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
Hello Apoorva,
Correct question uploaded here, Please read again.
Find the solution below, If you wants more explanation of this question then ask for detail explanation.
What is the remainder when 7^2017 is divided by 33?
Can you explain it without the use of Chinese Remainder Theorem.
Yes Apoorva,
Find the solution without the use of Chinese Remainder Theorem
What is the remainder of 38! / 41 ?
Hello ritika,
Please find the solution,
Using Wilson's theorem, we get
(41 - 1)! = 40! = -1mod41 = 40mod41
=> 39! = 1mod41
and let 38! = xmod41
=> 39x = 1mod41
=> -2x = 1mod41
=> 41y - 2x = 1
=> x = 20 and y = 1.
So Remainder when 38! is divided by 41 is 20.
Thanks sir
Sir,
I cant understand the step when
-2x=1mod41
Can you explain last three steps again
Hello Apoorva,
Find the solution in detail
As per Wilsons theorem when (n - 1)! is divided by p we will get a remainder of - 1 when p is a prime number.
As per the theorem when 40! is divided by 41 we will get a remainder of – 1
When (41 -1)! Is divided by 41, remainder is 1.
It can be written as 40 X 39 X 38! divided by 41 will give a remainder of -1
(41–1)(41–2) 38! / 41 will give a remainder of - 1 since 41 in the bracket will get divided by 41 we can write as -1*-2**Y /41 will give a remainder of -1 where Y is 38!
Rem 2Y / 41 = -1
or Rem 2Y = 41a - 1 where a is the constant factor
since the left hand side is a multiple of 2 the right hand side also will have to be multiple of 2. we will have to substitute values for a in such a way that 41a - 1 will is a multiple of 2.
Put a =1
2Y = 41 x 1 - 1
2Y =40 => Y=20
Where Y is the remainder when 38! Is divided by 41
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
1234123412341234.........upto 400 digits/999, then what is the remainder?
As 1000 = 103 ≡ 1 mod999.
So 12341234.... 400 digits ≡ 1 + 234 + 33(123 + 412 + 341 + 234) mod999 ≡ 1 + 234 + 33(111) mod999 ≡ 901 mod999
what is the remainder when 123123123............(300 digits) is divided by 504
hi vidisha ,
there is another topic by the name of remaineder plz post your question under that topic
I have a doubts in 23 and 24 th question
Find the remainder when 7^2008 +9^2008 is divided by 64?
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 .
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.
