Divisibility and Remainders

Take 1075^n common 

You will get 1075^(n-1) *1074

Now check options 15 will be divisible because 1075 contains 5 and 1074 contains 3 

1075 contains 43 

1074 contains 179 we can find it by prime factorisation

Hence 23 is not divisible 

Since digits of 10digits number n sum to 45 , n should be divisible by 9
n should be divisible by 99999

10 digits number will be of type
n = 100000x + y where x is 5digit number and y is 4 or 5 digit number
n = 99999x + (x+y)
means x+y should be divisible by 99999
and because neither x > 99999 nor y>99999
x+y = 99999

x=abcde
y=pqrst
----------------------
x+y=99999

also since max numbers is 9 and 8 we cannot carry to next digit (19)
so it has to be all 9s without carry

groups
0 9
1 8
2 7
3 6
4 5

means combination is for x only. y is related

digit 1 : {1-9} 9ways
digit 2 : {digit1,0} - {digit1,9-digit1} 8 ways
digit 3 : {digit2} - {digit2.9-digit2} 6 ways
digit 4 : {digit3} - {digit3.9-digit3} 4 ways
digit 5 : {digit4} - {digit4.9-digit4} 2 ways

total ways : 9 × 8 × 6 × 4 × 2 =3456

from 1sr statement there are two possibilities...2^2*3^6 or 2^5*3^3
(considering that from 2nd statement on multiplying n with 3 or n with 2 , factors don't double or get hugely increased hence 2 and 3 are the only prime factors)
now from 2nd statement we find that 1st case in not possible hence, our n= 2^5*3^3

now u can find 6n, (6+1)*(4+1) =35

25 pages consist of  25 odd numbers and 25 even numbers 

So there sum should always be odd 

Hence 1900 an even sum is never possible. 

a=10x+y

b=10y+x

a+b = 11(x+y)

x+y =11 for a+b to be a perfect sqaure and a and b to be a two digit nos.

Values can be 

(9,2) (8,3) (7,4) (6,5)

Hence 4 pairs

Hi Aniket, thankyou for the solution.

But since the question is asking for ordered pairs, shouldn't it be 8 pairs? (9,2) (2,9) (8,3) (3,8) (7,4) (4,7) and (6,5) (5,6)?

Please suggest if im going wrong somewhere

Yes ! You are right answer should be 8

put 17 as (2-19)   -19 to the power 36 and 19 to the power 36 cancels out hence the answer is 2

(Any odd number )^ (4k ) ends with 1 or 5 

*( odd multiples of 5 )^(Any natural number ) ends with 5. 

(Any even number )^4k ends with 6 or 0

*(Even multiples of 5)^ any natural number => 0

________________________________________

Here 2^2^2002 

2 ^ ( 2^2002) 

2^ ( 2 x 2 x 2 ..... 2002 times ) 

=> 2^4k 

Hence , the last digit of 2^2^2002 => 6 

 N = DQ + 52

=> 5N = D  (5Q) + 52 x 5 

5N = 5DQ + 260 

Also 5N = Dx + 4 

So , D divides (260 - 4) i.e 256 

hence , All factors of 256 > 52 . 

[ 256 , 128 and 64 ] 

PFA the solution 

this could only be possible if both digits must be same so nos. are 
33,44,55,66,77,88,99 

total =7 (answer)

answer is 0?

correct ans: 7

is that + or = bw 1/3 and 1/4

1 x 2 x 3 x 4 x 5x ...... x 13 i.e  13! is a multiple of 16 .

Hence 13! raised to any power would be a multiple of 16. 

Therefore the remainder will be 0. 

PFA the solution: 

Answer is 4?

How?

N*190/100 = N*19/10 

No. N should have 19 and 10 so that when it is increased by 90% it becomes a perfect square hence it will be multiple of 190

Possible values can be (190,380,570,760,950) 

Only 190 and 760 will give perfect sqaure when they are increased by 90% 

190/7 gives remainedr 1 and 760/7= gives remainder =4 

Sum of remainder= 4+1 =5 answer

done (Y)