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What is the remainder if 22 + 222 + 2222 +.....+ 22222......49 times2 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.
Sir, I still didnt get the last step! 4 × 49×50×99/6 mod 9 = 6 How did this came?
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 )
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
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
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?
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
Find the remainder when 199! is divided by 67^3 ?
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
Find the remainder when 1021^1022 is divided by 1023 ?
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
Find the remainder when 7^2008 +9^2008 is divided by 64?
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 .
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
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.
Find the remainder when 7^2008 + 9^2008 is divided by 64?
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 .
Yes we can by using Euler / Carmichael Theorem
It reduces to 7^8 + 9^8 mod 64. 🙂
(47^47) - (13^13) is divisible by
options:
a.)7
b.)9
c.)10
d.)4
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 )
The highest power of 12 that can divide 5^36 - 1 is
A) 1
B) 2
C) 3
D)4
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 .