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Divisibility and Remainders

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Post Questions based on Divisibility and Remainders in this forum 

This topic was modified 2 months ago by TathaGat
80 questions & discussions are there under this sub-topic
2

Q1.

q11
Shivangi Garg 11/02/2024 4:03 pm

@neraj-naiyar Sir,

 

Ans:-

Q11 ans

  

0

What is the remainder if 2+ 22+ 222+.....+ 22222......49 timesis divided by 9 ?

This post was modified 2 months ago by TathaGat
TG Team 04/07/2018 4:10 pm

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. 

MayankSri 09/06/2024 1:51 am

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

TathaGat 09/06/2024 1:52 am

@Mayank

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

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

0
Topic starter

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

TG Team 04/07/2018 4:09 pm

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 ) 

0

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

Anonymous 24/02/2018 6:33 pm

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

Hello sir, please explain 

p18
Anonymous 02/04/2018 11:32 am

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

0
Topic starter

If a number N, when divided by D, gives remainder 23. The number, when divided by 12D, gives remainder 104. What will be the remainder when the number is divided by 6D?

Anonymous 02/04/2018 11:45 am

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

0
Topic starter

Find the remainder when 199! is divided by 67^3 ?

TG Team 02/04/2018 11:45 am

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

0

Find the remainder when 1021^1022 is divided by 1023 ?

TG Team 02/04/2018 11:45 am

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

0

Find the remainder when 7^2008 +9^2008 is divided by 64?

TG Team 04/07/2018 2:37 am

Hello Nishant . 

 

Do you have any idea about binomial expansions   ? 

TG Team 04/07/2018 2:43 am

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 . 

0

Hello Sir 

Please Answer 

Question 1- What is the remainder when f(a) = a^15+a^14+a^13+........a^2+a is divided by a^2-1

A. 7a+8

B. 8a +7

C. 9a+6

D. 5a+8

 

TG Team 07/07/2018 7:51 pm

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. 

0

Find the remainder when 7^2008 + 9^2008 is divided by 64?

TG Team 11/07/2018 8:21 pm

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 . 

Samyak 11/07/2018 10:25 pm

Can we do it without using binomial expansion? 

TG Team 12/07/2018 11:37 am

Yes we can by  using Euler / Carmichael Theorem 

It reduces to 7^8 + 9^8 mod 64. 🙂 

0

(47^47) - (13^13) is divisible by

options:

a.)7

b.)9

c.)10

d.)4

TG Team 14/07/2018 5:09 pm

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 ) 

0

The highest power of 12 that can divide 5^36 - 1 is 

A) 1

B) 2

C) 3

D)4

TG Team 29/07/2018 12:42 pm

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.  

0

Hey Sir, kindly share the solution.

IMG 20180730 WA0039
TG Team 30/07/2018 7:02 pm

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 . 

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