@Aman.Malhotra said:If N = (63^1!+2!+3!+....+63! )+(18^1!+2!+3!+....+18!)+(37^1!+2!+3!+....+37!)than the unit digit of N is ...........??
8 ??
@Aman.Malhotra said:If N = (63^1!+2!+3!+....+63! )+(18^1!+2!+3!+....+18!)+(37^1!+2!+3!+....+37!)than the unit digit of N is ...........??
(1! + 2! + 3!) mod 4 = 1
[3^(4k + 1) + 8^(4k + 1) + 7^(4k + 1)] mod 4
=> (3 + 8 + 7) mod 4 = 2
@jain4444 said:If a car passes as many electric post in three minutes as it goes in kilometers per hour, how far apart are the posts?
50 meter Apart ????
@Aman.Malhotra said:If N = (63^1!+2!+3!+....+63! )+(18^1!+2!+3!+....+18!)+(37^1!+2!+3!+....+37!)than the unit digit of N is ...........??
8?
50 is correct
Is there an integer with the property that if you cross out its first digit, it decreased by 36 times?
@jain4444 said:(1! + 2! + 3!) mod 4 = 1 [3^(4k + 1) + 8^(4k + 1) + 7^(4k + 1)] mod 4 => (3 + 8 + 7) mod 4 = 2
Sir, 3^(4k+1) + (-3)^(4k+1) + 8^(4k+1) mod 4 = 8^(4k+1) = 8
I did like this, then why is the answer different?? :(
@jain4444 said:50 is correctIs there an integer with the property that if you cross out its first digit, it decreased by 36 times?
72?
@jain4444 said:50 is correctIs there an integer with the property that if you cross out its first digit, it decreased by 36 times?
72?
@ScareCrow28 said:Sir, 3^(4k+1) + (-3)^(4k+1) + 8^(4k+1) mod 4 = 8^(4k+1) = 8I did like this, then why is the answer different??
you need to divide 17 by 10 not by 4 for unit digit
@mailtoankit said:kaise hoga yaar?..
Let speed = x and distance between poles = y
Then, x = xy/ 3/60
=> y = 1/20 km = 50 mtrs
72 is correct for last one
I - 9
II - 27
III - 15
a) only I
b) I and II
c) only III
d) I , II and III
@jain4444 said:50 is correctIs there an integer with the property that if you cross out its first digit, it decreased by 36 times?
72 & 720 ??
Find a,b and c if the number "abc" in base 5 is equal to the number "cba" in base 8.
@jain4444 said:Find a,b and c if the number "abc" in base 5 is equal to the number "cba" in base 8.
331?
25a + 5b + c = 64c + 8b + a
24a = 63c + 3b
8a = 21c + b
b = 3
c = 1
a = 3
abc = 331
25a + 5b + c = 64c + 8b + a
24a = 63c + 3b
8a = 21c + b
b = 3
c = 1
a = 3
abc = 331
@jain4444 said:Find a,b and c if the number "abc" in base 5 is equal to the number "cba" in base 8.
a=3=b, c=1 ?
@jain4444 said:72 is correct for last one In a natural number a the digits were swapped round (put in different order), after which the number decreased by 3 times. (was 3 times less then the original) . initial number is divisible by I - 9II - 27 III - 15 a) only I b) I and II c) only III d) I , II and III
b?
142857 * 3 = 428571
428571 is divisible by 27
142857 * 3 = 428571
428571 is divisible by 27
@jain4444 said:what should be the general term for it ?
36(10x+y) = 700+10x+y
360x+36y-10x-y =700
350x+35y=700
70x+7y=140
X=2 Y=0 Satisfy
So 36*20
36*200
36*2000
36*20000 and so on......
360x+36y-10x-y =700
350x+35y=700
70x+7y=140
X=2 Y=0 Satisfy
So 36*20
36*200
36*2000
36*20000 and so on......
@jain4444 said:what should be the general term for it ?
Sir, thoda bhot general term aya hai..
Let the 1st digit of no be "x" and rest of the no be "y"
For ex:- In 245, x=2 and y=45
So, according to question;
10^n*x + y = 36y
=> 10^n * x= 35*y
So, x = 35y/10^n
We can put values of n=1,2...and so on and check what is possible value of y such that a is less than 9 and an integer.
Also from here we can see that x can only be 7...