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← Introduction to Number Theory levels

Introduction to Number Theory / LEVEL 3 · DIFFICULTY 3/5

Congruences

Solve problems using remainders.

3 stages · 24 practice problems · two 6-question assessment forms

Choose an island to read its lesson.

  1. 1MINI QUEST NT 3.1Cycles of Units DigitsRead the lesson
  2. 2MINI QUEST NT 3.2Linear CongruencesRead the lesson
  3. 3MINI QUEST NT 3.3Combining Remainder ConditionsRead the lesson
  4. LEVEL CHECKCastle challenge

    6 questions across this level.

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STAGE NT 3.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Cycles of Units Digits

Useful preparation: Computing with Remainders

The idea to remember

Powers modulo a fixed integer may repeat. Identify the cycle and reduce the exponent using its length.

Remember with 2 flashcards →

Goal: Understand and apply cycles of units digits.

Before you begin: Computing with Remainders

Understand the idea

Repeated powers often revisit the same remainder. For a fixed last digit, multiplying once more determines the next last digit, so a repeated state creates a cycle.

31\displaystyle 3^{1}31,32\displaystyle 3^{2}32,33\displaystyle 3^{3}33,34≡3\displaystyle 3^{4} \equiv 334≡3,9,7,1 (mod 10)

Choose and carry out a method

List successive powers modulo ten until the cycle closes. Reduce the exponent by the cycle length and match the correct position.

Check the reasoning

A remainder of zero refers to the last position of a cycle that begins at exponent one. Check a small exponent to avoid an index shift.

WORKED EXAMPLE 1

What is the units digit of 310\displaystyle 3^{10}310?

  1. Powers of 3 have units digits cycling through 3, 9, 7, 1.
  2. 10=4⋅2+2\displaystyle 10 = 4 \cdot 2 + 210=4⋅2+2; a zero remainder selects the fourth entry, otherwise use the remainder as the entry number.
  3. The units digit is 9; full groups of four powers repeat.

9

WORKED EXAMPLE 2

What is the units digit of 313\displaystyle 3^{13}313?

  1. Powers of 3 have units digits cycling through 3, 9, 7, 1.
  2. 13=4⋅3+1\displaystyle 13 = 4 \cdot 3 + 113=4⋅3+1; a zero remainder selects the fourth entry, otherwise use the remainder as the entry number.
  3. The units digit is 3; full groups of four powers repeat.

3

Common pitfalls

Possible mix-up: Exponent remainder zero means units digit zero.

It means use the last entry of the power cycle.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why can a units-digit calculation ignore all higher decimal digits?

Discuss together

Invite your learner to explain: Why can a units-digit calculation ignore all higher decimal digits? Use the worked examples to discuss their reasoning.

Preview the eight practice prompts
  1. Problem 1

    What is the units digit of 319\displaystyle 3^{19}319?

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  2. Problem 2

    What is the units digit of 322\displaystyle 3^{22}322?

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  3. Problem 3

    What is the units digit of 325\displaystyle 3^{25}325?

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  4. Problem 4

    What is the units digit of 328\displaystyle 3^{28}328?

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  5. Problem 5

    What is the units digit of 331\displaystyle 3^{31}331?

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  6. Problem 6

    What is the units digit of 334\displaystyle 3^{34}334?

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  7. Problem 7New context

    A branching design triples its count in each of 37 steps, starting with one piece. What is the units digit of the final count?

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  8. Problem 8New context

    A branching design triples its count in each of 40 steps, starting with one piece. What is the units digit of the final count?

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Open stage NT 3.1 in the student workspace →

STAGE NT 3.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Linear Congruences

Useful preparation: Cycles of Units Digits

The idea to remember

The congruence ax≡b\displaystyle ax \equiv bax≡b mod m is solvable exactly when gcd⁡ (a,m)\displaystyle \operatorname{gcd}\, \left(a, m\right)gcd(a,m) divides b. Division requires care.

Remember with 2 flashcards →

Goal: Understand and apply linear congruences.

Before you begin: Cycles of Units Digits

Understand the idea

Solving ax≡b\displaystyle ax \equiv bax≡b modulo m asks which residue classes give the desired remainder. A unique class exists when a and m are coprime, because multiplication by a permutes all residues.

ax≡b\displaystyle ax \equiv bax≡b (mod m), gcd⁡ (a,m)=1\displaystyle \operatorname{gcd}\, \left(a, m\right) = 1gcd(a,m)=1

Choose and carry out a method

Find an inverse of a modulo m or test the finite residue list. Multiply the target by the inverse and reduce the result to 0 through m−1\displaystyle m - 1m−1.

Check the reasoning

Substitute into the original congruence. Ordinary fraction division cannot replace a modular inverse.

WORKED EXAMPLE 1

Find the least nonnegative integer x satisfying 2 x≡6\displaystyle 2\, x \equiv 62x≡6 (mod 9).

  1. Solve the congruence using an inverse, or test numbers differing by the modulus.
  2. Since 9 is odd, 2 is invertible modulo 9. Here 2⋅3=6\displaystyle 2 \cdot 3 = 62⋅3=6 has remainder 6.
  3. The unique solution between 0 and 8 is 3.

3

WORKED EXAMPLE 2

Find the least nonnegative integer x satisfying 2 x≡8\displaystyle 2\, x \equiv 82x≡8 (mod 11).

