@techsurge said:@Estallar12 bhai isse s negative aa raha hai
Yeah. Thoda error tha in calculation of value of d.
Editing and aap bhi dekho. :)
@vbhvgupta said:Q
Is it 20??
a,b,c be the coins in the beginning
after 1st transaction -> A has a-2b-2c
B has 3b and C has 3c
after 2nd transaction -> B would have 3b - 3(a-2b-2c) - 6c = 3b - 3a+6b+6c - 6c = 7b - 2a - 2c
just manipulate the equation a little bit -> 9b - 2(a+b+c) = 9b - 2*80 = 9b - 160
9b - 160 = 20
=> b = 20
@vbhvgupta said:Q
a + b + c = 80
=> 2a + 2b + 2c = 160 .... (i)
A gives away x Coins => x = 2(b + c)
Coins left = a - x, 3b, 3c Resp.
Suppose B distributes y coins => y = 2(a - x + 3c) .
Coins Left = 3(a - x), 3b - y, 9c
Given 3b - y = 20
=> 3b - y = 3b - 2(a - x + 3c) = 20 = 3b - 2a + 4b + 4c - 6c
or, 7b - 2a - 2c = 20
Solve this with (i),
=> 9b = 180, or b = 20.
@vbhvgupta
Should be 20..
Initially A had 54 coins, B had 20 coins n C had 6 coins..
In the 1st Operation..A gave 40 coins to B n 12 coins to C..Thereby trippling their coins..Thus, he is left with 2 coins..
In the 2nd Operation..B gave 4 coins to A n 36 coins to C..Thereby trippling their coins..
Thus, It will look this :
=>[ Operation ] A B C
=>01: 54 20 06
=>02: 02 60 18
=>03: 06 20 54
what is the rightmost digit preceding the zero in the value of 20^53?
@vbhvgupta said:what is the rightmost digit preceding the zero in the value of 20^53?
2?
20^53 means 2^53...now cyclicity of 2 is 4
means last digit of 2^1=2
...........................2^2=4
...........................2^3=8
...........................2^4=6
...........................2^5=2
so it repeats after 4 cycles
therefore 53 mod 4=1
2^53=2^1=2
@vbhvgupta said:what is the rightmost digit preceding the zero in the value of 20^53?
20^53 = 2^53 * 10^53.
Zero(s) will be contirbuted by 10^53. So, you just need to find the last digit of 2^53.
Cyclicity of 2 = 2, 4, 8, 6
53 mod 4 = 1
Thus, Last Digit = 2. :)
@vbhvgupta
concentrate on 2^53 for the right most non zero digit
now the power of 2 has got a pattern.The units digit get repeated after every 4th power.
say 2^1=2
2^2=4
2^3=8
2^4=6
2^5=2 and from here the cycle repeats
now 53 mod 4=remainder 1
so only one power of 2 will be left after completing 13 cycles of 4 consecutive powers
answer=2
@vbhvgupta said:what is the rightmost digit preceding the zero in the value of 20^53?
20^53 has 53 zeroes
so we just need to find units digit of 2^53
53%4 = 1
so it is just 2^1 = 2
find remainder 2(8!) - 21(6!) divides 14(7!) + 14(13!)
@vbhvgupta said:find remainder 2(8!) - 21(6!) / 14(7!) + 14(13!)
Firstly, it should be the other way round division.
Now,
2(8!) - 21(6!) = 16*7! - 3*7! = 13*7! = X (suppose)
14*7! = 13*7! + 7! mod X = 7!
Also, 14*13! mod X = 0.
Thus, Remainder = 7!.
@vbhvgupta said:find remainder 2(8!) - 21(6!) / 14(7!) + 14(13!)
is this a self made question?? I can't figure out a way.... 😞
@Estallar12 said:Firstly, it should be the other way round division.Now, 2(8!) - 21(6!) = 16*7! - 3*7! = 13*7! = X (suppose)14*7! = 13*7! + 7! mod X = 7!Also, 14*13! mod X = 0.Thus, Remainder = 7!.
thats what....it should be the other way arnd. 7! it should be in that case........ 😃
@Estallar12 said:Firstly, it should be the other way round division.Now, 2(8!) - 21(6!) = 16*7! - 3*7! = 13*7! = X (suppose)14*7! = 13*7! + 7! mod X = 7!Also, 14*13! mod X = 0.Thus, Remainder = 7!.
if we consider that question posted by him answer is coming 3....yeah the question should be other way around..
@TONYMBA said:if we consider that question posted by him answer is coming 3....yeah the question should be other way around..
nah, in that case the answer would be 2(8!) - 21(6!) as the denominator > numerator 😃