Remainders and modular thinking

If it is $9$ am now, what time will it be $50$ hours later? The trick is not to track each hour but to remove all the complete $24$-hour days. The answer comes from a remainder. This is the heart of modular arithmetic , the maths of "what's left over".

Concept

For positive integers $a$ and $m$, the remainder of $a$ divided by $m$ is the integer $r$ with $0 \le r < m$ such that

$a = qm + r$

for some quotient $q$. Examples: $17 \div 5 = 3$ remainder $2$, so $17 = 3(5) + 2$. We write this as

$17 \equiv 2 \pmod{5}.$

"$a \equiv b \pmod m$" is read "$a$ is congruent to $b$ modulo $m$" and it means $a$ and $b$ leave the same remainder on division by $m$ , equivalently, $a - b$ is a multiple of $m$.

Why this matters. Many real-life cycles are modular:

• Clocks repeat every $12$ or $24$ hours.
• Calendars repeat every $7$ days for the days-of-week pattern.
• Months repeat every $12$.
• Compass directions repeat every $360^\circ$.

To answer "what day will it be $100$ days from Monday?", compute $100 \div 7 = 14$ remainder $2$. Two days after Monday is Wednesday.

Operations. Addition and multiplication respect remainders:

• If $a \equiv a' \pmod m$ and $b \equiv b' \pmod m$, then $a + b \equiv a' + b' \pmod m$.
• Similarly for multiplication.

So you can replace any number by its remainder before combining. To find $46 \times 38 \pmod 9$, replace each: $46 \equiv 1$, $38 \equiv 2$. So $46 \times 38 \equiv 1 \times 2 = 2 \pmod 9$.

Last digit problems. "Find the last digit of $7^{100}$." This is a mod-$10$ question. Last digits of $7^1, 7^2, 7^3, 7^4$ are $7, 9, 3, 1$ , then the pattern repeats every $4$. $100 \div 4 = 25$ remainder $0$, so $7^{100}$ ends in the same digit as $7^4$, i.e. $1$.

Check digits. Phone numbers, credit cards, and Aadhaar numbers include a final "check digit" computed by a modular formula. If you mistype a digit, the check fails. Computers use this billions of times a day.

Casting out nines. Recall the divisibility-by-$9$ rule. Why does it work? Because $10 \equiv 1 \pmod 9$, so a digit $d$ in any place contributes $d \cdot 10^k \equiv d \pmod 9$. Hence the number is congruent (mod $9$) to its digit sum. Same trick proves the divisibility-by-$3$ rule.

Worked examples

Example 1. Today is Wednesday. What day of the week will it be $200$ days from today?

• $200 \div 7 = 28$ remainder $4$.
• Counting $4$ days from Wednesday: Thu, Fri, Sat, Sun.
• Answer: Sunday.

Example 2. Find the remainder when $2{,}345{,}678$ is divided by $9$.

• Digit sum: $2+3+4+5+6+7+8 = 35$. Digit sum again: $3+5 = 8$.
• So remainder is $8$.

Example 3. Find the last digit of $3^{50}$.

• Last digits of $3^1, 3^2, 3^3, 3^4$ are $3, 9, 7, 1$ , cycle of length $4$.
• $50 \div 4 = 12$ remainder $2$.
• Answer: same last digit as $3^2$, which is $9$.

Example 4. A clock now shows $7$ pm. What will it show $250$ hours later?

• $250 \div 24 = 10$ remainder $10$.
• $7 + 10 = 17$ hours, i.e., $5$ am the next morning.

Try it yourself

1. Find the remainder when $93$ is divided by $7$.
2. Today is Friday. What day will it be after $1000$ days?
3. Find the last digit of $2^{30}$.
4. Find the last digit of $9^{99}$.
5. Compute $123 \times 124 \pmod 9$ using the digit-sum trick.
6. A clock at $10$ am , what time is it $4{,}000$ hours later?
7. Find the remainder when $1! + 2! + 3! + 4! + \dots + 10!$ is divided by $5$.
8. Show that if $a \equiv b \pmod m$ then $a^2 \equiv b^2 \pmod m$.

Activity / Insight

Test a check digit. Take an Aadhaar number (your own or a sample one) and look up its check-digit formula. Apply it to all digits except the last and verify that you recover the last digit. Now change one digit anywhere in the number and reapply , the check should fail. This is modular arithmetic protecting your identity.