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← Introduction to Number Theory levels

Introduction to Number Theory / LEVEL 4 · DIFFICULTY 4/5

Integer Equations and Counts

Combine integer constraints with exact counting.

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

STAGE NT 4.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Nonnegative Integer Solutions

Useful preparation: Combining Remainder Conditions

Goal: Understand and apply nonnegative integer solutions.

Before you begin: Combining Remainder Conditions

Understand the idea

Integer restrictions turn an equation into a finite count when coefficients and the target are positive. Bounds and parity often eliminate most candidate values before substitution.

x=N−3 y2\displaystyle x = \frac{N - 3\, y}{2}

Choose and carry out a method

For 2 x+3 y=N\displaystyle 2\, x + 3\, y = N, bound y between zero and floor(N3\displaystyle \frac{N}{3}). Keep only y values making N−3 y\displaystyle N - 3\, y even, then compute x.

Check the reasoning

Verify x and y are nonnegative integers. Each accepted y determines exactly one x, preventing duplicate counting.

WORKED EXAMPLE 1

How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=15\displaystyle 2\, x + 3\, y = 15?

  1. Use one variable to parameterize the possibilities, then enforce divisibility and nonnegativity.
  2. Try x=0\displaystyle x = 0 through 7; retain x only if 15−2 x\displaystyle 15 - 2\, x is divisible by 3.
  3. There are 3 pairs. Each retained x determines exactly one y.

3

WORKED EXAMPLE 2

How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=18\displaystyle 2\, x + 3\, y = 18?

  1. Use one variable to parameterize the possibilities, then enforce divisibility and nonnegativity.
  2. Try x=0\displaystyle x = 0 through 9; retain x only if 18−2 x\displaystyle 18 - 2\, x is divisible by 3.
  3. There are 4 pairs. Each retained x determines exactly one y.

4

Common pitfalls

Possible mix-up: Every y in the numerical range gives a solution.

The remaining numerator must also be divisible by two.

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

How does parity let you count the allowed y values without testing all of them?

Preview the eight practice prompts
  1. Problem 1

    How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=24\displaystyle 2\, x + 3\, y = 24?

    Practise this exact problem →
  2. Problem 2

    How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=27\displaystyle 2\, x + 3\, y = 27?

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  3. Problem 3

    How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=30\displaystyle 2\, x + 3\, y = 30?

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  4. Problem 4

    How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=33\displaystyle 2\, x + 3\, y = 33?

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  5. Problem 5

    How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=36\displaystyle 2\, x + 3\, y = 36?

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  6. Problem 6

    How many pairs of nonnegative integers (x,y) satisfy 2 x+3 y=39\displaystyle 2\, x + 3\, y = 39?

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  7. Problem 7New context

    A store sells packs containing 2 or 3 beads. How many choices of counts of the two pack sizes buy exactly 42 beads? Packs of the same size are indistinguishable.

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  8. Problem 8New context

    A store sells packs containing 2 or 3 beads. How many choices of counts of the two pack sizes buy exactly 45 beads? Packs of the same size are indistinguishable.

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Open stage NT 4.1 in the student workspace →

STAGE NT 4.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Repeating Decimals

Useful preparation: Nonnegative Integer Solutions

Goal: Understand and apply repeating decimals.

Before you begin: Nonnegative Integer Solutions

Understand the idea

Multiplying a repeating decimal by a power of ten shifts one complete block. Subtracting the original aligns and removes the infinite repeated tail.

0.overline(ab)=10 a+b99\displaystyle \frac{10\, a + b}{99}

Choose and carry out a method

For a two-digit repeating block, set x equal to the decimal and subtract x from 100x. Solve 99x equal to the block and simplify.

Check the reasoning

A proper decimal below one must give a fraction below one. Leading zeros within a block still count as positions.

WORKED EXAMPLE 1

Write 0.111111… as a fraction, where the two-digit block 11 repeats forever.

  1. Multiply by 100 to align the two-digit repeating tails.
  2. If x=0.1111\displaystyle x = 0.1111…, then 100 x−x=11\displaystyle 100\, x - x = 11, hence 99 x=11\displaystyle 99\, x = 11.
  3. The exact fraction is 19\displaystyle \frac{1}{9}; no rounding is needed.

19\displaystyle \frac{1}{9}

WORKED EXAMPLE 2

Write 0.121212… as a fraction, where the two-digit block 12 repeats forever.

