Bunuel
If positive integer x leaves the remainder 1 when divided by 3 and the remainder 3 when divided by 7, how many such numbers are there between 206 and 258?
(A) 0
(B) 1
(C) 2
(D) 3
(E) Cannot be determined from the information given.
In the given range the number of multiples of \(7\) will be less than the number of multiples of \(3\)
Hence the given criteria will be restricted by the values that satisfy the criteria of \(7\)
Lets check all the values in the range that leave remainder of \(3\) when divided by \( 7\)
Lowest multiple of \(7\) in the given range \(= 30* 7= 210 \) Hence \(210+3= 213\) will leave a remainder of \(3\) when divided by \(7 \)
Does \(213 \) leave a remainder of \(1\) when divided by \(3 \)?
NO Hence we can rule out \(213 \) as a possible answer
Next multiple of \(7\) in the range \(= 31*7= 217\) hence \(220 \) will leave a remainder of \(3\) when divided by \(7 \)
Does \(220\) leave a remainder of \(1\) when divided by 3 ?
YesHence \(220\) is one of the values that is the soln. to this question.
Similarly we can check other multiples of \(7\) .
We will see the other multiple of \(7 \) that meets the criteria is \(7*34= 238\) Hence \(238+3= 241\)
Hence there are only two values \(220 \) and \(241\) in the given range that meet the criteria given in the question.
Ans C
Hope it's clear.