( t + 5 ) = 0

t =9 gives t^3 - 6t^2 - 135t = 729- 486- 1215 = -972 minimum
t =-5 gives -125-150+ 675 = 400 maximum

2008/011) A 4 digit number a perfect square 1 added to each digit becomes a square.

A 4 digit number a perfect square 1 added to each digit becomes a square. Find the number

Let the number be x^2 none of digits = 9

adding 1111 makes it a perfect square say y^2

x^2+1111 = y^2

so y^2 – x^2 = 1111

(y-x) (y+x) = 11 * 101

As 11 and 101 are primes 1111 can be factored only as above other than 1 * 1111

So y – x = 11 and y+x = 101

Add 2 get 2y= 112 or y = 56 and x = 45

x^2 = 2025 and adding 1111 we get y^2(56^2) = 3036

2008/010) If P = arctan 1 + 1/2.arctan 2 + 1/3.arctan 3 and Q = arctan 1 + 2 arctan 1/2 + 3 arctan 1/3, then (P - Q)/5 = :

we know arctan 1/x = pi/2 - arc tan x

so Q = arctan 1 + 2(pi/2 - arc tan 2) + 3(pi/2 - arc tan 3)
= 1 + 5pi/2 - 2 arc tan 2 - 3 arc tan 3

so P-Q = 1/2 arc tan 2 + 1/3 arctan 3 -(5pi/2 - 2 arc tan 2 - 3 arc tan 3)
= 5/2 arctan 2 + 10/3 arc tan 3 - 5pi/2

devide by 5 to get
(P-Q)/5 = 1/2 arc tan 2 + 2/3 arc tan 3 - pi/2

= 1/6( 3 arc tan 2 + 4 arc tan 3) - pi/2

3 arctan 2+ 4 arctan 3
= 3 (arc tan 2 + arc tan 3 ) + arctan 3

we know arc tan 2 + arc tan 3 = arctan ( 2+3)/(1-2*3) = arc tan (-1) = - pi/4


so (P-Q)/5 = 1/6(arctan 3 - 3 pi/4)
= 1/6 arctan 3 - pi/8

2008/009) Prove; 1+ cos 56 + cos 58 - cos 66 = 4*cos 28. cos 29. sin 33 ?

1+ cos 56 = 2 cos^2 28 as cos 2t = 2 cos^2 t - 1
cos 58 = 2 cos^ 29 -1
cos 66 = 2cos ^2 33 -1

so 1+ cos 56 + cos 58 - cos 66
= 2 cos^2 28 + 2 cos^2 29 - 2 cos^2 33
= 2 ( cos^2 28 + cos ^2 29 - cos ^2 33)

now 28+29+33 = 90

if we prove cos ^2 A + cos^2 B - cos^2 c = 2 cos A cos B sin C when A+B+C = pi.2
then we are through

2008/008) Tough inequality

Prove that for all x>0, y>0, and all real a it holds true that

(x^((sin(a))^2))*(y^((cos(a))^2))
let us assume y > x

x^t y^(1-t) < x+ y putting (sin a)^2 = t we get (cos a)^2 = 1-t

devide by x on both sides

x^(t-1)y^(1-t) < (1+y/x)

or (y/x)^(1-t) < 1+ y/x

or m ^k < 1+m where k < 1 and m > 1
we know m ^k < m where k < 1

so m^ k < 1+m

if x > y then role of sin a and cos a are reversed and you get the same result

2008/007) Verify (tanx+secx)/(secx-cosx+tanx) = cscx

AS tan x + sec x is in both numerator and dnominator and denominator has one extra term I would take reciprocal of LHS

1/ LHS = 1 - cosx/(tan x + sec x)
= 1- cos x ( sec x- tan x)/ (tan x + sec x)( sec x- tan x)
= 1 - cos x ( sec x - tan x)/ (sec^2 x - tan ^2 x)
= 1- cos x( sec x - tan x) as sec^2x - tan ^2x = 1
= 1 - cos x(1/cosx - sinx /cos x)
= 1-1+ sin x = sin x
so LHS = 1/ sin x = csc x


proved

2008/006) If z = cos x + i sin x ,then simplify (z^2 -1)/(z^2 +1)

