They both leave the same remainder when divided by 3.

When dividing by 3: (t-4) + 3 → (t-1) - 0 → (t- 1)

sorry i dont get it

What’s the big idea? In your case, you want to test whether:

(t)(t - 4)(t + 1) is divisible by 3 for integer values of t.

I’m assuming you’re aware that all integers can be expressed as: 3k, 3k + 1, or 3k + 2.

Therefore, you can do casework in each case:

**t = 3k:**

\textcolor{lime}{(3k)}(3k - 4)(3k + 1) → is a multiple of 3 because green is a multiple of 3

**t = 3k + 1:**

(3k+1)\textcolor{red}{(3k + 1 - 4)}(3k + 1 + 1) → is a multiple of 3 because red is a multiple of 3

**t = 3k + 2**

(3k + 2)(3k + 2 - 4)\textcolor{orange}{(3k + 2 +1)} → is a multiple of 3 because orange is a multiple of 3

Therefore, the expression is divisible by 3 for all integers t.