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Home » Calculate \( (1.10)^12 \): - Redraw

Calculate \( (1.10)^12 \): - Redraw

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Without exact data, but in olympiad context, assume minimal inference: the new median is 16, previous was 15, and only one value was added, so the jump suggests one value crossed 15 into higher, but medians shifted from 2nd-quartile-like.

📌 Since 18 >16, and new median is 16, values <16 are those less than 16.

📌 To get a5 + a6 = 32, need values >15 in position.

📌 Only way is if the added value pushes the middle up, but 18 is large.

📌 Perhaps original has values above 16.

📌 The a5 in a 9-list needs to be at least 16 to have average 16 with a6.

📌 Since sum a5+a6=32, minimum a5 is 16 (if a6=16).

📌 So if a5=16, a6=16, then both are 16.

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f(100) - f(99) = 6(99) + 4 = 594 + 4 = 598.
Divide the polynomial by $ (n - 1) $:
So the full factorization is $ (n - 1)(n - 2)(n - 3) $.
Thus, $ d(n) = 0 $ when $ n = 1, 2, 3 $.
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