The word "SERENDIPITY" is taken, and the letters are arranged alphabetically. Each vowel is replaced with the next vowel in the aplhabetical series, and the consonants are replaced by next consonants in the alphabetical series. Find the number of letters between "P" and "Z".
Solution
SERENDIPITY arranged alphabetically becomes DEEIINPRSTY.
Doing the operation that is described in the question we now get, F I I O O P Q S T V Z.
We see that there are exactly 4 letters between and P and Z.
Hence, the answer is 4.
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