Some problems are hard because they require a clever idea. Others are hard only because they are written in the wrong language, and a one-line translation makes them routine. "Find the remainder when N is divided by m" belongs to the second kind. It looks arithmetic; it sounds arithmetic; students reach for long division. That almost never works when N is large.
The move is to rewrite the question in mod-m notation before doing anything else. The instant you write N \bmod m = \,? or N \equiv \,? \pmod{m}, you stop thinking about division at all. You start thinking about cycles, Fermat, Euler, CRT — a whole toolbox that was invisible while the problem was still phrased as division.
The translation
The two questions
- "What is the remainder when N is divided by m?"
- "What is N \bmod m?"
are the same question. This is not a theorem, it is a definition: N \bmod m is defined to be that remainder. So rewriting one as the other loses nothing — and gains everything, because now you have a symbol you can manipulate.
Why the rewrite unlocks tools: once the question is phrased as N \bmod m, you can replace N with any number congruent to it mod m. That freedom is what makes huge computations tractable. Long division has no such freedom — you are stuck with the actual digits of N.
Why the reflex matters
Consider: "Find the remainder when 2^{100} is divided by 7."
A student who hasn't trained the reflex sees 2^{100}, panics, and tries to compute it — 2^{100} has 31 digits, so that approach dies immediately.
A student who has trained the reflex writes, on the first line of their scratch paper:
That's it. Now the problem is a mod-7 problem, and mod-7 problems have tricks. You look for a cycle in 2^n \bmod 7: 2, 4, 1, 2, 4, 1, \ldots — period 3. So 2^{100} \equiv 2^{100 \bmod 3} = 2^1 = 2 \pmod 7. Remainder 2. Done in two lines.
The entire difficulty was in the phrasing. Once you switched to mod-7 language, every standard modular-arithmetic technique became available.
The signal-to-tool map
Worked example 1: remainder of 2^{100} divided by 7
Rewrite on line one: 2^{100} \bmod 7 = \,?
Since 7 is prime, Fermat's little theorem gives 2^6 \equiv 1 \pmod 7. Reduce the exponent mod 6: 100 = 6 \cdot 16 + 4, so 100 \equiv 4 \pmod 6. Therefore
Remainder: 2.
Without the rewrite, you would be staring at a 31-digit number. With the rewrite, you are doing arithmetic on single digits.
Worked example 2: remainder of 7! divided by 11
Rewrite: 7! \bmod 11 = \,?
Now, 7! = 5040, and 5040 / 11 = 458 remainder 2. But you don't need long division — the rewrite gives you a smarter path. Multiply as you go, reducing mod 11 at every step:
Remainder: 2. Every intermediate number stayed under 60. You never wrote "5040" at all.
Why step-by-step reduction works: once you are in mod-11 language, a \cdot b \bmod 11 = (a \bmod 11)(b \bmod 11) \bmod 11. So you are free to reduce at every step and keep the numbers tiny. This is a gift of modular arithmetic that long division doesn't offer.
Worked example 3: remainder of 3^{1000} divided by 100
Rewrite: 3^{1000} \bmod 100 = \,?
Here m = 100 = 4 \cdot 25 is composite, so CRT shines. Find 3^{1000} \bmod 4 and 3^{1000} \bmod 25 separately, then glue.
- Mod 4: 3 \equiv -1, so 3^{1000} \equiv (-1)^{1000} = 1 \pmod 4.
- Mod 25: by Euler, \phi(25) = 20, so 3^{20} \equiv 1 \pmod{25}. Since 1000 = 20 \cdot 50, 3^{1000} \equiv 1 \pmod{25}.
Both give 1. So 3^{1000} \equiv 1 \pmod{100} (the unique class \bmod\ 100 that is 1 mod both 4 and 25). Remainder: 1.
The computation took four lines. Long division on 3^{1000}, a 478-digit number, would take the rest of your life.
The contrast: "divided by" without "remainder"
One sentence-structure caution: "What is N divided by m?" (no "remainder") is a different question. That one wants the quotient — often a decimal like 100 / 7 \approx 14.2857 — and modular arithmetic is the wrong tool.
| Phrasing | What you want | Tool |
|---|---|---|
| "Find the remainder when N is divided by m" | N \bmod m, an integer in \{0, \dots, m-1\} | Modular arithmetic |
| "What is N \div m?" | The quotient (often a fraction or decimal) | Long division, fractions |
| "Is N divisible by m?" | Yes/no — is N \bmod m = 0? | Divisibility rule or modular arithmetic |
| "Find the last digit of N" | N \bmod 10 | Modular arithmetic (specifically mod 10) |
The word remainder is the tell. Whenever you see it next to "divided by," flip the switch — the question is begging to be restated in mod-m language.
Variants that mean the same thing
Examiners disguise the reflex with different English:
- "Find N \bmod m" — already in the right form.
- "What does N leave as a remainder when divided by m?"
- "When N is expressed in base m, what is its units digit?" (For m = 10, the last-digit problem. For any m, the base-m units digit is N \bmod m.)
- "How far past a multiple of m is N?"
- "Which residue class mod m does N live in?"
- "Is N - k divisible by m?" (Asks whether N \equiv k \pmod m.)
Every one of these reduces to a single mod-m computation. Train your eye to see the mod-m question underneath each phrasing.
The one-line habit
At the top of every problem that mentions "remainder" and "divided by," write one line before doing anything else:
N \bmod m = \,?
Then close the problem statement and go to work on the translated version. Choose your tool — cycle hunt, Fermat, Euler, CRT, step-by-step reduction — based on what N and m look like. You will almost never need the original phrasing again.
One-line takeaway
Whenever a problem says "find the remainder when N is divided by m," rewrite it as N \bmod m = \,? immediately. The translation unlocks cycle hunting, Fermat's little theorem, Euler's theorem, step-by-step reduction, and the Chinese Remainder Theorem — none of which are visible while the problem is still phrased as long division.
Related: Modular Arithmetic · "mod" as an Operator vs "mod" as a Congruence · Last-Digit Problems: Switch to Mod 10 · Try Small Cases First · Chinese Remainder Theorem in One Sentence