Math Olympiad Lecture 1: Trailing Zeroes

Main Ideas and Lessons

• Purpose of the lecture series: Train students for math competitions where calculators are not allowed, so problems must be solved by reasoning.
• Goal of this lesson: Learn how to count the number of trailing zeros in large numbers, including factorials.

Trailing Zeros: Definition

Trailing zeros are the consecutive zeros at the end of a number (nothing else follows them).

Counting Trailing Zeros in Simpler Examples

• Example: (4200 \rightarrow 2) trailing zeros
• Example: (109000 \rightarrow 3) trailing zeros
  • The internal zero (the one not at the end) is not counted because digits follow it.
• Example: Comparing 500 million and 200,000
  • The count given: 5 trailing zeros
  • Extra zeros in the middle (e.g., between 5 and 2 in that comparison) are not counted.

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Key Methodology for Factorial Trailing Zeros (Main Technique)

Trailing zeros come from factors of 10, and each [ 10 = 2 \times 5. ]

In factorials (like (2017!)), there are typically more 2s than 5s, so the number of trailing zeros is determined by the number of factors of 5 in the factorial.

Strategy Used in the Video (Step-by-Step)

1. Start with an easier case to build intuition

   • [ 10! = 10 \cdot 9 \cdot 8 \cdots 1 ]

   • Each factor of 10 contributes trailing zeros, and since (2 \times 5 = 10), (10!) has at least 1 factor of 10; in fact it has 2 trailing zeros.

   • This motivates counting how many 5s are available.

2. Count multiples of 5 in (2017!)

   • Every group of 5 numbers contributes at least one factor of 5.

   • Compute: [ \left\lfloor \frac{2017}{5} \right\rfloor = 403 ]

   • This gives 403 factors of 5 → 403 trailing zeros initially.

3. Account for extra 5 factors from multiples of 25

   • Numbers divisible by 25 contribute an extra factor of 5 (since (25 = 5^2)).

   • Compute: [ \left\lfloor \frac{403}{5} \right\rfloor = \left\lfloor \frac{2017}{25} \right\rfloor = 80 ]

   • Add 80 more trailing zeros.

4. Continue with higher powers of 5 (125, 625, …)

   • Multiples of (125 = 5^3) add one more factor of 5 each: [ \left\lfloor \frac{2017}{125} \right\rfloor = 16 ] Add 16 zeros.

   • Multiples of (625 = 5^4) add one more factor of 5 each: [ \left\lfloor \frac{2017}{625} \right\rfloor = 3 ] Add 3 zeros.

   • After a certain power, the contribution becomes zero for this problem size.

5. Add contributions

   • Total trailing zeros: [ 403 + 80 + 16 + 3 = 502 ]

   • Conclusion: (2017!) has 502 trailing zeros.

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Problems Given for Extension

• Problem 1: Find the number of trailing zeros for [ 20^{5} \times 50^{25} \times 75^{13}. ]

• Problem 2: Find the number of trailing zeros for (2017!) when written in binary representation.

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Speakers / Sources Featured

• Speaker: A single instructor (no name provided in the subtitles).