2016/098) Let $a = 5^{1000}\sin(1000\alpha)$ where $\sin(\alpha) = \frac{3}{4}$ prove that $a\in Z$

because $\sin \alpha = \frac{3}{5}$ we have $\cos \alpha = \frac{\pm 4}{5}$
we have $\sin(1000\alpha) = Im(\cos(1000\alpha) + i\sin (1000\alpha)) = Im(e^{1000\alpha i})$
$=Im((e^{\alpha i})^{1000})  = Im((\cos \alpha + i\sin\alpha)^{1000}) = Im ( \frac{\pm 4}{5} + i \frac{3}{5})^{1000}$
hence $a = Im (\pm 4 + 3i)^{1000} = Im(\sum_{n=0}^{1000}{1000 \choose b} (\pm 4)^n * (3i)^{1000-n}$
A is imaginary part of the above and as each element is integer we have a is integer

2016/097) Find all integers n such that $\{1,2,3,4, ...,n\}$ can be written as the disjoint union of the subsets, $A$, $B$ and $C$ -whose sums of elements are equal.

for 3 sets the sum of elements to be same sum upto n must be divisible by 3
or $\frac{n(n+1)}{2}$ should be divsible by 3
so n or n+ 1 must be divisible by 3
let us taken n divisible by 3
for n =3 we cannot divide into 3 equal groups
for 6 successive numbers say m+1 to m+ 6 we can devide into 3 groups $\{m+1,m+6\}, \{m+2,m+5\}.\{m+3,m+4\}$
so if n is multiple of 6 we can divide the numbers into groups as above
but if n is not divsible by 6 we can have a group of 9(1-9)  and multiple groups of 6
the multiple groups of 6 can be broken into 3 groups and we can combine them into 3 groups
and 9 can be grouped as $\{4,9,2\},\{3,5,7\}, \{(1,6,8\}$ and take the elements and add to the above groups to get the result.
Let us take n + 1 divisible by 3 so n is of the form 3p + 2 where p >= 1
for p = 1 or n= 5 we can group as $\{4,1\},\{2,3\}, \{5\}$
for p = 2 or n= 8 we can group is $\{4,8\},\{1,5,6\}, \{(2,3,7\}$
for p > 2 we have number of term 5 + 6k or 8 + 6k and we can choose A,B,C accordingly as above

2016/096) Consider the set of integers 1000,1001,1002,...1998,1999,20001000,1001,1002,...1998,1999,2000.

There are times when a pair of consecutive integers can be added without "carrying": 1213+12141213+1214 requires no carrying, whereas 1217+12181217+1218 does require carrying.

For how many pairs of consecutive integers is no carrying required when the two numbers are added?

Solution
Overflow shall not occur in the following cases
1) No digit is greater than 4 and next number digit  is < 5. The lower numbers are 1xxx number of numbers = 125
2) unit digit is 9 so next number unit digit is zetro and no other dight > 4. number of numbers = 25
3) unit and ten's digit is 9 so next number unit and tensdigit is zetro and no other dight > 4. number of numbers = 5
4) 1999 so next number 2000
this gives us 156 numbers

2016/095) If $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots = \frac{\pi^4}{90}$

then  $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots = ...$

Solution
Let $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots = x$
$\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots + \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots$
or $\frac{\pi^4}{90} = x + \frac{1}{2^4}(\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots)$
$\frac{\pi^4}{90} = x + \frac{1}{16}\frac{\pi^4}{90}$
or $y = \frac{\pi^4}{96}$

2016/094) Prove that in a sequence of numbers 49,4489, .... where each number is made by inserting 48 in the middle of previous number each number is a perfect square.

