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 maximum of n such that 45*57*92*91*52*62*63*64*65*66*67 is perfectly divisible by 30 raised to power n.
Something is missing in the question since,
30n=2nx3nx5n
Number - 45*57*92*91*52*62*63*64*65*66*67 = 211x33x52x7x132….
So n=2 is the answer.
1260=23x32x52x7
For 1260n to divide 50! Perfectly, n should be such that, 23n ≤ maximum power of 2 in 50! and similarly for powers of 3, 5 and 7.
Now 50! = 247x322x512x78….
Thus maximum value of n is min(47/3,22/2,12/2,8) = 6
For how many +ve int between 1 to 1000, both inclusive, 4x6+x3+5 is divisible by 7
Let us assume, 4x6+x3+5 = 0mod7
=> 4x6+x3 = 2mod7
=> x3(4x3+1) = 2mod7
Let x3 = 1mod7, then x3(4x3+1) = 5mod7.
Let x3 = 2mod7, then x3(4x3+1) = 4mod7
Let x3 = 3mod7, then x3(4x3+1) = 4mod7
Let x3 = 4mod7, then x3(4x3+1) = 5mod7
Let x3 = 5mod7, then x3(4x3+1) = 0mod7
Let x3 = 6mod7, then x3(4x3+1) = 3mod7
Let x3 = 0mod7, then x3(4x3+1) = 0mod 7.
That means x3(4x3+1) is never 2mod7 in turn 4x6+x3+5 is never divisible by 7 for any value of x.
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.
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 )
Find all positive integers n such that n^2 + 3n+ 1 is a multiple of 3n+ 10.
Hello, Ritika
Please find the soloution.
n2 + 3n + 1 is a multiple of 3n + 10.
=> (n2 + 3n + 1)/(3n + 10) is an integer.
=> [(3n-1)(3n+10) + 19]/9*(3n+10)
=> 3n + 10 = -19 or -1 or 1 or 19
=> 3n = -29 or -11 or -9 or 9
=> n = -3 or 3. As n is an integer.
Sir I still didnt get the last step! 4 × 49×50×99/6 mod 9 = 6 How did this came?
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
Hi Apoorva !
Sum of squares of first n natural number = n ( n +1 )( 2n +1 ) /6
so (1+2²+3²+....+49²) = 49 ( 50) ( 99)/6 .
11!^12! can be written as 11^12! X 10!^12!.
10!^12! = 10!^11! X 10!^11x11!
Therefore, it can be written as 11^12! X 10!^11! X 10!^11x11! / 10!^11!
So, the remainder is zero.
Hello sir, please explain
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
