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If the final digit of the number is 0, then some carrying takes place when 1 is subtracted. Otherwise, no carrying happens.

If no carrying happens, call the three digits of today's number \(A\), \(B\), and \(C\). We know that \(A^3+B^3+C^3\) is one more than \(A^3 + B^3 + (C-1)^3\).
This implies that \(C^3=(C-1)^3+1\), which is only possible if \(C\) is 1.

Therefore either the final digit of today's number is 0 or the final digt of one less that today's number is 0. In both cases, we need to find a number
with the desired property whose final digit is 0: we are looking for digit \(A\) and \(B\) such that \(A^3+B^3\) is a multiple of 10.

Looking at all the cube numbers, there are a few combinations that add up to multiple of 10:

$$0^3+0^3=0$$
$$1^3+9^3=730$$
$$2^3+8^3=520$$
$$3^3+7^3=370$$
$$4^3+6^3=280$$
$$5^3+5^3=250$$

The only one of these that has the required property is 370. By checking 369 and finding it doesn't have the property, we see that the two numbers must be
370 and **371**.