Video summary

Math Olympiad Lecture 2: Last Digits

Main summary

Key takeaways

Educational

Main ideas and lessons

1) Goal of the lesson

  • Learn how to determine the last digit (and later the last two digits) of very large powers.
  • Target problems are numbers written in index notation of the form:
    • (a^n) = “(a) to the power (n)”

2) Key concept: only the last digits matter for the last digit of a product

  • To find the last digit of a product:
    • Use only the last digits of the factors.
  • Examples:
    • Last digit of (2012^2): compute from (2 \times 2 = 4)
    • Last digit of (2012^3): use (4 \times 2 = 8)

Result: The original problem reduces to finding the last digit of:

  • (2012^{2017}) → the last digit depends on (2^{2017})

3) Method for last digit of powers: find a repeating pattern (cycle)

For (2^n), the last digits cycle:

  • (2^1 \to 2)
  • (2^2 \to 4)
  • (2^3 \to 8)
  • (2^4 \to 6)
  • Then the pattern repeats every 4 powers.

Specific instructions applied

  • Determine where (2017) falls in the cycle using the cycle length 4.
  • Find a multiple of 4 close to 2017:
    • (2016) is divisible by 4.
  • Then:
    • (2^{2016}) has last digit 6
    • Multiply one more time by 2 to get (2^{2017}):
      • (6 \times 2 \to 2) (last digit 2)

Alternative general rule using (n \bmod 4)

Let (n) be the exponent:

  • If (n \bmod 4 = 1) → last digit 2
  • If (n \bmod 4 = 2) → last digit 4
  • If (n \bmod 4 = 3) → last digit 8
  • If (n \bmod 4 = 0) → last digit 6

✅ Final answer for the example:

  • Last digit of (2012^{2017}) is (\boxed{2})

4) General cycle patterns for last digits (periodicity)

The lesson summarizes how last digits cycle depending on the base’s last digit:

  • If the base ends with 1, 5, 6, or 0:
    • The last digit never changes (constant for any exponent).
  • If the base ends with 2, 3, 7, or 8:
    • The cycle has period 4.
  • If the base ends with 4 or 9:
    • The last digit cycle has period 2
    • (Example mentioned: (4^2) ends with 6, and (4^3) returns to 4.)

5) Extension: last two digits of powers

Problem considered

  • Find the last two digits of (2007^{2007}).

Method adapted from last digit case

  • Only the last two digits of the base matter.
  • Since (2007) ends in 07, reduce to:
    • last two digits of (7^{2007})

Stated pattern

  • For powers of 7 (mod 100), the last two digits repeat with period 4.
  • Conclusion:
    • last two digits of (7^{2007}) are (\boxed{43})

Caveats about periodicity

  • The speaker warns:
    • the repeating cycle for last-two-digits may appear after some steps, and
    • the original ending (like “07” for base 7) may not reappear even if a cycle exists.
  • Example cautions:
    • For base 2: the last two digits cycle may not return to the starting “02”
    • For powers of 14: cycle behavior is described as repeating after some later point

Practical support

  • An Excel spreadsheet is mentioned for detailed patterns of last two digits (placed in the info section).

6) Summary rules for last-two-digit periodicity (as given)

  • Numbers ending with 5 (like 5, 15, 25, 35):
    • Period is 1 or 2
    • Depends on whether the number before 5 is even or odd:
      • even before 5 → period 1
      • odd before 5 → period 2
  • Numbers ending with 2, 3, 7, 8:
    • last two digits repeat after a multiple of 4 (period 4 or 20 mentioned)
  • Some bases like:
    • 11, 14, 19, 29, 31, etc.
    • repeat after 10 numbers (period 10)

7) Instructional link between “last digit” and “final answer digit” (family of problems)

  • The lesson notes an “interesting family” of problems where:
    • the last digit of the exponent/power structure influences the last digit of the final result.

Detailed methodologies / instructions

Method A: Last digit of (a^n) (only last digits matter)

  1. Identify the last digit of the base (a).
  2. Replace the base with its last digit (e.g., (2012 \to 2)).
  3. Build a small list of last digits of (a^1, a^2, a^3, \dots) until a cycle appears.
  4. Use the cycle length (period) and compute:
    • (n \bmod (\text{period}))
  5. Map the remainder to the corresponding last digit and output it.

Method B: Last two digits of (a^n)

  1. Reduce the base to its last two digits (mod 100).
  2. Determine the cycle for the last two digits of the powers of the reduced base.
  3. Use (n) modulo the cycle length to select the correct last-two-digit result.
  4. Apply caution:
    • the original ending (e.g., “07”) may not reappear even inside the cycle.

Method C: Periodicity quick checks

  1. Identify the base’s last digit.
  2. Use provided period patterns:
    • last digit 0, 1, 5, 6 → constant
    • last digit 4, 9 → period 2
    • last digit 2, 3, 7, 8 → period 4
  3. For last two digits, apply the more specific periodicity notes (period 1/2/4/10/20 depending on endings).

Extension problems assigned

  • Extension Problem 1

    • Find the last digit of [ 2009^{2007^{2005^{2003^{\cdots^{3^1}}}}} ]

    • (nested exponents continuing down until (3^1))

  • Extension Problem 2

    • Find the last two digits of [ 2019^{2019^{2019}} ]

    • (A solution link is promised in the info section later.)

Speakers / sources featured

  • Primary speaker/teacher: Unnamed instructor hosting the “Math Olympiad Lecture” series (the voice presenting all content).
  • Source references: None explicitly named beyond the instructor and the mention of an Excel spreadsheet placed in the video’s info section.

Original video