**Find the maximum value of n such that 50! is perfectly divisible by 12600 raised to the power n.**

10?

**Find the maximum of n such that 45*57*92*91*52*62*63*64*65*66*67 is perfectly divisible by 30 raised to power n.**

Something is missing in the question since,

30^{n}=2^{n}x3^{n}x5^{n}

Number - 45*57*92*91*52*62*63*64*65*66*67 = 2^{11}x3^{3}x5^{2}x7x13^{2}….

So n=2 is the answer.

1260=2^{3}x3^{2}x5^{2}x7

For 1260^{n} to divide 50! Perfectly, n should be such that, 2^{3n} ≤ maximum power of 2 in 50! and similarly for powers of 3, 5 and 7.

Now 50! = 2^{47}x3^{22}x5^{12}x7^{8}….

Thus maximum value of n is min(47/3,22/2,12/2,8) = 6

**For how many +ve int between 1 to 1000, both inclusive, 4x ^{6}+x^{3}+5 is divisible by 7 **

Let us assume, 4x^{6}+x^{3}+5 = 0mod7

=> 4x^{6}+x^{3 }= 2mod7

=> x^{3}(4x^{3}+1) = 2mod7

Let x^{3 }= 1mod7, then x^{3}(4x^{3}+1) = 5mod7.

Let x^{3} = 2mod7, then x^{3}(4x^{3}+1) = 4mod7

Let x^{3} = 3mod7, then x^{3}(4x^{3}+1) = 4mod7

Let x^{3} = 4mod7, then x^{3}(4x^{3}+1) = 5mod7

Let x^{3} = 5mod7, then x^{3}(4x^{3}+1) = 0mod7

Let x^{3} = 6mod7, then x^{3}(4x^{3}+1) = 3mod7

Let x^{3} = 0mod7, then x^{3}(4x^{3}+1) = 0mod 7.

That means x^{3}(4x^{3}+1) is never 2mod7 in turn **4x ^{6}+x^{3}+5 is never divisible by 7 for any value of x.**

**What is the remainder if 2**^{2 }**+ 22**^{2 }**+ 222**^{2 }**+.....+ 22222......**_{49 times}^{2 }**is divided by 9 ?**

Hello Vidisha !

Take 2² common from each term ,

2²( 1²+11²+111²+....1111....11²) mod 9

4(1+2²+3²+....+49²) mod 9

4 × 49×50×99/6 mod 9 = 6.

the number 888....M999.... is divisible by 7, such that there are fifty 8's before M and fifty 9's after M (M is an integer). Then the value of M is:

A. 3 B. 4 C. 5 D. 6

Hello Ritika !

We know any digit repeated 6 times is always divisible by 7.

so 8888......written 48 and 99999.... written 48 times is always divisible by 7.

Now , 88M99 should be divisible by 7 .

M99 - 88 should be divisible by 7

M = 5 .

Option (C )

Find all positive integers n* *such that n^2 + 3n*+ *1 is a multiple of 3n+ 10.

Hello, Ritika

Please find the soloution.

n^{2} + 3n + 1 is a multiple of 3n + 10.

=> (n^{2} + 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.

Sir I still didnt get the last step! 4 × 49×50×99/6 mod 9 = 6 How did this came?

The highest power of 12 that can divide 5^36-1? Please explain in detail.

Hi Apoorva,

Thanks for asking question

Please find the solution below.

Using binomial theorem,

(a + b)^{n} = a^{n} + (^{n}C_{1})a^{n-1}b + (^{n}C_{2})a^{n-2}b^{2} + … + (^{n}C_{n-1})ab^{n-1} + b^{n}

5^{36} - 1 = 25^{18} - 1 = (24 + 1)^{18} - 1 = 24^{18} + 18 x 24^{17} + ... + 18 x 24 + 1 - 1 = 12^{2}k.

You can see here that after expanding only 18 x 24 ^{ }is possible to take common. So 18 x 24 is divided by 12^{2}

So maximum required the power of 12 is 2.

OR

5^{36} - 1 = (4 + 1)^{36} - 1 = 12^{2}k.

So highest power of 12 that will divide the expression is 2

Hi Apoorva !

Sum of squares of first n natural number = n ( n +1 )( 2n +1 ) /6

so (1+2²+3²+....+49²) = 49 ( 50) ( 99)/6 .

11!^12! can be written as 11^12! X 10!^12!.

10!^12! = 10!^11! X 10!^11x11!

Therefore, it can be written as 11^12! X **10!^11!** X 10!^11x11! **/** **10!^11!**

So, the remainder is zero.

Hello Kinshuk,

Please find the solution below,

Two digit numbers which leave a remainder of 6 when divided by 8 are 14, 22, 30, 38, ……94.

Let the number of terms in the series 14, 22, 30, 38, ……94 is n

T_{n} =a + (n-1) d

94 = 14 +(n-1) 8

94 =14 + 8n-8

8n =88 => n=11

Sum of the series 14, 22, 30, 38, ……94

= (n/2) (2a + (n-1)d)

=(11/2) (2 x 14 + (11-1)8)

=(11/2)(28+80)

=(11/2)(108) =594