We have the $n^{th}$ number is n 4's followed by n 8's + 1
n 4's = $4 *(\frac{10^n-1}{9})$ n '8s = $8 *(\frac{10^n-1}{9})$
it may kindly be noted that the numerator  is divisible by 9
so the number = $4 *(\frac{10^-1}{9}) * 10^n + 8 *(\frac{10^-1}{9}) + 1$
$=\frac{4 * (10^n-1) * 10 ^n + 8(10^n-1) + 1}{9}$
$=\frac{ 4 * 10^{2n} + 4 * 10^n + 1}{9}$
$= \frac{(2 * 10^n)^2 + 2 * (2 * 20^n) + 1}{9}$
$= \frac{(2 * 10^n + 1)^2}{3^2}$
$= (\frac{2 * 10^n+1}{3})^2$
Beacuse numerator is divisible by 3 so it is a perfect square

2016/093) Suppose $a_1,a_2,...,a_n$ are poistive real numbers satisfying \(a_1\cdot a_2\cdots a_n=1\).

Show that $(a_1+1)(a_2+1)\cdots(a_n+1)>=2^n$

Solution 
We have using AM GM $\frac{a_k+1}{2} >= \sqrt{a_k}$ or $a_k + 1>=2\sqrt{a_k}$
multiplying over k from 1 to n we have
$\prod_{k=1}^n(a_k+1) >= 2^n \sqrt{\prod_{k=1}^na_k} = 2^n$
or $\prod_{k=1}^n(a_k+1) >= 2^n$

2016/092) a,b,c, are 1st 3 terms of a geometric series . if harmonic mean of a,b is 12 and of b,c is 36 find first 5 terms of the series.

let common ratio be t
then terms are a, at
the harmonic mean is 12
so $\frac{1}{a} + \frac{1}{at} = \frac{1}{6}\cdots(1)$
and
$\frac{1}{at} + \frac{1}{at} = \frac{1}{18}$
deiding (1) by (2) we get $t = 3$
now from (1) $\frac{1}{a} + \frac{1}{3a} = \frac{1}{6}$ or a = 8
giving 5 terms 8,24,72,216,648.

2016/91) if $a_1,a_2,\cdots,a_n$ are in HP show that $a_1a_2+a_2a_3+\cdots + a_{n-1}a_n = (n-1)a_1a_n$

we have
  $\frac{1}{a_1},\frac{1}{a_2}\cdots,\frac{1}{a_n}$ are in AP hence
or $(n-1)(\frac{1}{a_2} - \frac{1}{a_1})  = \frac{1}{a_n} - \frac{1}{a_1}$
or $(n-1)(a_1-a_2)(a_n a_1) = a_2a_1(a_1-a_n) \cdots(1)$
similarly
 $(n-1)(a_2-a_3)(a_n a_1) = a_3a_2(a_1-a_n) \cdots(2)$\
...
   $(n-1)(a_{n-1}-a_n)(a_n  a_1) = a_{n-1}a_m(a_1-a_n) \cdots(n-1)$\
adding above n eqautions we get
$(n-1)(a_1-a_n)(a_n a_1) = (a_1a_2+a_2a_3+\cdots + a_{n-1}a_n) (a_1-a_n)$
cancelling $a_1-a_n$ from both sides we get the result

2016/090) Find the value of $\frac{18^3+7^3+3.18 *7 *25}{3^6+6*243 *2 + 15 *81 *4+20 *27 *8 + 15 *9 *16 + 6*3*32 + 64}$

We have numerator = $18^3+7^3+3.18*7*25 = 18^3+7^3 + 3 *18 *7 (18+7) = (18+7)^3 = 25^3 = 5^6$
denominator = $3^6+6*243*2 + 15*81*4+20*27*8 + 15*9*16 + 6*3*32 + 64$*
$= 3^6 + 6 * 3^5 * 2 + 15 * 3^4 * 2^2  + 20 * 3^3 *2^3 + 15 * 3^2 * 2^4 + 6 * 3 * 2^5 + 2^6$
$= {6 \choose 0}3^6 + {6 \choose 1} * 3^5 * 2 + {6 \choose 2} * 3^4 * 2^2  + {6 \choose 3} * 3^3 *2^3 + {6 \choose 4} * 3^2 * 2^4 + {6 \choose 5} * 3 * 2^5 + {6 \choose 6}2^6$
$= ( 3+2)^6 = 5^6$
so given expression is 1