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Question: Let $[c]_{37}$ be the multiplicative inverse of $[15]_{37}$, where c is natural number less that 37. What is c?

My attempt:

First I found the gcd of (15, 37),

37 = 2 . 15 + 7

15 = 2 . 7 + 1

7 = 1 . 7 + 0

So gcd (15, 37) = 1 (which shows there is an inverse), the calculation is 1 = 15 - 2 . 7, therefore the inverse of $[15]_{37}$ is $[-2]_{37}$, so for it to be a natural number less that 37 the element must be 35, so it's $[35]_{37}$

However I am told this is incorrect, c is not equal to 35, where did I go wrong?

Bill Dubuque
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rh1710
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    $-2\cdot7$ is $1$ mod $15$. You have to move one step backwards the chain of divisions to find the Bezout identity between $37$ and $15$. – Andrea Mori Jul 23 '22 at 12:16
  • As Andrea said, you need to work backwards up the chain, substituting remainders to get an identity in Bézout form. $$1=15-2(37-2\cdot15)$$ – PM 2Ring Jul 23 '22 at 12:28
  • Cheers @AndreaMori and PM 2 Ring, I went from 15 - 2 . 7 = 1 -> 15 - 2 . (37 - 2 . 15) =1 -> 5(15) - 2(37) =1, so s and t in Bezout form is 5 and -2. So my answer is 5, which is correct, thanks for the help! – rh1710 Jul 23 '22 at 16:55
  • @rh1710 It's far less error-prone, and much easier to use (and remember) the forward (vs. above backward) method described in the dupe. Once you learn that way you will never go back again. – Bill Dubuque Jul 23 '22 at 17:07
  • Thank you for letting me know and linking the dupe! @BillDubuque – rh1710 Jul 23 '22 at 17:41
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    Well really, in a case like this, where the numbers are small, why not play around a little and see what’s what? Can’t you see a multiple of $15$ that’s one more than a multiple of $37$? – Lubin Jul 23 '22 at 18:27
  • Thank you for bringing this to my attention, 5 . 15 is one more than 2. 37, much, MUCH easier way of viewing it, I feel silly I didn't resort to that – rh1710 Jul 24 '22 at 11:05

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