Divisor 20173
- Prime Number:
- Yes!
- Divisibility test:
- The "Eric Sexton Test"
- Test Discovered by:
- Matt Parker
- Date:
- 11/11/2024
The "Eric Sexton Test" for Divisibility by 20173
To determine if any number is divisible by 20173, apply the "Eric Sexton Test":
- If your number ("X") has 6 digits or more, separate the last (smallest) 5 digits from the rest. This makes two smaller numbers, call them Left and Right (note: don't add in trailing zeros to L). If there are fewer than 6 digits, L = 0 and therefore R = X.
- Multiply L by 865 and subtract this from R.
- Take that result and cross off its final digit (units). Take this new number and add 6052 times the digit you just crossed off. Call this final result "Y".
- Y will be much smaller than X, but we have preserved divisibility by 20173. That is, your original number is divisible by 20173 if (and only if) Y is. Now that it's much smaller, with basic knowledge of your 20173-times tables, it should be easy to visually see if Y is divisible by 20173. If the Y is still much larger than 20173, the above process can be repeated until it does reduce to within small multiples of 20173.
Easy!