Let H1 be the hash function that computes the mod of the sum of every odd digit of an integer (ones, hundreds, tens hundreds, etc.). Let H2 be the hash function that computes the mod of the sum of every even digit of an integer (tens, thousands, etc.). for example if the mod was 10 and the number given is 867, then H1(867) would be 5 and H2(867) would be 6. Using H1 and H2 what does the bloom filter of size 10 look like after adding the values 561, 2018, 12345, and 42.

Index 0 Bit 3 4 List 3 distinct integer that would be "contained" in the bloom filter but were not in the original set