Divisibility and Remainders

We can write the number as 

N = 195k + 47 

195k is divisible by 15 , so net remainder when N is divided by 15 will be same as the remainder obtained when 47 is divided by 15 i.e 2 

by euler Re(47^6/18) =1
hence (47^(6k+R))/18 = (47^R)/18 ........(1)
Where R is remainder when 57^67 is divided by 6
so R is Re(57^67)/6 = (3^67)/6 = 3
so now
putting value of R is equation (1)
answer is (47^3)/18 = (11^3)/18 = 121*11/18 = 13.*11/18 = 143/18 = 17

@mayanksri 

12600= (2^3)* (3^2)*(5^2)*(7)

powers of 2 in 50! by successive division = 50+25+12+6+3+1= 97, hence power of 2^3 will be 97/3 = 32
same way powers of 3 = 16+5+1 = 22; powers of 3^2 = 11
powers of 5 = 10+2 = 12; powers of 5^2 = 6
powers of 7 = 7+1 = 8
from the above set since there are only 6 powers of 5^2 are available, hence maximum value of n can be 6 only

@anna 

30ⁿ=2ⁿx3ⁿx5ⁿ

Number - 45*57*92*91*52*62*63*64*65*66*67 = 2¹¹x3³x5²x7x13²….

So n=2 is the answer.

@anna 

Let us assume, 4x⁶+x³+5 = 0mod7

=> 4x⁶+x³ = 2mod7

=> x³(4x³+1) = 2mod7

Let x³ = 1mod7, then x³(4x³+1) = 5mod7.

Let x³ = 2mod7, then x³(4x³+1) = 4mod7

Let x³ = 3mod7, then x³(4x³+1) = 4mod7

Let x³ = 4mod7, then x³(4x³+1) = 5mod7

Let x³ = 5mod7, then x³(4x³+1) = 0mod7

Let x³ = 6mod7, then x³(4x³+1) = 3mod7

Let x³ = 0mod7, then x³(4x³+1) = 0mod 7.

That means x³(4x³+1) is never 2mod7 in turn 4x⁶+x³+5 is never divisible by 7 for any value of x.

@anna 

Please find the soloution.

n² + 3n + 1 is a multiple of 3n + 10.
=> (n² + 3n + 1)/(3n + 10) is an integer.
=> [(3n-1)(3n+10) + 19]/9*(3n+10)
=> 3n + 10 = -19 or -1 or 1 or 19
=> 3n = -29 or -11 or -9 or 9
=> n = -3 or 3. As n is an integer.

@anna 

Please find the solution,

Using Wilson's theorem, we get
(41 - 1)! = 40! = -1mod41 = 40mod41
=> 39! = 1mod41
and let 38! = xmod41
=> 39x = 1mod41
=> -2x = 1mod41
=> 41y - 2x = 1
=> x = 20 and y = 1.
So Remainder when 38! is divided by 41 is 20.

Find the solution in detail

As per Wilsons theorem when (n - 1)! is divided by p we will get a remainder of - 1 when p is a prime number.

As per the theorem when 40! is divided by 41 we will get a remainder of – 1

When (41 -1)! Is divided by 41, remainder is 1.

It can be written as 40 X 39 X 38! divided by 41 will give a remainder of -1

(41–1)(41–2) 38! / 41 will give a remainder of - 1 since 41 in the bracket will get divided by 41 we can write as -1*-2**Y /41 will give a remainder of -1 where Y is 38!

Rem 2Y / 41 = -1

or Rem 2Y = 41a - 1 where a is the constant factor

since the left hand side is a multiple of 2 the right hand side also will have to be multiple of 2. we will have to substitute values for a in such a way that 41a - 1 will is a multiple of 2.

Put a =1

2Y = 41 x 1 - 1

2Y =40 => Y=20

Where Y is the remainder when 38! Is divided by 41

@anna 

As 1000 = 10³ ≡ 1 mod999.

So 12341234.... _{400 digits} ≡ 1 + 234 + 33(123 + 412 + 341 + 234) mod999 ≡ 1 + 234 + 33(111) mod999 ≡ 901 mod999

@mayanksri 

ABCD – DBCA

1000A + 100B + 10C + D – ( 1000D + 100B + 10C + A)

999A – 999D

999( A-D)

999 is not divisible by 7 . so ( A – D) should be a multiple of 7.

(A- D) =   ( 9 – 2) , ( 8 -1 ) , ( 7-0), ( 2 – 9) ,( 1-8) , and ( 1 -1) , ( 2-2) ,….,  ( 9 – 9) total 9 + 5 = 14 cases

Now for each of these 14 cases there are 15 multiples of 7 .
[ 1. 9002 , 9072, 9172, ……., 9982 ( 15 numbers )
  2. 8001 , 8071, ……….., 8981  ( 15 numbers ) similarly others ]

Hence Total : 14  15 = 210 such numbers .

@mayanksri 

111111.... (27 ) times 

111 × 10^24 + 111 × 10^21 + 111 × 10^18 +...... +111 × 10^3 + 111 

As 1000 = 10^3 ≡ 1 mod 999.

111 × 10^24 + 111 × 10^21 + 111 × 10^18 +...... +111 × 10^3 + 111 

≡

111 × 1 + 111 × 1 +...... + 111 × 1  mod 999 

≡ 999 mod 999 

≡ 0 mod 999. 

Hence divisible by 999.