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
a=10x+y
b=10y+x
a+b = 11(x+y)
x+y =11 for a+b to be a perfect sqaure and a and b to be a two digit nos.
Values can be
(9,2) (8,3) (7,4) (6,5)
Hence 4 pairs
Hi Aniket, thankyou for the solution.
But since the question is asking for ordered pairs, shouldn't it be 8 pairs? (9,2) (2,9) (8,3) (3,8) (7,4) (4,7) and (6,5) (5,6)?
Please suggest if im going wrong somewhere
Yes ! You are right answer should be 8
solution please
put 17 as (2-19) -19 to the power 36 and 19 to the power 36 cancels out hence the answer is 2
solution please
(Any odd number )^ (4k ) ends with 1 or 5
*( odd multiples of 5 )^(Any natural number ) ends with 5.
(Any even number )^4k ends with 6 or 0
*(Even multiples of 5)^ any natural number => 0
________________________________________
Here 2^2^2002
2 ^ ( 2^2002)
2^ ( 2 x 2 x 2 ..... 2002 times )
=> 2^4k
Hence , the last digit of 2^2^2002 => 6
solution please
N = DQ + 52
=> 5N = D (5Q) + 52 x 5
5N = 5DQ + 260
Also 5N = Dx + 4
So , D divides (260 - 4) i.e 256
hence , All factors of 256 > 52 .
[ 256 , 128 and 64 ]
solution please
PFA the solution
What is the remainder when 7190 is divided by 199 ?
How many two digit numbers are there such that last two digits of this number remain unchanged when it is cubed?
this could only be possible if both digits must be same so nos. are
33,44,55,66,77,88,99
total =7 (answer)
1 + 1/2 + 1/3 = 1/4 +....... + 1/23 = n/23!
Find the remainder when n is divided by 23.
answer is 0?
correct ans: 7
is that + or = bw 1/3 and 1/4
What is the remainder when 13!^14!^15! Is divided by 16?
1 x 2 x 3 x 4 x 5x ...... x 13 i.e 13! is a multiple of 16 .
Hence 13! raised to any power would be a multiple of 16.
Therefore the remainder will be 0.
If n=539x2^18 and m=9x2^13 then the remainder when n is divided by m is?
PFA the solution:
What are the last non zero digit of 36!-24!?
Answer is 4?
How?
N*190/100 = N*19/10
No. N should have 19 and 10 so that when it is increased by 90% it becomes a perfect square hence it will be multiple of 190
Possible values can be (190,380,570,760,950)
Only 190 and 760 will give perfect sqaure when they are increased by 90%
190/7 gives remainedr 1 and 760/7= gives remainder =4
Sum of remainder= 4+1 =5 answer
done (Y)
9. Find the largest positive integer for which
If A = 71421 .......... 98 105 112 .........189 196,
what is the remainder when A is divided by 9?
(A) 1 (B) 3 (C) 7 (D) 5
Can we use the concept of digital sum of an A.P. here?
If we add 7+14+21...196, will it be same as the digital sum of A? And since we have to find the remainder when divided by 9, we can anyway eliminate the 9s while adding.
To obtain the 9’s remainder of any number, we can proceed in a number of ways.
At one extreme, we can consider the given number itself as a single quantity.
As the other extreme, we can focus on each individual digit and add them all up.
As an intermediate approach, we can ‘fragment’ the number in any convenient way and consider the fragments.
In the given question, the third approach is the most convenient.
i.e. instead of A itself (or the individual digits of A) we consider B = 7 + 14 + 21 + … + 189 + 196 = 7(1 + 2 + 3 + … + 27 + 28) = 7(14)(29)
The 9’s remainder of the product of several numbers is the product of the 9’s remainders of the numbers.
∴ The required remainder is the remainder of 7(5) (2) or 7
Sir , can you please share the solution to ques 9?
