M7MBA wrote: ↑Thu Aug 20, 2020 8:42 am
When positive integer \(k\) is divided by 5, the remainder is 2. When \(k\) is divided by 6, the remainder is 5. If \(k\) is less than 40, what is the remainder when \(k\) is divided by 7?
A. 2
B. 3
C. 4
D. 5
E. 6
Answer:
B
Solution:
We are given that k < 40. Since, when positive integer k is divided by 5, the remainder is 2:
k = 5Q + 2
So k can be 2, 7, 12, 17, 22, 27, 32, or 37.
Since, when k is divided by 6, the remainder is 5:
k = 6P + 5
So k can be 5, 11, 17, 23, 29, or 35.
Thus, we see that k must be 17, and 17/7 = 2 remainder 3.
Alternate Solution:
Since, when positive integer k is divided by 5, the remainder is 2:
k = 5Q + 2
Since, when k is divided by 6, the remainder is 5:
k = 6P + 5
We see that k - 2 = 5Q = 6P + 3 is divisible by both 5 and 3; therefore, k - 2 must be divisible by 15.
The only numbers divisible by 15 and which would produce a k-value less than 40 are 0, 15 and 30. If k - 2 is 0, 15 or 30; then k is 2, 17 or 32, respectively. We see that only k = 17 produces a remainder of 2 when divided by 5 and a remainder of 5 when divided by 6. The remainder when k = 17 is divided by 7 is 3.
Answer: B
