Quote:
If n is a positive integer less than 400, what is the number of n such that when n is divided by 4, the remainder is 3 and when n is divided by 7, the remainder is 1?
I found this question really tricky and spent some time thinking about the concept.
What we are given:
n is a positive integer
n > 0 and
n < 400 n / 4 = X + 3n / 7 = Y + 1We need to find a number of possible "n"s.
From what we are given,
n is divisable both by
4 and
7 with some remainder in both cases. Thus, the maximum number which satisfies condition
n / 4 = x + 3 is
n = 399 and for
n / 7 = Y + 1 it is
n = 394.
Since the number is divisable both by 4 and 7, we need to find out the LCM (least common multiple) of 4 and 7.
LCM of 4 and 7 is the least number smallest positive integer that is divisible by both 4 and 7, which is
4 * 7 = 28.
Now let us find out how many times the number 28 is met in 400 to find the number of "n"s. Since 394 is the smallest of two possible numbers, we need to use it in the calculation:
394 / 28 = 14 and R (remainder) is 2.
Thus, the answer is
D.