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What integer \(n\) satisfies \(0\le n<{101}\) and \(100n\equiv 72\pmod {101}~?\)

 Jan 31, 2021
 #1
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Using the Euclidean Algorithm, I found n = 31.

 Jan 31, 2021
 #2
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101 - 72 = 29.

 

\(\displaystyle 29 \equiv -72 \bmod101,\\ 100 \equiv -1 \bmod 101, \\ \text{so}\\ 2900 \equiv 72 \bmod 101.\)

 Feb 2, 2021

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