A more direct check-by-listing approach works well here since the numbers are small. We list numbers that leave remainder 8 when divided by 25, then check each one against the second condition (remainder 22 when divided by 28).
So the smallest total number of sweets satisfying both conditions is 358, confirming option C. Any smaller common solution would require going below the first term of the sequence, which isn't possible since sweets can't be negative.