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When positive integer n is divided by 5, the remainder is 1.
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24 Aug 2020, 09:51
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When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?
Re: When positive integer n is divided by 5, the remainder is 1.
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24 Aug 2020, 10:08
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Carcass wrote:
When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?
(A) 3 (B) 4 (C) 12 (D) 32 (E) 35
There's a nice rule that says, If, when N is divided by D, the remainder is R, then the possible values of N include: R, R+D, R+2D, R+3D,. . . For example, if k divided by 6 leaves a remainder of 2, then the possible values of k are: 2, 2+6, 2+(2)(6), 2+(3)(6), 2+(4)(6), . . . etc.
When n is divided by 5, the remainder is 1. So, possible values of n are 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, etc.
When n is divided by 7, the remainder is 3. So, possible values of n are 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, etc.
So, we can see that n could equal 31, or 66, or an infinite number of other values.
Important: Since the Least Common Multiple of 7 and 5 is 35, we can conclude that if we list the possible values of n, each value will be 35 greater than the last value. So, n could equal 31, 66, 101, 136, and so on.
Check the answer choices....
Answer choice A: If we add 3 to any of these possible n-values, the sum is NOT a multiple of 35. ELIMINATE A
Answer choice B: if we take ANY of these possible n-values, and add 4, the sum will be a multiple of 35.
So, the smallest value of k is 4 such that k+n is a multiple of 35.
Now, only those values of "b" which will give "a" also as an integer will give us the common values of n which satisfy both n = 7b + 3 = 5a + 1 the conditions