Question

find the no of solutions for x if 100<=x<=1000 such that[ x/6] +[x/5] +[x/10] =7x/15 where [x] indicates the greatest integer less than or equal to y.

Answer 1

Answer:

Step-by-step explanation:

[x+  

2

1

​  

][x−  

2

1

​  

]=prime

Hence possible solutions can be 2×1=2

and −2×−1=2

First case-

[x+  

2

1

​  

][x−  

2

1

​  

]=2

where [x+  

2

1

​  

]=2  and [x−  

2

1

​  

]=1

⟹x+  

2

1

​  

∈[2,3) and x−  

2

1

​  

∈[1,2)

⟹x∈[  

2

3

​  

,  

2

5

​  

)

Second case-

[x+  

2

1

​  

][x−  

2

1

​  

]=2

where [x+  

2

1

​  

]=−1  and [x−  

2

1

​  

]=−2

⟹x+  

2

1

​  

∈[−1,0) and x−  

2

1

​  

∈[−2,−1)

⟹x∈[  

2

−3

​  

,  

2

−1

​  

)

Hence solution set is x∈[  

2

3

​  

,  

2

5

​  

)∪[  

2

−3

​  

,  

2

−1

​  

)

⟹x  

1

2

​  

+x  

2

2

​  

+x  

3

2

​  

+x  

4

2

​  

=  

4

3  

2

+5  

2

+3  

2

+1  

2

 

​  

 

=  

4

44

​  

=11