Learn with Amar
Teaching video
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Undergraduate chapters and video availability01 · Read and understand
What you will learn
- Find a modular inverse through a Bézout identity rather than trial division.
- Justify the conclusion "x≡6·38≡13 mod 43" using the stated assumptions.
Before you start
Euclid's algorithm and congruences.
Keep paper nearby. Read the question once for the context, then again to identify what is known and what you need to find.
Start with a question
Solve 17x≡6 modulo 43.
Why this math matters
Find a modular inverse through a Bézout identity rather than trial division. This worked micro-lesson connects a precise mathematical condition to a conclusion you can check. The transfer task asks you to change the setting and decide which parts of the reasoning still apply.
Set up the model
A useful answer starts with clear assumptions:
- Solutions are residue classes modulo 43.
- The modulus is nonzero and divisibility is over integers.
02 · Work through the example
Follow the reasoning, one step at a time.
Try to predict the next step before reading it. After each calculation, explain why the operation makes sense and how it helps answer the original question.
Work through it with Amar
See this example unfold.
The complete worked example, one idea at a time.
Undo multiplication modulo a prime
PausedQuestion: Start with the question. Paused.
Question
Start with the question
Solve 17x≡6 modulo 43.
Before you calculate
Read what is known and what you need to find. Make a prediction before moving to the first calculation.
Starts paused. Play advances through the full text at a reading pace; pause whenever you need more time. Previous, Next, and the phase buttons let you set your own pace. Playback pauses when this walkthrough leaves the screen or you switch tabs.
Your device’s reduced-motion setting keeps each phase still. Manual controls remain available. The full written solution stays below.
Build the model
43=2·17+9; 17=9+8; 9=8+1
The gcd is one, so an inverse exists.
Work through the mathematics
1=2·43−5·17, hence 17⁻¹≡−5≡38
Back substitution identifies the coefficient of seventeen.
Check the conclusion
x≡6·38≡13 mod 43
Checking 17·13=221=5·43+6 verifies the solution class.
The result
x≡6·38≡13 mod 43
Checking 17·13=221=5·43+6 verifies the solution class.
Common mistakes to catch
- A reciprocal in real arithmetic is not a modular inverse.
- Cancellation requires the cancelling element to be invertible.
03 · Practice independently
Try it before revealing the answer.
Use paper or a calculator as needed. Write your units and reasoning, then open the hint or explanation to check your approach.
Practice 1
Solve 3x≡1 mod 7.
Show a hint
Find a product one more than a multiple of seven.
Reveal answer and explanation
x≡5
3·5=15≡1 modulo seven.
Practice 2
Does 6 have an inverse modulo 15?
Show a hint
Compute the gcd.
Reveal answer and explanation
No
gcd(6,15)=3, so 6x−15k cannot equal one.
Take the idea with you
Explain why some multipliers destroy information in a remainder-based encoding.
04 · Reflect and continue
Can you explain it in your own words?
Before moving on, explain the main idea without looking at the worked example. Try both practice questions, check your reasoning, and name one mistake you now know how to avoid. Return to a step if you still need support.
Up next: Combine two independent remainder schedules
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