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kubler-ross 5 stages - AdVision eCommerce

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 So $ 7^{-1} \equiv 2 \pmod{13} $

📌 k \equiv 2 \pmod{13} \Rightarrow k = 13m + 2

📌 Then $ L = 7k = 7(13m + 2) = 91m + 14 $

📌 For $ m = 0 $, $ L = 14 $ (two-digit)

📌 Check: Is 14 $ \equiv 1 \pmod{13} $?

📌 $ 14 \div 13 = 1 $ remainder 1 â yes.

📌 And 14 is a multiple of 7 â valid.

📌 Is there a smaller two-digit value? Next smaller would be $ m = -1 $: $ L = 91(-1) + 14 = -77 $, not positive.

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