A number with three remainders
The challenge contract
Find the integer n with 0 ≤ n < 105 that leaves remainder 2 when divided by 3, remainder 3 when divided by 5, and remainder 2 when divided by 7. Give the number and explain why no other integer in the stated interval works.
Division here means ordinary integer division with a nonnegative remainder smaller than the positive divisor. There is no hidden base change, digit trick or alternative interpretation of “remainder.” The interval includes zero but excludes 105. Those details are part of the mathematical content: remove the bound and the problem has infinitely many answers.
You can solve this by listing candidates, but first choose a representation that makes the shared structure visible. Two clues ask for the same remainder. That repeated feature is more useful than the order in which the clues were written. Resist the urge to start with the divisor that looks easiest merely because it appears first.
Translate sentences into constraints
Write the conditions as n ≡ 2 modulo 3, n ≡ 3 modulo 5 and n ≡ 2 modulo 7. Congruence means that subtracting the stated remainder leaves a multiple of the divisor. The first and third conditions therefore say that n − 2 is divisible by both 3 and 7. Since those divisors are coprime, n − 2 is a multiple of 21.
This is already a compressed search. Instead of checking 105 integers independently, write n = 21k + 2. The bound restricts the possible values of k. The middle condition will distinguish among those possibilities. Notice the direction of the reasoning: every valid answer must belong to this family, but not every member of the family has yet passed all three tests.
A common mistake is to add the remainders or multiply the divisors and then adjust until a number “looks right.” Such experimentation can reveal a candidate, but it does not preserve the logical relationship between clue and conclusion. A useful line of algebra should either retain all current possibilities or explicitly explain why a discarded possibility is impossible.
Help in controlled increments
Hint 1 — Combine the equal remainders
The first and third conditions force n = 21k + 2 for some integer k. Use 0 ≤ n < 105 to identify the short list of relevant k values.
Hint 2 — Reduce the remaining equation
Modulo 5, the expression 21k + 2 has the same remainder as k + 2. Therefore k + 2 must be congruent to 3 modulo 5.
Hint 3 — Use the bound, not a guess
The permitted k values are 0, 1, 2, 3 and 4. Exactly one of them is congruent to 1 modulo 5.
Full solution — 23, with exclusion
The bound gives k ∈ {0, 1, 2, 3, 4}. The modulo-5 condition gives k ≡ 1 modulo 5, so k = 1 and n = 23. Direct checks are 23 = 3 × 7 + 2, 23 = 5 × 4 + 3 and 23 = 7 × 3 + 2.
For a second uniqueness proof, suppose x and y both satisfy the three clues. Their difference is divisible by 3, 5 and 7. These are pairwise coprime, so their difference is divisible by 105. Two integers in the interval from 0 inclusive to 105 exclusive differ in absolute value by less than 105. The only possible multiple is zero, so x = y.
Generalize only after verifying
The unrestricted answer family is n = 23 + 105t for integer t. In an expanded interval, more than one family member may qualify. This is why a solution that ignores the interval has not finished the original task. Conversely, a brute-force scan of all permitted integers can prove uniqueness if its range and checking rule are complete; it is simply less revealing about the underlying structure.
The Chinese remainder theorem formalizes this behavior for pairwise coprime moduli. Its historical context includes the Sunzi Suanjing. Euclid’s Elements supplies foundational number-theoretic ideas, while Donald Knuth’s The Art of Computer Programming places modular arithmetic within a broader computational setting. These are routes for further study, not credentials a reader needs before attempting the riddle.
For a harder variation, replace one modulus with a number that shares a factor with another. Now compatibility matters before uniqueness. Conditions such as n ≡ 0 modulo 2 and n ≡ 1 modulo 4 cannot both hold: the first requires an even number and the second an odd number. The obstruction is structural, not a failure to search long enough.
Finding one witness proves existence. Ruling out every rival proves uniqueness.
Carry that distinction into sequence puzzles, probability questions and counting arguments. State whether order matters, whether replacement is allowed, and whether equally likely outcomes have actually been justified. Claude Shannon’s information theory and John Conway’s recreational mathematics reward the same habit: make the model explicit before extracting an impressive-looking answer. For a less numerical test of assumptions, try the escape and lateral guide.