Divisibility and Re...
 

Divisibility and Remainders  

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

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

94 Answers
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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.

0

Something is missing in the question since,

30n=2nx3nx5n

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

So n=2 is the answer.

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1260=23x32x52x7

For 1260n to divide 50! Perfectly, n should be such that, 23n ≤ maximum power of 2 in 50! and similarly for powers of 3, 5 and 7.

Now 50! = 247x322x512x78….

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

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For how many +ve int between 1 to 1000, both inclusive, 4x6+x3+5 is divisible by 7 

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Let us assume, 4x6+x3+5 = 0mod7

=> 4x6+x= 2mod7

=> x3(4x3+1) = 2mod7

Let x= 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.

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What is the remainder if 2+ 22+ 222+.....+ 22222......49 timesis 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. 

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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 ) 

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Find all positive integers n such that n^2  + 3n1 is a multiple of 3n+ 10.

0

Hello, Ritika

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.

0

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

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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 = an + (nC1)an-1b + (nC2)an-2b2 + … + (nCn-1)abn-1 + bn

 

536 - 1 = 2518 - 1 = (24 + 1)18 - 1 = 2418 + 18 x 2417 + ... + 18 x 24 + 1 - 1 = 122k.

You can see here that after expanding only 18 x 24  is possible to take common. So 18 x 24 is divided by 122 

So maximum required the power of 12 is 2.

 

OR

536 - 1 = (4 + 1)36 - 1 = 122k.

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

 

0

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 .

0

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.

0

Hello sir, please explain 

p18

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

Tn =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

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