  1. Multiply by 100 to align the two-digit repeating tails.
  2. If x=0.1212\displaystyle x = 0.1212…, then 100 x−x=12\displaystyle 100\, x - x = 12, hence 99 x=12\displaystyle 99\, x = 12.
  3. The exact fraction is 433\displaystyle \frac{4}{33}; no rounding is needed.

433\displaystyle \frac{4}{33}

Common pitfalls

Possible mix-up: Divide a repeating two-digit block by 100.

A repeating block uses 99; division by 100 gives a terminating decimal.

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

How would the denominator change for a three-digit repeating block?

Preview the eight practice prompts
  1. Problem 1

    Write 0.141414… as a fraction, where the two-digit block 14 repeats forever.

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  2. Problem 2

    Write 0.151515… as a fraction, where the two-digit block 15 repeats forever.

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  3. Problem 3

    Write 0.161616… as a fraction, where the two-digit block 16 repeats forever.

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  4. Problem 4

    Write 0.171717… as a fraction, where the two-digit block 17 repeats forever.

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  5. Problem 5

    Write 0.181818… as a fraction, where the two-digit block 18 repeats forever.

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  6. Problem 6

    Write 0.191919… as a fraction, where the two-digit block 19 repeats forever.

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  7. Problem 7New context

    A display’s exact value repeats the block 20 forever after the decimal point, starting immediately: 0.2020… . Express this reading as a fraction.

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  8. Problem 8New context

    A display’s exact value repeats the block 21 forever after the decimal point, starting immediately: 0.2121… . Express this reading as a fraction.

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Open stage NT 4.2 in the student workspace →

STAGE NT 4.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Counting Coprime Residues

Useful preparation: Repeating Decimals

Goal: Understand and apply counting coprime residues.

Before you begin: Repeating Decimals

Understand the idea

An integer is coprime to n exactly when it avoids every prime factor of n. Removing multiples of these primes by inclusion and exclusion counts the permitted residue classes.

ϕ (n)=n\displaystyle \phi\, \left(n\right) = n∏(1−1p)\displaystyle \left(1 - \frac{1}{p}\right) over distinct prime factors p

Choose and carry out a method

For n with prime factors two and three, subtract their multiple counts and restore multiples of six. Express the result as n (1−12) (1−13)\displaystyle n\, \left(1 - \frac{1}{2}\right)\, \left(1 - \frac{1}{3}\right).

Check the reasoning

A step size traverses a full circular cycle precisely when it is coprime to the cycle length. Check this on a small circle.

WORKED EXAMPLE 1

How many integers from 1 through 24 are relatively prime to 24?

  1. A step size visits the whole cycle exactly when it is coprime to its length.
  2. Exclude multiples of 2 or 3:ϕ (24)=24 (1−12) (1−13)\displaystyle 3 : \phi\, \left(24\right) = 24\, \left(1 - \frac{1}{2}\right)\, \left(1 - \frac{1}{3}\right).
  3. The count is 8; multiples of both primes were excluded twice and must be accounted for once.

8

WORKED EXAMPLE 2

How many integers from 1 through 48 are relatively prime to 48?

  1. A step size visits the whole cycle exactly when it is coprime to its length.
  2. Exclude multiples of 2 or 3:ϕ (48)=48 (1−12) (1−13)\displaystyle 3 : \phi\, \left(48\right) = 48\, \left(1 - \frac{1}{2}\right)\, \left(1 - \frac{1}{3}\right).
  3. The count is 16; multiples of both primes were excluded twice and must be accounted for once.

16

Common pitfalls

Possible mix-up: Apply the factor once for every repeated prime.

Use each distinct prime only once.

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 a noncoprime step size revisit its starting point too early?

Preview the eight practice prompts
  1. Problem 1

    How many integers from 1 through 192 are relatively prime to 192?

    Practise this exact problem →
  2. Problem 2

    How many integers from 1 through 384 are relatively prime to 384?

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  3. Problem 3

    How many integers from 1 through 768 are relatively prime to 768?

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  4. Problem 4

    How many integers from 1 through 1536 are relatively prime to 1536?

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  5. Problem 5

    How many integers from 1 through 3072 are relatively prime to 3072?

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  6. Problem 6

    How many integers from 1 through 6144 are relatively prime to 6144?

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  7. Problem 7New context

    A 12288-position cycle is traversed using a fixed positive step size at most 12288. How many step sizes visit every position before returning to the start?

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  8. Problem 8New context

    A 24576-position cycle is traversed using a fixed positive step size at most 24576. How many step sizes visit every position before returning to the start?

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Open stage NT 4.3 in the student workspace →
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