Divisibility and Remainders

@neraj-naiyar Sir,

Ans:-

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. 

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

@Mayank

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

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

Hi Apoorva,

Thanks for asking question

Please find the solution below.

Using binomial theorem,

(a + b)ⁿ = aⁿ + (ⁿC₁)a^n-1b + (ⁿC₂)a^n-2b² + … + (ⁿC_n-1)ab^n-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

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

Hi Ritika,

N = 23modD                  
N = 104mod12D              
=> 104 = 23modD   
=> 81 = 0modD
=> D = 27 or 81
Now Nmod6D = 104mod6D [Because 12d is divisible by 6D]
And 27*6 = 162<104.
=> N = 104mod6D

Hello , Ritika 

199 ! = 1× 2 × 3 × ......67 × 68 × 69 ×........132 × 133×134×......199.

 1× 2 × 3 × ......67 × 68 × 69 ×........132 × 133×134×......199 mod 67³

1×2×3×......66×68×69×.....133×2×135×.......199 mod 67

66! × 66! × 65! × 2 mod 67

We know , ( P-1)! mod P = ( P-1 ) or -1 [ Wilson's Theorem ] 

-1 × -1 × 1× 2= 2

Hence , net remainder 2× 67² = 8978

Hello Abhi , 

1021^1022 mod 1023 

Using Negative remainder concept  

(-2)^1022 mod 1023 

2^1023 mod 1023 

 2^10 mod 1023 = 1024 mod 1023 = 1

so (2^10)^102 * 2^2  mod 1023 = 4

Hello Nishant . 

Do you have any idea about binomial expansions   ? 

Approach : Using binomial expansions . 

If you have any doubts  feel free to write back 

7^2008 + 9^2008 mod 64

( 8 -1) ^2008 + (8 +1)^9 ^2008 mod 64

Now the binomial expansion of ( 8 -1)^2008 = 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (-1)^2007 + a multiple of 64

and the binomial expansion of (8 + 1)^2009 = 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (+1)^2007 + a multiple of 64

Hence ( 7 - 1)^2009 + ( 7 +1)^2009 mod 64

= 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (-1)^2007 + 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (1)^2007 mod 64

1 + 1 mod 64 = 2 . 

Hi Mayank ! 

We know, the remainder when f(x) = a + bc + cx^2 + dx^3 +...... is divided by  ( x - a ) is f(a).

( Check NS - 4 Class Sheet ) 

Hence , Substitute a^2 = 1 in the expression  a^15+a^14+a^13+........a^2+a to get the remainder . 

(a^2)^7. a + (a^7)^2 + ( a^2)^6 . a + (a^2)^6 + ...... + (a^2)a + a + 1 

(1)^7. a + (1)^2 + (1)^6 . a + (1)^6 + ...... + (1)a + a + 1 

= 8a + 7. 

Option B. 

Hi Samyak , 

Approach : Using binomial expansions . 

If you have any doubts  feel free to write back 

7^2008 + 9^2008 mod 64

( 8 -1) ^2008 + (8 +1)^9 ^2008 mod 64

Now the binomial expansion of ( 8 -1)^2008 = 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (-1)^2007 + a multiple of 64

and the binomial expansion of (8 + 1)^2009 = 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (+1)^2007 + a multiple of 64

Hence ( 7 - 1)^2009 + ( 7 +1)^2009 mod 64

= 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (-1)^2007 + 2008C1 × 8^0 × (1)^2008 + 2008C1 × 8 × (1)^2007 mod 64

1 + 1 mod 64 = 2 . 

Can we do it without using binomial expansion? 

Yes we can by  using Euler / Carmichael Theorem 

It reduces to 7^8 + 9^8 mod 64. 🙂 

Hello Harsh , 

Check the last digit of the expression (47^47) - (13^13)

47^47 ends with 3 and 13^13 also ends with 3 

Hence the unit digit of (47^47) - (13^13) is 0. So the expression is divisible by 10 . ( Option C ) 

Hello Rhythm, 

5³⁶ - 1 = 25¹⁸ - 1 = (24 + 1)¹⁸ - 1

Now just expand binomially and the last term has the lowest power of 12 i.e 

24¹⁸ + 18 × 24¹⁷ + ... + 18 × 24 + 1 - 1 

Thus 2 is the answer.  

Hello Richa , 

( n + 32)² is divisible by n +4

=> ( n + 32)² = ( [n + 4] + 28)²

so ,   28²    should be divisible by n + 4 

(n is a natural number so n + 4 >  5)

Hence all the factors of 28² > 5 will be the answer . 

28² = (2²7)² = 2⁴7² => 15 factors

remove 1 , 2 and 4 . hence 15 - 3 = 12 values .