if 1001 divides a 6-digit number of the form xxyyzz where x y z leaves the remainder R what is

@nishant
i think u skip one condition i.e.x!=y!=z.
9 years agoHelpfull: Yes(3) No(1)
how is 1!=0!=9 if 110099 is xxyyzz?
9 years agoHelpfull: Yes(3) No(1)
@nishant how u have start to solve this question... pls reply
9 years agoHelpfull: Yes(1) No(0)
ans (B)
xxyyzz= 110011 or 110000
for case 110000 we get remainder 891
for case 110011 we get remainder 902
9 years agoHelpfull: Yes(1) No(2)
Ans Will be option B. 891
9 years agoHelpfull: Yes(1) No(0)
case 1 x! = y! = z
1! = 0! =1
output is 1 = 1 = 1 condition true make a number 110011

case 2 x!=y!=z
0!=1!=1
output is 1=1=1 condition is true but make a number 001111 so that's why this is not 6 digit number

110011 / 1001 = reminder is 902
3 years agoHelpfull: Yes(1) No(0)
nishant you are ryt totally..
9 years agoHelpfull: Yes(0) No(6)
what is this condition x!=y!=z.????? I cann't understand it. Explain it clearly @ Nishant hw can u take 1001*109+990 there are various other numbers????/
9 years agoHelpfull: Yes(0) No(0)
X!=y!=z means x,y and z are not equal to each other.
9 years agoHelpfull: Yes(0) No(2)

if 1001 divides a 6-digit number of the form xxyyzz where x!=y!=z leaves the remainder R...what is max value of the remainder R?

A)990
B)891
C)561
D)121