Find the smallest positive integer N having the following properties:
I. When
N
is divided by 3 the remainder is 1;
II. When
N
is divided by 5 the remainder is 2;
III. When
N
is divided by 7 the remainder is 3;
IV. When
N
is divided by 9 the remainder is 4.
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Great solution; just to note, you have said "To satisfy conditions II and III" twice by accident.
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Thanks for catching that. The edit has been made. :)
Let the natural number which will be multiplied by each number, 3 , 5 , 7 , 9 to a .
Then N = 3 a + 1 = 5 a + 2 = 7 a + 3 = 9 a + 4 .
If N is the multiplier of the upper expressions or can be divided by 3 , 5 , 7 , 9 , then it cannot satisfy the problem.
Then let's just find the number.
9 a + 4 always satisfies that the expression, 3 a + 1 because 9 × 5 + 1 = 4 6 = 3 × 1 5 + 1 , always satisfies.
Then we can just ignore it.
N = 5 a + 2 = 7 a + 3 = 9 a + 4 .
The smallest positive integer N is 1 5 7 because 1 5 7 = 5 × 3 1 + 2 = 7 × 2 2 + 3 = 9 × 1 7 + 4 .
Hence, the answer is 1 5 7 .
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The numbers that satisfy condition IV form a proper subset of those that satisfy condition I, so we can just focus on the last three conditions.
Now to satisfy conditions II and III we are looking for integers m , n such that
5 m + 2 = 7 n + 3 ⟹ 5 m − 7 n = 1 ,
the least of which are m = 3 , n = 2 . Thus N = 1 7 + 3 5 k for some integer k .
To satisfy conditions III and IV we are looking for integers m , n such that
7 m + 3 = 9 n + 4 ⟹ 7 m − 9 n = 1 ,
the least of which are m = 4 , n = 3 . Thus N = 3 1 + 6 3 j for some integer j .
we are thus looking for j , k that satisfy
1 7 + 3 5 k = 3 1 + 6 3 j ⟹ 3 5 k − 6 3 j = 1 4 ⟹ 5 k − 9 j = 2 ,
the least of which is k = 4 , j = 2 , yielding N = 1 7 + 3 5 ∗ 4 = 1 5 7 .