(z^2-1)/(z^2+ 1) = (z - 1/z)/ (z+ 1/z) ...1

now z = cos x + i sin x ... 2

so z = e^ix

so 1/z = e ^-ix = cos x - i sin x ... 3

add 2 and 3 to get z+ 1/z = 2 cos x

subtract 3 from 2 to get

(z-1/z) = 2i sin x

from 1 by deviding (z^2-1)/z^2+1)= 2i sin x/(2 cos x) = i tan x

2008/005) rational no. p,q and r satisfying the property that pq+qr+rp=1

rational no. p,q and r satisfying the property that pq+qr+rp=1
prove that (p^2+1)(q^2+1)(r^2+1) square of a rational number

so p = (1-qr)/(q+r)
if we chose q = tan A and r = tan B

we get
1/p = (q+r)/(1-qr) = (tan A + tan B)/(1- tan A tan B) = tan (A+B)

or p = cot (A+B)

now
(p^2+1) (q^2+1)(r^2+ 1) = sec^2 A sec ^2 B cosec^2 (A+B)
= (sec^2 A sec^B)/ sin^2 (A+B)

this is square of reciprocal of sin (A+B) cos A cos B
sin (A+B) cos A cos B
=( sin A cos B + cos A sin B)cos A cos B
= sin A cos A cos ^2B + cos^2 A sin B cos B
= tan A cos ^2 A cos ^2 B + tan B cos ^2 A cos ^2 B
= (tan A + tan B)/(sec^2 A sec ^2B)
= (tan A + tan B)/(1+ tan ^2 A)(1+ tan ^2B)

so (p^2+1) (q^2+1)(r^2+ 1) = ((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B))^2

if tan A and tan B that is q and r are rational then

((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B)) is rational and so (psquare +1)(qsquare +1)(rsquare +1) is the square of a rational number

2008/004) If α + β + γ = π/2, show that,

If α + β + γ = π/2, show that,

[(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan α/2)(1 + tan α/2)]
= (sin α + sin β + sin γ - 1)/(cos α + cos β + cos γ)

proof:
We should start with the RHS as it is more complex

Before we proceed let us use/deduce certain information as they shall be used
(actually they should be derived as required but I am deriving to keep the flow
α + β + γ = π/2
so α + β = π/2 – γ
or α + β = π/2 – γ …1

cos a + cos b = 2 cos (a+b)/2 cos(a-b)/2 … 2

taking sin of both sides of 1 we get
sin (α + β) = sin (π/2 – γ) = cos γ … 3
sin 2a = 2 sin a cos a …. 4

from 1
(α + β)/2 = (π/2 – γ)/2
So sin (α + β)/2 = sin (π/2 – γ)/2 = cos (π/2 + γ)/2 … 5

Again as
α + β + γ = π/2
so α - β + γ = π/2 - 2 β
so α - β + γ + π/2 = π - 2 β
so (α - β + γ + π/2)/ 4 = π/4 - β/2 …6

Again as
α + β + γ = π/2
so α - β - γ = π/2 - 2 β – 2 γ
so α - β - γ - π/2 = - 2 β – 2 γ
so (α - β - γ - π/2)/ 4 = - (β + γ)/2
so cos (α - β - γ - π/2)/ 4 = cos (β + γ)/2 as cos - A= cos A
= cos (π/2 – α)/2
= cos (π/4 – α/2) .. 7
And sin (α - β - γ - π/2)/ 4 = - sin (β + γ)/2 = - sin (π/4 – α/2)
Or sin (-α + β +γ + π/2)/ 4 = sin (π/4 – α/2) … 8


Now let us find he denominator
cos α + cos β + cos γ
= 2 cos (α + β)/2 cos(α - β )/2 + cos γ (using 2)
= 2 cos (α + β)/2 cos(α - β )/2 + sin (α + β) (using 3)
= 2 cos (α + β)/2 cos(α - β )/2 + 2 cos (α + β)/2 sin (α + β)/2 ( using 4)
= 2 cos (α + β)/2 (cos(α - β )/2 + sin (α + β)/2)
= 2 cos (π/2 - γ)/2 (cos(α - β )/2 + cos (π/2 + γ)/2)) (using 1 and 5)
= 2 cos (π/4 – γ/2)*2 cos (α - β + π/2 + γ)/2 cos (α - β - π/2 - γ)/2 (using 2)
= 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (α - β - π/2 - γ)/2 (using 6)
= 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2) using 7

So cos α + cos β + cos γ = 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2) … (A)
Now numerator

For the additional identities

Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2 … 9
sin (α + β + γ) = sin π/2 = 1 …. 10

sin A – sin B = 2 sin (A-B)/2 cos (A+B)/2 … 11


cos A – cos B =2 sin (A+B)/2 sin (B-A)/2 … 12

tan (π/4- A)= (1- tan π/4 tan A) / (tan π/4 + tan A) = ( 1- Tan A)/(1+tan A) … 13

