2.2 Unit Digits, Cyclicity Rules & Power Modulo Calculations

The Mathematical Principle of Cyclicity

The unit digit of any product $A \times B$ depends solely on the unit digits of $A$ and $B$:

$$(A \times B) \bmod 10 = ((A \bmod 10) \times (B \bmod 10)) \bmod 10$$

When an integer with base unit digit $u$ is repeatedly raised to successive integer powers ($u^1, u^2, u^3, u^4, \dots$), the resulting unit digits repeat in a fixed periodic pattern called cyclicity.

+-----------------------------------------------------------------------------+
|                          CYCLICITY CLASSIFICATION                           |
+-----------------------------------------------------------------------------+
| FAMILY 1: Cyclicity Period = 1                                              |
|   Base Digits: {0, 1, 5, 6}                                                 |
|   Behavior: Unit digit remains identical for any positive integer power.    |
|   Examples: 5^1=5, 5^2=25, 5^89 -> 5; 6^1=6, 6^2=36, 6^142 -> 6.            |
+-----------------------------------------------------------------------------+
| FAMILY 2: Cyclicity Period = 2                                              |
|   Base Digits: {4, 9}                                                       |
|   Behavior: Unit digit alternates strictly based on odd/even parity.        |
|   Examples: 4^odd -> 4, 4^even -> 6; 9^odd -> 9, 9^even -> 1.               |
+-----------------------------------------------------------------------------+
| FAMILY 3: Cyclicity Period = 4                                              |
|   Base Digits: {2, 3, 7, 8}                                                 |
|   Behavior: Unit digit cycles through four distinct values.                 |
|   Universal period for all base digits divides into 4 (LCM of 1, 2, 4 = 4). |
+-----------------------------------------------------------------------------+

Comprehensive Cyclicity Reference Table

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The Universal Unit Digit Algorithm

To find the unit digit of any expression $X^N$:

1. Isolate the Base Unit Digit: Identify $u = X \bmod 10$.
2. Examine Only the Last Two Digits of the Exponent: A number is divisible by 4 if and only if its last two digits are divisible by 4. Let $n$ be the number formed by the last two digits of $N$.
3. Compute Exponent Remainder Modulo 4:

$$r = n \bmod 4$$

1. Evaluate the Power:
   • If $r > 0$, the unit digit is the unit digit of $u^r$.
   • If $r = 0$, the unit digit is the unit digit of $u^4$.

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Multi-Term Expressions, Differences & Borrowing

In complex arithmetic questions featuring multiple terms combined through addition and subtraction:

1. Calculate the unit digit of each term independently.
2. Perform the indicated operations modulo 10.
3. Handling Negative Results (Borrowing): If the subtracted digit exceeds the leading digit, borrow 10 from the tens place before subtracting.

$$\text{Unit Digit} = (u_1 - u_2) + 10 \quad \text{when } u_1 < u_2$$

For example, if an expression reduces to unit digits $3 - 8$, the final unit digit is not $-5$, but $(13 - 8) = 5$.

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Factorials and Factorial Series Unit Digits

The factorial of a non-negative integer $n$, denoted $n!$, is the product of all positive integers less than or equal to $n$:

$$n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$$

Factorial Values and Their Unit Digits

• $1! = 1$
• $2! = 2$
• $3! = 6$
• $4! = 24 \implies \text{Unit digit is } 4$
• $5! = 120 \implies \text{Unit digit is } 0$
• $6! = 720 \implies \text{Unit digit is } 0$

Sum of Factorials: $S = 1! + 2! + 3! + \dots + n!$

When evaluating the unit digit of $S = 1! + 2! + 3! + \dots + 100!$, all terms from $5!$ onward end in 0:

$$S \equiv 1! + 2! + 3! + 4! + 0 + 0 + \dots + 0 \pmod{10}$$ $$S \equiv 1 + 2 + 6 + 24 = 33 \equiv 3 \pmod{10}$$

Thus, the unit digit of $1! + 2! + 3! + \dots + n!$ is always 3 for any $n \ge 4$.

Factorial Exponents: $x^{n!}$

When the exponent is a factorial $n!$ where $n \ge 4$:

• Since $4! = 24$, $n!$ is always a multiple of 4 for all $n \ge 4$.
• Therefore, $n! \equiv 0 \pmod 4$, meaning $r = 0$.
• Hence, the unit digit of $x^{n!}$ is simply $u^4$ for all $n \ge 4$:
  • If base $u$ is even ($2, 4, 6, 8$), $u^4$ ends in 6.
  • If base $u$ is odd ($1, 3, 7, 9$), $u^4$ ends in 1.
  • If base $u = 5$, $5^{n!}$ ends in 5.
  • If base $u = 0$, $0^{n!}$ ends in 0.

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Tens Digit Determination

While unit digit problems test cyclicity modulo 10, higher-level CBT questions occasionally ask for the tens digit (or the last two digits, which is arithmetic modulo 100).

Numbers Ending in 1: $(\dots a1)^n$

By the Binomial Theorem:

$$(10a + 1)^n = 1 + \binom{n}{1}(10a) + \binom{n}{2}(10a)^2 + \dots \equiv 1 + 10(a \cdot n) \pmod{100}$$

This yields a rapid formula for numbers ending in 1:

• Unit digit: Always $1$.
• Tens digit: $(a \times [n \bmod 10]) \bmod 10$.

