Add the digits, then test the total
A whole number is divisible by 3 if the sum of its digits is divisible by 3. Divisible means that division leaves no remainder. This is one of the standard divisibility tests listed by OpenStax.
For 123, add 1 + 2 + 3 = 6. Six divides exactly by 3, so 123 does too. A full check confirms it: 123 ÷ 3 = 41.
For 124, the digit sum is 1 + 2 + 4 = 7. Seven is not divisible by 3, so 124 cannot be split into equal whole groups of three without something left over.
Why place value makes it work
Look at the hundreds, tens and ones in 123: 100 + 20 + 3. Now rewrite 100 as 99 + 1, and 20 as 18 + 2.
That gives 99 + 18 + 1 + 2 + 3. The 99 and 18 already divide into groups of three. Only 1 + 2 + 3 remains to be checked: the digit sum.
The same idea continues with larger place values, because 10 = 9 + 1, 100 = 99 + 1, and 1,000 = 999 + 1. NRICH explains the place-value reason. If you need the groundwork, return to tens and ones.
Passing the test for 3 does not mean passing for 9
For 9, add the digits too, but the total must be divisible by 9. The sum for 123 is 6: that passes the test for 3 and fails the test for 9. Indeed, 123 = 9 × 13 + 6.
The order of the digits does not change their sum. So 321 also passes the test for 3: 3 + 2 + 1 = 6, and 321 ÷ 3 = 107. The test tells you whether division is exact; it does not tell you the answer to the division.
Use it as a quick check when sharing a whole number into equal groups. Looking only at the final digit would miss the point.