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Yash Agarwal

#2:

Here are some of the mod rules needed for this problem:

If a is equal to b (mod m), then:

a + c = b + c (mod m)

a^c = b^c (mod m)

In this case we are using mod 13, so we know that m would equal 13.

Now, let’s start with some of the easy things. The easiest square mod 13 equals 1 would be 1. 1^2 = 1. And, -1^2 is also equal to 1. So, by using the addition mod rule above, we can get:

1 + 0 = 1 + 13 (mod 13)

1 = 14 (mod 13)

Now, you might be thinking, this is a, not a^2. But, it would be the same thing for a^2. But, using the exponential rule above, we get:

1^2 = 14^2 (mod 13)

1 = 196 (mod 13)

It is the same thing for a^2. So, you just have to keep on adding 13 until you get a value of a which exceeds 100. So, those numbers would be:

From -1: 12, 25, 38, 51, 64, 77, 90

From 1: 1, 14, 27, 40, 53, 66, 79, 92

Since -1 is not from 0 to 100, -1 would not be included in this case. So, there is a total of 15 possible numbers for a.

Any prime or composite mod would word since the mod doesn’t matter. The rule states … (mod m). M can be prime or composite.

** This response is written by Yash Agarwal.**