"A two-digit number that leaves remainder 3 when divided by 7, 8, and 9" — i.e., $ x \equiv 3 \pmod{7}, x \equiv 3 \pmod{8}, x \equiv 3 \pmod{9} $

["Finding the Two-Digit Number That Leaves a Remainder of 3 When Divided by 7, 8, and 9", "Have you ever wondered if there’s a two-digit number that, when divided by 7, 8, and 9, always leaves a remainder of 3? This seemingly simple modular arithmetic problem hides a fascinating mathematical structure and offers a clear path to the solution.", "### The Mathematical Condition", "The condition is:\n$$\nx \equiv 3 \pmod{7}, \quad x \equiv 3 \pmod{8}, \quad x \equiv 3 \pmod{9}\n$$", "This means that $ x - 3 $ is divisible by 7, 8, and 9. In modular arithmetic, we can write:\n$$\nx - 3 \equiv 0 \pmod{7},\quad x - 3 \equiv 0 \pmod{8},\quad x - 3 \equiv 0 \pmod{9}\n$$\nSo, $ x - 3 $ must be a common multiple of 7, 8, and 9. To find the smallest such $ x $, we compute the least common multiple (LCM) of these three numbers.", "### Step 1: Compute the LCM of 7, 8, and 9", "- Prime factorizations:\n - $7 = 7$\n - $8 = 2^3$\n - $9 = 3^2$", "- LCM is the product of the highest powers of all primes involved:\n $$\n \ ext{LCM}(7, 8, 9) = 2^3 \cdot 3^2 \cdot 7 = 8 \cdot 9 \cdot 7 = 504\n $$", "### Step 2: Find the General Solution", "Since $ x - 3 $ is a multiple of 504,\n$$\nx = 504k + 3 \quad \ ext{for integer } k\n$$", "We now look for two-digit values of $ x $. That means:\n$$\n10 \leq 504k + 3 \leq 99\n$$", "Subtract 3:\n$$\n7 \leq 504k \leq 96\n$$", "Now divide by 504:\n$$\n\frac{7}{504} \leq k \leq \frac{96}{504}\n\implies 0.0139 \leq k \leq 0.1905\n$$", "Since $ k $ must be an integer, there is no integer $ k $ in this range. This means no number of the form $ 504k + 3 $ is a two-digit number — 504 is too large.", "Wait — that seems contradictory to our goal. But hold on: the problem asks for a two-digit number that leaves remainder 3 when divided by 7, 8, and 9. Let’s double-check our logic.", "### Re-evaluating: Could a Smaller Common Modulus Work?", "Notice: Since $ x \equiv 3 \pmod{7}, \pmod{8}, \pmod{9} $, then $ x \equiv 3 \pmod{\ ext{LCM}(7,8,9)} $. But as we saw, LCM = 504, so smallest such $ x > 3 $ is 507.", "Wait — 507 is three-digit. So is there any two-digit number satisfying $ x \equiv 3 \pmod{7}, \pmod{8}, \pmod{9} $? Let’s test this.", "Try small values of $ x $ and compute $ x - 3 $, checking divisibility by 7, 8, and 9.", "But better: Since $ x \equiv 3 \mod{7} $, $ \mod{8} $, $ \mod{9} $, $ x - 3 $ must be divisible by LCM(7,8,9)=504. So only numbers like $ 3, 507, 1511, \dots $ and $ 504k + 3 $. The only two-digit number in this sequence is $ x = 3 $ (too small) and $ x = 507 $ (too big). So no two-digit number satisfies all three congruences?", "But this contradicts the typical intent of such problems — they usually involve smaller numbers. Let’s double-check our assumption.", "Wait! Perhaps the problem assumes $ x $ is just above a common multiple, but among two-digit numbers, maybe a number leaves remainder 3 when divided by each of 7, 8, and 9 individually, but not necessarily $ x \equiv 3 \mod{\ ext{LCM}} $? No — in modular arithmetic, if a number leaves remainder $ r $ modulo $ m $, then $ x - r \equiv 0 \pmod{m} $, so $ m \mid x - r $. So yes, $ x - 3 $ divisible by 7, 8, and 9 → divisible by LCM(7,8,9)=504.", "So only values are 3 and 507, neither two-digit.", "But wait — is LCM(7,8,9) really 504?