rx10 wrote:
Range : 13 to 217
A no. with remainder \(4\) after divided by \(7\)
Nos will be 18, 25, 32, ... , 214
In an arithmetic series : T_n \(= a + (n-1)d\)
where,
T_n = \(n^{th}\) term
\(a =\) first term
\(d =\) common difference
\(214 = 18 + (n-1)7\)
\(n = 29\)
Answer C
Accepted. Your solution may be a bit cumbersome, if I enter additional operation, say divided by 8 with the remainder 3. Then we are looking for two sets of integers with a common subset.
My take on this question is universal and follows
\(7x + 4 =< 217\), where \(x\) is integer and \(4\) is a remainder.
\(7x =< 213\), and \(x =< 30\) \(3/7\) not including the case with \(x=0\). Hence 31 integers are between 0 ... 217, and reducing the number of integers by 2 for \(x = {0, 1}\) result in \(7x + 4\) such as \(7*0+4=4\) and \(7*1+4=11\). Answer is \(31-2=29\) integers. As I said with additional operation and more remainders we are looking into a common subset and need to simplify the solution, IMH