Example: Find the last two digits of $41^{87}$:

• Unit digit $= 1$.
• Tens digit $= (4 \times 7) \bmod 10 = 28 \bmod 10 = 8$.
• Last two digits $= 81$.

Numbers Ending in 3, 7, 9

Transform the base into a number ending in 1 by exploiting powers:

• $3^4 = 81$ (ends in 1)
• $7^4 = 2401$ (ends in 01)
• $9^2 = 81$ (ends in 1)

Example: Find the last two digits of $7^{84}$:

• Rewrite as $(7^4)^{21} = (2401)^{21} \equiv (01)^{21} \pmod{100} = 01$.
• The last two digits are $01$ (tens digit is 0).

Powers of 2 and Even Bases Modulo 100

Key building block: $2^{10} = 1024 \equiv 24 \pmod{100}$.

• $24^{\text{even power}} \equiv 76 \pmod{100}$
• $24^{\text{odd power}} \equiv 24 \pmod{100}$
• Furthermore, $76 \times 76 \equiv 76 \pmod{100}$, and $76 \times 2^k \equiv 2^k \pmod{100}$ (for $k \ge 2$).

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Worked Examples

Worked Example 1: Multi-Term Compound Unit Digit

Problem: Find the unit digit of the expression:

$$E = 378^{126} \times 643^{185} - 287^{143} + 514^{226}$$

• Step 1: Evaluate $378^{126}$: Base unit digit $u = 8$. Exponent $126$: last two digits $26$. Divide by 4: $26 = 4 \times 6 + 2 \implies r = 2$. Unit digit $= 8^2 = 64 \implies 4$.
• Step 2: Evaluate $643^{185}$: Base unit digit $u = 3$. Exponent $185$: last two digits $85$. Divide by 4: $85 = 4 \times 21 + 1 \implies r = 1$. Unit digit $= 3^1 \implies 3$.
• Step 3: Evaluate product of first two terms:

$$(4 \times 3) \bmod 10 = 12 \bmod 10 = 2$$

• Step 4: Evaluate $287^{143}$: Base unit digit $u = 7$. Exponent $143$: last two digits $43$. Divide by 4: $43 = 4 \times 10 + 3 \implies r = 3$. Unit digit $= 7^3 = 343 \implies 3$.
• Step 5: Evaluate $514^{226}$: Base unit digit $u = 4$. Cyclicity is 2. Exponent $226$ is even $\implies 4^{\text{even}}$ ends in $6$.
• Step 6: Combine terms with borrowing:

$$\text{Unit Digit} = (2 - 3 + 6) \bmod 10$$

Since $2 < 3$, perform $(2 - 3) = -1 \implies 10 - 1 = 9$. Then add $6$: $9 + 6 = 15 \implies 5$. (Equivalently: $(2 + 6) - 3 = 8 - 3 = 5$.)

• Conclusion: The unit digit of $E$ is 5.

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Worked Example 2: Factorial Exponents Series

Problem: Find the unit digit of the summation:

$$S = (1!)^1 + (2!)^2 + (3!)^3 + (4!)^4 + (5!)^5 + \dots + (100!)^{100}$$

• Step 1: Analyze terms for $n \ge 5$: For every $n \ge 5$, $n!$ ends in 0. Therefore, $(n!)^n$ ends in $0^n = 0$. These terms contribute 0 to the unit digit.
• Step 2: Evaluate the first 4 terms individually:
  • Term 1: $(1!)^1 = 1^1 = 1$.
  • Term 2: $(2!)^2 = 2^2 = 4$.
  • Term 3: $(3!)^3 = 6^3 = 216 \implies \text{Unit digit is } 6$.
  • Term 4: $(4!)^4 = (24)^4$. Base unit digit is 4, exponent 4 is even $\implies 4^4 = 256$, unit digit is $6$.
• Step 3: Sum the non-zero unit digits:

$$\text{Unit Digit} = (1 + 4 + 6 + 6) \bmod 10 = 17 \bmod 10 = 7$$

• Conclusion: The unit digit of the summation is 7.

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Worked Example 3: Tens Digit Calculation

Problem: Find the last two digits (tens and units) of $31^{146}$.

• Step 1: Check form: The base ends in 1: $31 = 10(3) + 1$, so $a = 3$. The exponent is $n = 146$.
• Step 2: Apply the binomial unit and tens shortcut:
  • Unit digit is always $1$.
  • Tens digit is given by: $(a \times [n \bmod 10]) \bmod 10$.
  • Here $a = 3$, and $n \bmod 10 = 6$.
  • Tens digit $= (3 \times 6) \bmod 10 = 18 \bmod 10 = 8$.
• Step 3: Verify via binomial expansion:

$$(30 + 1)^{146} = 1 + 146(30) + \binom{146}{2}(30)^2 + \dots$$

Modulo 100, all terms from $(30)^2$ onward have factors of $900 \equiv 0 \pmod{100}$.

$$1 + 146 \times 30 = 1 + 4380 = 4381 \equiv 81 \pmod{100}$$

• Conclusion: The last two digits are 81 (tens digit is 8, unit digit is 1).