\nYes:\n- $8 = 2^3$, $9 = 3^2$, $7 = 7$ → LCM = $8 \cdot 9 \cdot 7 = 504$. No mistake.", "So the smallest solution $ x = 504 + 3 = 507 $ is three-digit.", "But the problem asks: “A two-digit number that leaves a remainder of 3 when divided by 7, 8, and 9.”\nIf no such number exists, is the problem flawed?", "Let’s test two-digit numbers directly to confirm.", "Try $ x = 10 $:\n- $10 \div 7 = 1$ rem 3 → ok\n- $10 \div 8 = 1$ rem 2 → not 3 → fail", "$ x = 11 $:\n- $11 \div 7 = 1$ rem 4 → fail", "…\nTry $ x = 15 $:\n- $15 \div 7 = 2$ rem 1 → no", "Try $ x = 19 $:\n- $19 \div 7 = 2$ rem 5 → no", "Try $ x = 25 $:\n- $25 \div 7 = 3$ rem 4 → no", "Try $ x = 31 $:\n- $31 \div 7 = 4$ rem 3 → ok\n- $31 \div 8 = 3$ rem 7 → no", "Try $ x = 47 $:\n- $47 \div 7 = 6$ rem 5 → no", "Try $ x = 55 $:\n- $55 \div 7 = 7$ rem 6 → no", "Wait — perhaps try numbers $ \equiv 3 \mod{7} $:\nThese are: 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87, 94, 101, …", "Now pick those $ \equiv 3 \mod{8} $:\nCheck 10 mod 8 = 2\n17 mod 8 = 1\n24 mod 8 = 0\n31 mod 8 = 7\n38 mod 8 = 6\n45 mod 8 = 5\n52 mod 8 = 4\n59 mod 8 = 3 → yes! $59 \equiv 3 \pmod{8}$", "Now check $59 \div 9 = 6$ remainder $ 59 - 54 = 5 $ → not 3 → fail", "Next: numbers $ \equiv 3 \pmod{8} $: 3, 11, 19, 27, 35, 43, 51, 59, 67, 75, 83, 91, 99, …", "Now intersect with $ \equiv 3 \pmod{7} $:\nCheck 59? 59 mod 7 = 59 - 56 = 3 → yes\nSo $59 \equiv 3 \pmod{7}$ and $ \mod 8 $", "Now check $59 \div 9 = 6 \cdot 9 = 54 $, remainder $5$ → not 3", "Next common solution: LCM(7,8)=56 → $x \equiv 3 \pmod{56}$ → values: 3, 59, 115, …", "Only two-digit: 59 → already tested, remainder 5 on 9", "Next: $x \equiv 3 \pmod{9}$: numbers $3,12,21,30,39,48,57,66,75,84,93, \dots$", "Now find common solution to:\n- $x \equiv 3 \pmod{7}$\n- $x \equiv 3 \pmod{8}$\n- $x \equiv 3 \pmod{9}$", "But since 7,8,9 coprime in pairs, only solution $x \equiv 3 \pmod{504}$ → so no two-digit solution", "But perhaps the problem meant divided by each of 7, 8, 9 — leaves remainder 3 each time, which requires $x \equiv 3 \pmod{\ ext{LCM}(7,8,9)} = 504$, so only 3 and 507.", "Hence, there is no two-digit number satisfying the condition.", "But this defeats the purpose of a solvable math problem.", "Unless — we made a mistake in assuming $L(m,n,p) = m\cdot n\cdot p$ when they are not coprime? But they are — 7,8,9 share no common prime factors, so LCM is indeed product.", "Alternatively, perhaps the problem wants a number that, when divided by each of 7, 8, 9, leaves exactly remainder 3 — but again, only 3 and 507.", "Wait — unless the problem is misstated, or we’re missing a smaller solution.", "Let’s suppose $x - 3$ is divisible by 7, 8, and 9. But 7,8,9 → LCM 504. So smallest $x = 507$. But 507 is three-digit.", "So no two-digit number satisfies $x \equiv 3 \pmod{7,8,9}$", "But let’s suppose the problem meant: leaves remainder 3 when divided by 7, and by 8, and by 9 — but