if 76 < x< 92 and 142 < x< 2003; then how many solutions are possible for the following equation: [x/143] + [2x/91] + [3x/77] = 68x/1001 where [x] represents greatest integer less than equal to x.
RHS is same as LHS but without gint ..... so only their common multiples will hold the equality there lcm is 1001 , 2002 only these two values fall in given range of x ..
v knw gint is always integer...so output on LHS is integer....so,RHS wil also be integer.so, x should be a multiple of 1001 i.e. 13*7*11. consequently x/143, 2x/91, 3x/77 all will be integers....so, gint can be removed.nw x/143 + 2x/91 + 3x/77 = 68x/1001 is always true..so, x= 1001, 2002 in d givn range....so, ans is 2
6 comments:
is it 2?
yup answr is 2...cn u explain it...
RHS is same as LHS but without gint .....
so only their common multiples will hold the equality
there lcm is 1001 , 2002
only these two values fall in given range of x ..
hu can't remove gint ds way...well u r going on d right track...gint can be removed in ds particular question....bt hwcmmm??????
v knw gint is always integer...so output on LHS is integer....so,RHS wil also be integer.so, x should be a multiple of 1001 i.e. 13*7*11.
consequently x/143, 2x/91, 3x/77 all will be integers....so, gint can be removed.nw
x/143 + 2x/91 + 3x/77 = 68x/1001 is always true..so, x= 1001, 2002 in d givn range....so, ans is 2
i was also tryin to say the same...
he he bt smhw cud nt xplain that well
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