based on above numerator

sin α + sin β + sin γ – 1
= 2 sin (α + β)/2 cos(α - β )/2 + sin γ – sin (α + β + γ) (using 9 and 10)
= 2 sin (π/4 – γ/2) cos(α - β )/2 –(sin (α + β + γ) - sin γ) (using 5)
= 2 sin (π/4 – γ/2) cos(α - β )/2 - 2 sin (α + β)/2 cos (α + β + γ) + γ)/2 (using 11)
= 2 sin (π/4 – γ/2) cos(α - β )/2 - 2 sin (π/4 – γ/2) cos (π/2+ γ)/2 (using 5)
= 2 sin (π/4 – γ/2) (cos(α - β )/2 - cos (π/2+ γ)/2)
= 2 sin (π/4 – γ/2) (2 sin (α - β + π/2+ γ)/4 sin (-α + β +π/2-+γ)/4 (using 12)
= 4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2) (using 6 and 8)

sin α + sin β + sin γ – 1 = 4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2) …(B)

from A and B we get
RHS =
(sin α + sin β + sin γ – 1)/ (cos α + cos β + cos γ)
= (4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2))/ (4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2))
= tan (π/4 – γ/2) tan (π/4 – β/2) tan (π/2- α/2)
= tan (π/4- α/2) tan (π/4 – β/2) tan (π/4 – γ/2)
= [(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan β/2)(1 + tan γ/2)]/[( (using 13 and rearranging the terms)

= LHS

2008/003) Prove that the sum of

1/(sinA + sin2A) + 1/(sin2A + sin3A) + 1/(sin3A + sin4A)........to n terms

IS

cosecA[cotA - cot(n+1)A]

proof:
We know
1/( sin ka sin (k+1) a)
= cosec a sin a /( sin ka sin (k+1) a)
= cosec a ( sin ((k+1) a - ka))/( sin ka sin (k+1) a)
= cosec a ( sin (k+1) a cos ka - cos(k+1)a sin ka) / sin (k+1) a sin ka
= cosec a ( cos ka/ sin ka - cos (k+1)a / sin (k+1) a)
= cosec a ( cot ka - cot (k+1) a)

addding from k = 1 to n we get the result.

2008/002) Prove the identity.

Prove the identity.

csc(20°) - cot(40°) = tan(60°)

Proof:
LHS= 1/sin (20°) – cos (40°) /sin (40°)
= 1/sin (20°) - cos(40°) /(2 sin (20°) cos (20°)
= ( 2 cos (20°) – cos (40°) )/ (2 sin (20°) cos (20°))
= ( 2 cos (20°) – cos (40°) )/ ( sin (40°))

Now numerator
= 2 cos (20°) – cos (40°)
= 2 cos (30°-10°) – (cos (30°+10°)
= 2 (cos (30°) cos (10°)+ sin (30°) sin (10°)) – (cos (30°) cos (10°)- sin (30°) sin (10°))
= cos (30°) cos (10°)+ 3 sin (30°) sin (10°)
= sqrt(3)/2 cos (10°) + 3/2 sin (10°)
= sqrt(3)( cos (10°)/2 + sqrt(3)/2 sin (10°) (take sqrt(3) common as it is tan 60)
= sqrt(3)( cos (10°) sin (30°) + sin (30°) sin (10°))
= sqrt(3)(sin (30°+10°))
= sqrt(3)(sin (40°)

so LHS = ( 2 cos (20°) – cos (40°) )/ sin (40°)
= sqrt(3)(sin (40°)/sin (40°)
= sqrt(3)

2008/001) What is the 2500th digit (from right side) in 10000!

First we need to compute the number of trailing zeroes in 10000!

Based on that we should find out which digit to look for.

As 10 is divisible by 2 and 5 ( 10= 2*5) the trailing zeros shall come from power of
2 and 5.

So we should factor out the 5’s and 2’s from all the numbers from 1 to 10000 as factors and count them.

As there are much more number of 5’s then the 2’s we need to count the number of 5’s. 25 shall contribute to 2 5’s so we need to find out how may numbers are divisible by 5^2 and so on.

Number of numbers divisible by 5 = 2000
Number of numbers divisible by 5^2 = 400
Number of numbers divisible by 5^3 = 80
Number of numbers divisible by 5^4 = 16
Number of numbers divisible by 5^5 = 3

They contribute to 2000+ 400+80+16+3 = 2499 powers of 5.

As there are more number of 2’s than 5 we have 2499 trailing zeroes.

That is we need to find the right most non zero digit.

Before that we need to find out the how many powers of 2 and out of which 2499 shall go into multiplication with 5 to give the trailing zeroes.

