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n ranges over the positive integers between 100 and 200, inc

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by Max@Math Revolution » Tue Jul 30, 2019 10:54 pm

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[GMAT math practice question]

n ranges over the positive integers between 100 and 200, inclusive. Find the number of values of 7n+2 which are multiples of 5

A. 18
B. 20
C. 22
D. 24
E. 26
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by Max@Math Revolution » Fri Aug 02, 2019 12:46 am
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If n = 5k, then 7n + 2 = 7(5k) + 2 = 5(7k) + 2 is not a multiple of 5.
If n = 5k+1, then 7n + 2 = 7(5k+1) + 2 = 5(7k) + 7 + 2 = 5(7k) + 9 = 5(7k+1)+4 is not a multiple of 5.
If n = 5k+2, then 7n + 2 = 7(5k+2) + 2 = 5(7k) + 14 + 2 = 5(7k) + 16 = 5(7k+3)+1 is not a multiple of 5.
If n = 5k+3, then 7n + 2 = 7(5k+3) + 2 = 5(7k) + 21 + 2 = 5(7k) + 23 = 5(7k+4)+3 is not a multiple of 5.
If n = 5k+4, then 7n + 2 = 7(5k+4) + 2 = 5(7k) + 28 + 2 = 5(7k) + 30 = 5(7k+6) is a multiple of 5.
Thus, n has remainder 4 when it is divided by 5.

The possible values of n are 104, 109, ..., 199.
The number of possible values of n is (199-104)/5 + 1 = 20.

Therefore, B is the answer.
Answer: B
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by deloitte247 » Sun Aug 04, 2019 8:52 am
$$if\ n=5k,\ then\ 7n+2=7\left(5k\right)+2=5\left(7k\right)+2$$
This is not a multiple of 5, it remains 2.
$$if\ n=5k+1,\ then\ 7n+2=7\left(5k+1\right)+2=5\left(7k\right)+7+2=5\left(7k\right)+5+4$$
This is not a multiple of 5; it remains 4.
$$if\ n=5k+2,\ then\ 7n+2=7\left(5k+2\right)+2=5\left(7k\right)+14+2=5\left(7k\right)+15+1$$
This is not a multiple of 5; it remains 1.
$$if\ n=5k+3,\ then\ 7n+2=7\left(5k+3\right)+2=5\left(7k\right)+21+2=5\left(7k\right)+20+3$$
This is not a multiple of 5; it remains 3.
$$if\ n=5k+4,\ then\ 7n+2=7\left(5k+4\right)+2=5\left(7k\right)+28+2=5\left(7k\right)+30$$
This is a multiple of 5.
Therefore, n can be said to have remainder 4 when it is divided by 5.
So, the possible values of n = 104, 109, 114, ..., 199
$$possible\ value\ of\ n=\frac{\left(199-104\right)}{5}+\frac{1}{1}=\frac{95}{5}+1$$
$$possible\ value\ of\ n=19+1=20\ \ \ \ \ \left(option\ B\right)$$
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