how to remove ambiguity that whether should take lower or upper shield in counting problem

Question

how many positive integer between 50 and 100 ,

(a) divisible by 7
(b) divisible by 11
(c) divisible by 7 and 11 ?

people generally answer this question as

(a) lowershield [((100-50)-1 )/7] = 7 , yess its true the no is 56 , 63 , 70, 84, 91 , 98
(b) lowershield [((100-50)-1 )/11] = 4 , oops its not true the no is 55,66,77,88,99

(c) lowershield [((100-50)-1 )/lcm(7,11)] = 0 , oops its not true the no is 77

if u will say to take uppershield then lets come

(a) uppershield[((100-50)-1)/7]=7 , yess its true

(b) uppershield[((100-50)-1)/11]=5 , yess its true

(c) uppershield[((100-50)-1)/lcm(7,11)]=1 , yess its true

now come to another question

how many no is divisible between 5 to 31 is divisible by 4 ?

case1 : loweshield [((31-5)-1)/4]=6 , yess its true 8,12,16,20,24,28

case 2 : uppershierld [((31-5)-1/4)]=7 , oops its wrong

how can we remove this ambiguity ?

if u will argue that check the no ? but we have to check for large no like 50 to 300000 then how can u count ?

my doubt is , is any way to remove this ambiguity ?

Answer 1

a) first no which is divisible by 7 = 56 and last no =98 so total no =((98-56)/7) + 1 = (42/7) +1 = 6+1=7 divisible by 7
b)  first no which is divisible by 11 = 55 and last no =99 so total no =((99-55)/11)+1 = (44/11)+1 = 4+1 =5
c)  first no which is divisible by 77 = 56 and last no =97 so total no =((77-77)/77)+1 =(0/77)+1 =0+1=1

Answer 2

simply think if question is asked how many no between 1 to 10 divisible by 2 .lets count { no are 2,4,6,8,10}..we got it by 10/2=5.. ok fine . next question how many no between 1 to 11 divisible by 2 .lets count  {again  no are 2,4,6,8,10}.how i got it 11/2=5.5. answer comes 5.5 but our result gives only 5.
by analyzing we find that   everything after decimal is neglected.this means we have used greatest integer concept here.
now when question is how many no between 5 and 31 divisible by 4.first of all we will break it in two parts 1 to 31 and 1 to 4.
 no between 5 and 31 divisible by 4= (no between 1 to 31 divisible by 4)-( no between 1 to 4 divisible by 4)
                                                              =greatest integer{31/4} -greatest integer{4/4}
                                                               =7-1
                                                                =6