joyjitpal
February 14, 2013, 9:06am
19646
@fireatwill said: Maximum number of square possible that can be constructed using 31 pencils of equal length on the table? PLEASE explain the method
the square will be of grid 4*3
gnehagarg
February 14, 2013, 9:13am
19650
@pavimai said: A 50 digit number has all 7 s .find the remainder when the number is divided by 74
2x+1=37y+3
saurav205
February 14, 2013, 9:17am
19652
@pavimai said: A 50 digit number has all 7 s .find the remainder when the number is divided by 74
3 Shoul be the remainder...cyclicity rule se...
saurav205
February 14, 2013, 9:21am
19653
@maroof10 said: what should be the values of a and b such that 30a0b03 is divisible by 13?
I think the condition.will be b-a should be 7.....
@maroof10 said: what should be the values of a and b such that 30a0b03 is divisible by 13?
Apply rule of divisibility of 13,you would get the answer:-
(3*1)+(b*-4)+(a*3)+(3*1)=multiples of 13
=>3-4b+3a+3=0 or 3-4b+3a+3=13 or 3-4b+3a+3=26 or 3-4b+3a+3=39
1st Case:
3a-4b=-6 =>a=2,b=3 and a=6,b=6
2nd Case:
3a-4b=7 =>a=5,b=2 and a=9,b=5
3rd Case:
3a-4b=20 =>a=8,b=1
4th case:
3a-4b=33 =>no solution
joyjitpal
February 14, 2013, 9:37am
19655
@saurabhlumarrai said: Apply rule of divisibility of 13,you would get the answer:-
what is the divisibility rule of 13?
@joyjitpal said: what is the divisibility rule of 13?
follow this link:
You would be amazed to find the divisibilty rule of 7 and 13
joey_sharma
February 14, 2013, 9:44am
19657
@maroof10 said: what should be the values of a and b such that 30a0b03 is divisible by 13?
there will be total 7 sets of (a,b) which are (0,8) , (2,3) , (3,7) , (5,2) , (6,6) , (8,1) , (9,5) .
@maroof10 said: follow this link and you will understand by yourself:-
it contains the divisibility rule of numbers from 1 to 20...you can follow the divisibility rule of 13...if u dont understand,you can revert back to me
pavimai
February 14, 2013, 11:22am
19661
remainder when 1^3+2^3+3^3...+1000^3 is divided by 13 is???
subhashdec2
February 14, 2013, 11:28am
19662
@pavimai said: what is the sum of all 5 digit numbers that are formed using 1,3,5,7,9 taken one at a time
numbers at 10000's position= 1*10^4*4! +3*10^4*4!+.....+9*10^4*4!=25*10^4*4!
1000's position = 25*10^3*4!
.
.
.
unit position =25*4!
25*4!(10^4+10^3+...+1)=16*24*9999/9 = 16*24*1111=666600
subhashdec2
February 14, 2013, 11:31am
19663
@pavimai said: remainder when 1^3+2^3+3^3...+1000^3 is divided by 13 is???
a^3+b^3+c^3+.... is divisible by a+b+c+...
pavimai
February 14, 2013, 11:34am
19664
@Subhashdec2 said: numbers at 10000's position= 1*10^4*4! +3*10^4*4!+.....+9*10^4*4!=16*10^4*4!
there is an easy funda to find it
sum of all numbers formed from given digits=(n-1)!*{sum of digits}*{1111...n times}
so
sum of all numbers formed from digits {1,3,5,7,9} will be= 4!{1+3+5+7+9}*{11111}
= 24*25*11111
=6666600
