How many natural numbers m such that: for any integers a,b such that a-=b(mod120) the following congruence is also true:a-=b(mod m)

Answer 1

Hi Emma,

Since a = b (mod 120), this means b = 120*n₁ + a, where n₁ is a natural number.

We need to find natural numbers m such that a = b (mod m), which means b = m*n₂ + a.

Thus, 120*n₁ + a = m*n₂ + a, leaving us with: 120*n₁ = m*n_{2 .}

We next decompose 120 into its prime factors and thus: 2*3*4*5 * n₁ = m*n_{2 .}

Now, how many possible m natural numbers to satisfy the condition above?

We do not know anything about n₁, but we can start making assumptions.

If n₁= 1:

m could be : n choose k as follows:

4 choose 3: this will give us 4 possibilities for m (4!/(3!*1!): m = 2*3*4, or m = 2*3*5, or m = 2*4*5, or m = 3*4*5

4 choose 2: this will give us 6 possibilities for m (4!/(2!*2!): m = 2*3, or m = 2*4, or m = 2*5, or m = 3*4, or m = 3*5, or m = 4*5

4 choose 1: this will give us 4 possibilities for m (4!/(1!*3!): m = 2, or m = 3, or m = 4, or m = 5

This adds up to 14 possibilities for m. However, we will not keep all of them, as the modulo calculation will yield something different if any of these m's are <a. So we will only keep the m numbers that are >a.

If n₁ is prime, we will include it when we write out the possible m values.

m could be:

5 choose 4: this will yield...

You could follow the same strategy and figure out how many m's.

Lastly, if n₁ can be decomposed into factors, you will get even more possible m values. I wonder if you have any more constraints/ information about this problem. Regardless, this strategy should work. Let me know if you have more questions.