Divisibility and Remainders

10?

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

2B^2 will give unit digit 0 

hence 3A^2= C^2  (unit digit will be same)

A^2 = (C^2)/3

since C is a odd no. C^2 can have unit digit 1,5,9 

if we put 9 as C^2 then on dividing it with 3 we will get unit digit 3 which is not a square's unit digit and if we put 1 then on dividing with 3 we get 7 which is also not a square no.'s unit digit so only left is 5 if we put that we will get unit digit of A as 5 hence answer

y² + 8y - 857 = x² 

y² + 2• y • 4 + 16 - 857 -16 = x² 

( y + 4)² - 873 = x² 

( y +4)² - x² = 873

p² - x² = 873

( p - x) ( p +x) = 1 • 873 

( p - x) ( p +x) = 3 • 291 

or 9 × 97 

p = 437 or 147 or 53 

y = 433 or 143 or 49

Now find sigma f(y) .

Thank you sir for an instant reply 🙂

If four numbers are in AP .

Then,

their product + ( Common difference )⁴ is always a perfect square. 

Hence , 2⁴ => 16 

Option ( D) 

PFA the solution: 

How 50^60^70 is of form 3k+1?

We can write 50 as ( 51- 1)

( 51 -1)^60 ^70 divided by 3 

( -1)^60^70 => (-1)^even = 1 

Hence , 3k + 1 

5! = 120 
last two digits of 6! => 6 x 20 = ___20 
last two digits of 7! => 7 x 20 = ___ 40 
last two digits of 8! => 8 x 40 = ___ 20 
last two digits of 9! => 9 x 20 = ___ 80 
last two digits of 10! => 00

Last two digits of N : 

1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 + 00 = __13 

N^N /4 => 13/4 => 1 

got it. thanks

(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

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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 

take 10! common 

N = 10! ( 1 + 11 x 12 + 11 x 12 x 13 x 14 +.........) 

= 10! x ( an odd number ) 

hence the  highest power of 2 in N = highest power of 2 in 10! 

i.e [10/2 ] + [5/2] + [2/2]= 8 

33. 

In order to divisible by 99, 

123N321 should be a multiple of 9 and 11 . 

1 + 2 + 3 + N + 3 + 2 + 1 = 12 + N should be a multiple of 9 

Only possibility N = 6 

When N = 6

1236321 

(1 + 3 + 3 + 1) - ( 2 + 6 + 2) = -2 
So 1236321 is not divisible by 11. 

Hence , no solution. 

The remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively. Hence, the problem reduces to finding the remainder when (4^{) 2222} + (3)⁵⁵⁵⁵ is divided by 7.

Now (4)²²²² + (3)⁵⁵⁵⁵ = (4²)¹¹¹¹ + (3⁵)¹¹¹¹ = (16)¹¹¹¹ + (243)¹¹¹¹. Now (16)¹¹¹¹ + (243)¹¹¹¹ is divisible by 16 + 243 or it is divisible by 259, which is a multiple of 7.

Hence the remainder when (5555)²²²² +(2222)⁵⁵⁵⁵ is divided by 7 is zero.

3¹ => 3 unit digit 3
3² => 9 unit digit 9
3³ => 27 unit digit 7
3⁴ => 81 unit digit 1
3⁵ => 243 unit digit 3
3⁶ => 729 unit digit 9
…………………………………

……………………………….

3⁴ always ends with 1
unit digit of 1273^122! is same as the unit digit of 3^122!
3^4k  => 1

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