Refer to the given series and answer the question that follows. All numbers are single-digit numbers only. Counting to be done from left to right only. (Left) 2 8 7 5 6 9 7 6 8 4 8 2 1 8 2 8 6 8 1 3 1 1 3 (Right) How many such odd digits are there, each of which is immediately preceded by an odd digit and also immediately followed by an even digit?
Two
We need each odd digit whose left neighbour is odd and whose right neighbour is even; the pattern to find is odd–odd–even, and we count the middle odd digit.
Write the series with the required triplets checked from left to right: 2 8 7 5 6 9 7 6 8 4 8 2 1 8 2 8 6 8 1 3 1 1 3.
Consider 7(pos3) 5(pos4) 6(pos5): 7 is odd, 5 is odd, 6 is even → 5 qualifies.
Consider 9(pos6) 7(pos7) 6(pos8): 9 is odd, 7 is odd, 6 is even → 7 qualifies.
Check the other odd digits: the last 3(pos23) has no right neighbour; 1 1 3 near the end gives 1(odd)–1(odd)–3(odd), where the follower 3 is odd, so it fails; no further triplet satisfies odd–odd–even.
So exactly two digits (the 5 and the 7 identified above) meet all three conditions.
Hence, the number of such odd digits is two.
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