Then We need to find the product of all the odd numbers(existing ones),
powers of 2 that are there in excess of power of 5 and all the odd numbers which we get after dividing by power of 2.
For example 6 = 2 * 3 and we have taken care of 2 then we need to multiply by 3 as for multiplication by 6 we need to multiply by 2 * 3.
Secondly we should not count by 5 as they are counted along with 2 for the trailing zeros as they have been accounted for.
Let us find out how many powers 2
5000 numbers are divisible by 2, 2500 out of which are divisible by 2^2 or 4, 1250 out of which divisible by 2^3 or 8, 625 out of which divisible by 2^4 or 16, 312 out of which divisible by 2^5 or 32, 156 out of which divisible by 2^6 or 64, 78 out of which divisible by 2^7 or 128,39 out of which divisible by 2^8 or 256,19 out of which divisible by 2^9 or 512,9 out of which divisible by 2^10 or 1024,4 out of which divisible by 2^11 or 2048,2 out of which divisible by 2^12 or 4096,1 out of which divisible by 2^13 or 8192.

So power of 2 = 5000 + 2500+1250+625+312+156+78+39+19+9+4+2+1 = 9995
So extra power of 2 after 5 = 9995-2499 or 7496

We know 2^4 = 6 mod 10 and 6^any number = 6 mod 10

So 2^7496 mod 10 = (2^4)^1874 = 6^1874 mod 10 = 6

Now computation of multiple of odd number (except 5) as it is considered below.

For counting odd numbers we successively divide all even numbers by 2 and take odd numbers only ignoring the numbers that ar 5 ending. We keep dividing each time by 2 so that a number that is multiple of power of 2 is eventually divided by power of 2 after successive divisions and odd number is left divide by highest power of 2 possible and get the odd digit at the end. So after each division we take care of odd number. This is so because for multiplying by 6 we need to multiply 3 and 2. Further for multiplily by 12 we need to multiply by 3 and 4. As we have to keep track of multiples of 2 separately we need to take care of numbers after multiplying by the odd number 3 that is left by successively by 2.
Additionally each set of 10 numbers from 10n to 10(n+1) we have 4 numbers 1*3*7*9 one time.

We realise that 1*3*7 *9 = -1 mod 10 and now we go multiplying as below, This is mentioned as this shall simplify the steps for calculation.
We give the data in tabular form what to multiply after successive division by 2.

S. no. Upto Product Product with previous result
6
10000 (1*3*7*9) ^1000 = (-1)^1000 = -1 6
10000/2 = 5000 -(1)^500 = 1 6
10000/2^2 = 2500 (-1) ^ 250 = 1 6
10000/2^3 = 1250 (-1)^125 = - 1 -6 or 4
10000/2^4 = 625 (-1)^*62 *1 *3 = 3 2
10000/2^5 = 312 (-1)^*31 *1 = -1 -2
10000/2^6 = 156 (-1)^*15 *1*3 = -3 6
10000/2^7 = 78 (-1)^*15 *1*3*7 = 1 6
10000/2^7 = 39 (-1)^*3 *1*3*7*9 = 1 6
10000/2^9 = 19 (-1)^*1 *1*3*7*9 = 1 6Upto 10000 are (-1) ^ 1000 = 1 (there are 1000) of each
1* 6 = 6
Devide by 2 and we get numbers upto 5000
Now odd numbers upto 5000 (-1) ^ 500 = 1
1* 6 = 6
Devide by 2 and we get numbers upto 2500
Now odd numbers upto 2500 (-1) ^ 250 = 1
1*6 = 6
Devide by 2 and we get numbers upto 1250
Now odd numbers upto 1250 (-1) ^ 125 = -1 or 9
9*6 = 54 = 4 mod 10
Devide by 2 and we get numbers upto 625
Now odd numbers upto 625 (-1) ^ 62*1*3 = 3
4*3 = 12 = 2 mod 10
Devide by 2 and we get numbers upto 312
Now odd numbers upto 312 (-1) ^ 31*1 = -1
2*(-1) = -2 or 8 mod 10
Devide by 2 and we get numbers upto 156
Now odd numbers upto 156 (-1) ^15 *1*3 = -3
8*3= 24 = 4 mod 10
Devide by 2 and we get numbers upto 78
Now odd numbers upto 78 (-1) ^7 *1*3*7 = -1
4*-1 = -4 or 6 mod 10
Devide by 2 and we get numbers upto 37
Now odd numbers upto 37 (-1) ^3 *1*3*7 = -1
6*-1 = -6 or 4 mod 10

Devide by 2 and we get numbers upto 19
Now odd numbers upto 19 (-1) ^1 *1*3*7*9 = 1
4*1 = 4 mod 10

Devide by 2 and we get numbers upto 9
Now odd numbers upto 9 (-1) = -1
4*-1 = -4 = 6 mod 10


Devide by 2 and we get numbers upto 4
Now odd numbers upto 4 (1*3) = 3
6*3 = 8 mod 10

Devide by 2 and we get numbers upto 2
Now odd numbers upto 2 (1) = 1
8*1 = 8 mod 10

Devide by 2 and we get numbers upto 1
Now odd numbers upto 1 (1) = 1
8*1 = 8 mod 10

So the required digit is 8 .