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
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 .
Can we do it without using binomial expansion?
What will be the remainder when 11111111- - - -(27times) is divided by 999?
111111.... (27 ) times
111 × 10^24 + 111 × 10^21 + 111 × 10^18 +...... +111 × 10^3 + 111
As 1000 = 10^3 ≡ 1 mod 999.
111 × 10^24 + 111 × 10^21 + 111 × 10^18 +...... +111 × 10^3 + 111
≡
111 × 1 + 111 × 1 +...... + 111 × 1 mod 999
≡ 999 mod 999
≡ 0 mod 999.
Hence divisible by 999.
(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.
Hi sir, please help with the problem below:
Hey Sir, kindly share the solution.
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 .
How many numbers N are there such that sum of N and its digits is 1000003?
Sir, I did not understand the solution to this question. Can you please explain using some other method?
Sir Help!
Find the digit A if the number 888….888A999….999 is divisible by 7, where both the digits 8 and 9 are 50 in number.
Hello Richa ,
Any digit repeated (p-1) times is always divisible by p if p is a prime number greater than 5.
So 888888 written 6 times is divisible by 7
[ or 888888 = 888 × ( 1001) = 888 × 7 × 11 × 13 so any digit repeated 6 times will be divisible by 7 , 11 and 13. ]
Hence , 88888.... written 48 times ( a mutiple of 6) and 9999.... written 48 times will be divisible by 7
So we need to find the value of A for which 88A99 is divisible by 7 .
88 × 1000 + A99 should be divisible by 7
-4 + A × 100 + 99 should be divisible by 7
-4 + 2A + 1 should be divisble by 7
A = 5
Thankyou sir!
A doubt from copy cat 03 -
How many four digit positive integers divisible by 7 have a property that , when the first and the last digit is interchanged , the result is a(not necessarily four digit) positive integer divisible by 7?
ABCD – DBCA
1000A + 100B + 10C + D – ( 1000D + 100B + 10C + A)
999A – 999D
999( A-D)
999 is not divisible by 7 . so ( A – D) should be a multiple of 7.
(A- D) = ( 9 – 2) , ( 8 -1 ) , ( 7-0), ( 2 – 9) ,( 1-8) , and ( 1 -1) , ( 2-2) ,…., ( 9 – 9) total 9 + 5 = 14 cases
Now for each of these 14 cases there are 15 multiples of 7 .
[ 1. 9002 , 9072, 9172, ……., 9982 ( 15 numbers )
2. 8001 , 8071, ……….., 8981 ( 15 numbers ) similarly others ]
Hence Total : 14 15 = 210 such numbers .
Hi Tarishi
We Know,
an + bn is divisible by a + b when n is ODD.
an – bn is divisible by a + b when n is EVEN.
an – bn is ALWAYS divisible by a – b.
So ,
182000 + 122000 – 52000 -12000
182000 – 52000 is divisible by (18 -5) i.e 13
122000 - 12000 is divisible by (12 + 1) i.e 13
so the whole expression is divisible by 13.
Again ,
182000 –12000 is divisible by ( 18 – 1) i.e 17
122000 – 52000 is divisible by ( 12 + 5) i.e 17
So the whole expression is divisible by 17.
Hence 182000 + 122000 – 52000 -12000 is divisible by ( 17 x 13) i.e 221.
let b be a positive integer and a =b^2-b . If b >= 4 , then a^2-2a is divisible by :
a) 15
b)20
c) 24
d) All of the above
a = b2 – b
a2 – 2a = a (a – 2)
= (b2 – b)( b2 – b – 2)
= (b -1)(b-2) b (b+1)
Product of four consecutive integers is always divisible by 4! i.e = 24 .
hence Option C .
What is the remainder when 7777777.....201 digits is divisible by 41? Can we solve it using Euler's theorem?
Hi Aarushi ,
Any digit repeated ( p -1) times is always divisible by p if p is a prime number greater than 5. ( an application of Euler's theorem )
So 77777..... written 40 times is divisible by 41.
Hence 77777.... Written 200 times is a multiple of 41.
Net remainder : 7
