Question

How many integers from -100 to 100 are congruent to 3 (mod 11)?

Answer 1

• 18 integers

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The integers that are congruent to 3 (mod 11) have the form:

• N = 11k + 3, so they all give remainder of 3 when divided by 11.

We are looking for the number k for which:

• -100 < 11k + 3 < 100

Solve it for whole numbers of k:

• -103 < 11k < 97
• - 103/11 < k < 97/11
• - 10 < k < 9

The number of integer solutions of k is:

• 9 (negative values of k) + 1 (k = 0) + 8 (positive values of k) =
• 18