  1. Solve the congruence using an inverse, or test numbers differing by the modulus.
  2. Since 11 is odd, 2 is invertible modulo 11. Here 2⋅4=8\displaystyle 2 \cdot 4 = 82⋅4=8 has remainder 8.
  3. The unique solution between 0 and 10 is 4.

4

Common pitfalls

Possible mix-up: Divide both sides as ordinary fractions.

Use an integer inverse modulo m, when one exists.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

What changes when the coefficient and modulus share a common factor?

Discuss together

Invite your learner to explain: What changes when the coefficient and modulus share a common factor? Use the worked examples to discuss their reasoning.

Preview the eight practice prompts
  1. Problem 1

    Find the least nonnegative integer x satisfying 2 x≡12\displaystyle 2\, x \equiv 122x≡12 (mod 15).

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  2. Problem 2

    Find the least nonnegative integer x satisfying 2 x≡14\displaystyle 2\, x \equiv 142x≡14 (mod 17).

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  3. Problem 3

    Find the least nonnegative integer x satisfying 2 x≡16\displaystyle 2\, x \equiv 162x≡16 (mod 19).

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  4. Problem 4

    Find the least nonnegative integer x satisfying 2 x≡18\displaystyle 2\, x \equiv 182x≡18 (mod 21).

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  5. Problem 5

    Find the least nonnegative integer x satisfying 2 x≡20\displaystyle 2\, x \equiv 202x≡20 (mod 23).

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  6. Problem 6

    Find the least nonnegative integer x satisfying 2 x≡22\displaystyle 2\, x \equiv 222x≡22 (mod 25).

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  7. Problem 7New context

    A machine moves two positions per step on a 27-position dial. Starting at zero, how many steps first land it at position 24?

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  8. Problem 8New context

    A machine moves two positions per step on a 29-position dial. Starting at zero, how many steps first land it at position 26?

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Open stage NT 3.2 in the student workspace →

STAGE NT 3.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Combining Remainder Conditions

Useful preparation: Linear Congruences

The idea to remember

Remainders with coprime moduli specify a unique residue modulo their product.

Remember with 2 flashcards →

Goal: Understand and apply combining remainder conditions.

Before you begin: Linear Congruences

Understand the idea

Two congruences describe overlapping arithmetic progressions. For coprime moduli, exactly one residue class modulo their product meets both conditions.

x≡r\displaystyle x \equiv rx≡r (mod m), x≡s\displaystyle x \equiv sx≡s (mod n)

Choose and carry out a method

List numbers satisfying one condition, or write x=r\displaystyle x = rx=r+km and use the other to determine k. Reduce to the least nonnegative solution.

Check the reasoning

Check both remainders. Adding the product of the moduli produces more solutions but not the least one.

WORKED EXAMPLE 1

Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 4) and x≡1\displaystyle x \equiv 1x≡1 (mod 5).

  1. Numbers satisfying the first condition form an arithmetic progression.
  2. Test 2, 6, 10, … in the second condition. The first match is 6.
  3. The moduli are consecutive and coprime, so the pattern repeats every 20 indices.

6

WORKED EXAMPLE 2

Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 5) and x≡1\displaystyle x \equiv 1x≡1 (mod 6).

  1. Numbers satisfying the first condition form an arithmetic progression.
  2. Test 2, 7, 12, … in the second condition. The first match is 7.
  3. The moduli are consecutive and coprime, so the pattern repeats every 30 indices.

7

Common pitfalls

Possible mix-up: Add the two remainders to obtain x.

Both conditions must hold simultaneously; addition alone does not enforce them.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why does adding mn preserve both remainder conditions?

Discuss together

Invite your learner to explain: Why does adding mn preserve both remainder conditions? Use the worked examples to discuss their reasoning.

Preview the eight practice prompts
  1. Problem 1

    Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 7) and x≡1\displaystyle x \equiv 1x≡1 (mod 8).

    Practise this exact problem →
  2. Problem 2

    Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 8) and x≡1\displaystyle x \equiv 1x≡1 (mod 9).

    Practise this exact problem →
  3. Problem 3

    Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 9) and x≡1\displaystyle x \equiv 1x≡1 (mod 10).

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  4. Problem 4

    Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 10) and x≡1\displaystyle x \equiv 1x≡1 (mod 11).

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  5. Problem 5

    Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 11) and x≡1\displaystyle x \equiv 1x≡1 (mod 12).

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  6. Problem 6

    Find the smallest nonnegative x with x≡2\displaystyle x \equiv 2x≡2 (mod 12) and x≡1\displaystyle x \equiv 1x≡1 (mod 13).

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  7. Problem 7New context

    A repeating marker leaves remainder 2 when its index is divided by 13, and remainder 1 when divided by 14. What is the first nonnegative index satisfying both rules?

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  8. Problem 8New context

    A repeating marker leaves remainder 2 when its index is divided by 14, and remainder 1 when divided by 15. What is the first nonnegative index satisfying both rules?

    Practise this exact problem →
Open stage NT 3.3 in the student workspace →
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∑ Numerive

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HomeTry a questionMy questsLearning SearchHomeworkUnderstand & rememberCurriculumFlashcardsUniversity

Exam preparation

Mock examsSATPSAT/NMSQTAMC 8 / 10 / 12

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Family & help

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Congruences: lessons and practice | Numerive Math