Question

Find the integer a such that

a) a ≡ −15 (mod 27) and −26 ≤ a ≤ 0.
b) a ≡ 24 (mod 31) and −15 ≤ a ≤ 15.
c) a ≡ 99 (mod 41) and 100 ≤ a ≤ 140.

Answer 1

To find the integer a that satisfies the given congruence and inequality conditions, we can start by listing the numbers that are congruent to -15 (mod 27) and fall within the range of -26 to 0. So, for part a), the numbers are: -26, -53, -80, -107, ..., -2, -29, -56, -83, -110. We can check each number to see if it satisfies the inequality. Out of the list, the only number that satisfies the inequality is a = -2.

Step-by-step explanation:

To find the integer a that satisfies the given congruence and inequality conditions, we can start by listing the numbers that are congruent to -15 (mod 27) and fall within the range of -26 to 0.

So, for part a), the numbers are: -26, -53, -80, -107, ..., -2, -29, -56, -83, -110.

We can check each number to see if it satisfies the inequality. Out of the list, the only number that satisfies the inequality is a = -2.

Answer 2

To find the required integers a, we use modular arithmetic to determine values within the specified ranges. The solutions satisfying both the congruence and the range conditions are a) −−6, b) −6, and c) 123.

Step-by-step explanation:

To solve for the integer a in each scenario, we need to find a value of a that satisfies both the modular equation and the given range. The modular arithmetic condition a ≡ b (mod n) means that a leaves the same remainder when divided by n as b does.

1. a ≡ −15 (mod 27) means that when a is divided by 27, the remainder is the same as when −15 is divided by 27. Since −15 and 27 add up to 12, a can be 12 less than any multiple of 27 that is within the range −26 ≤ a ≤ 0. The correct value in this range is −−6.
2. For a ≡ 24 (mod 31), we find when divided by 31, 24 has a remainder of 24. We need to find a number that has the same remainder when divided by 31 and in the range −15 ≤ a ≤ 15. In this scenario, a is −6 because it's the multiple of 31 that falls in the given range and leaves a remainder of 24.
3. Lastly, a ≡ 99 (mod 41) implies that a has the same remainder as 99 when divided by 41. Since 99 ≡ 17 (mod 41), we find that the number satisfying the conditions within the range 100 ≤ a ≤ 140 is 123, which is 99 plus a multiple of 41.

The integers that satisfy these conditions for each part are a) −−6, b) −6, and c) 123.