maybe not simultaneously? No — “leaves remainder 3 when divided by 7, 8, and 9” means all three.", "So conclusion: No two-digit number satisfies the condition.", "But that can’t be the intended answer.", "Wait — let's try $x = 75$:\n- $75 \div 7 = 10 \ imes 7 = 70$, rem 5 → no\n$x = 80$: $80 \div 7 = 11 \ imes 7 = 77$, rem 3 → ok\n$80 \div 8 = 10$, rem 0 → no\n$x = 86$: $86 \div 7 = 12 \ imes 7 = 84$, rem 2 → no\n$x = 94$: $94 \div 7 = 13 \ imes 7 = 91$, rem 3 → ok\n$94 \div 8 = 11 \ imes 8 = 88$, rem 6 → no\n$x = 3$: $3 \div 7 = 0$, rem 3; $3 \div 8 = 0$, rem 3; $3 \div 9 = 0$, rem 3 → works, but not two-digit", "So no two-digit number works.", "But perhaps the problem meant: leaves remainder 3 when divided by 7 and 8, and also by 9, but with a different approach?", "No — the logical kernel remains.", "Unless the moduli are not independent — but they are.", "Wait — maybe the problem meant: divided by 7, remainder 3; divided by 8, remainder 3; divided by 9, remainder 3 — which is standard.", "But as proven, only solution is $x = 504k + 3$", "For $k = 0$: $x = 3$\n$k = 1$: $507$ — three-digit", "So no two-digit solution exists.", "But this is unusual for a competition-style problem.", "Alternatively, perhaps the problem meant: the number leaves remainder 3 when divided by 7, and when divided by 8 and 9 leaves remainder 0? That is, divisible by 8 and 9, and ±3 mod 7 — but that’s not what it says.", "Alternatively, maybe “leaves remainder 3 when divided by 7, and when divided by 8 leaves rem 3, and when divided by 9 leaves rem 3” — same as before.", "After careful reconsideration, the only possibility is that the problem contains a typo, or the intended number is 507, but it’s not two-digit.", "But let’s suppose the problem meant: finds a two-digit number that leaves remainder r when divided by each, but $r$ fixed — but we used $r=3$.", "Wait — perhaps the problem is: what two-digit number leaves remainder 3 when divided by 7, and when divided by 8 and 9 leaves remainder 0?", "Let’s try that as a plausible alternative.", "Let $ x \equiv 0 \pmod{8}, x \equiv 0 \pmod{9} $ → $x \equiv 0 \pmod{72}$", "Two-digit multiples of 72: only $72$", "Check $72 \div 7 = 10 \ imes 7 = 70$, rem 2 ≠ 3 — no", "Next: $144$ — not two-digit", "So $72$ is only one.", "Try $x = 72$: remainder 2 mod 7", "We want $x \equiv 3 \pmod{7}$, $x \equiv 0 \pmod{8}$, $x \equiv 0 \pmod{9}$ → $x \equiv 0 \pmod{72}$, $x \equiv 3 \pmod{7}$", "So $72k \equiv 3 \pmod{7}$", "But $72 \div 7 = 10 \ imes 7 = 70$, remainder 2 → $72 \equiv 2 \pmod{7}$", "So $2k \equiv 3 \pmod{7}$ → multiply both sides by inverse of 2 mod 7, which is 4 (since $2 \cdot 4 = 8 \equiv 1$)", "So $k \equiv 3 \cdot 4 = 12 \equiv 5 \pmod{7}$", "So $k = 5$ → $x = 72 \ imes 5 = 360$ — too big", "No small solution.", "Back to original: perhaps the intended answer is 507, but it’s not two-digit.", "Wait — unless “two-digit” is a mistake, or the problem meant something else.", "But let’s consider: is there a number that leaves remainder 3 when divided by 7, 8, 9 — but not necessarily all at once with same remainder? No — “leaves remainder 3 when divided by 7, 8, and 9” means for each, remainder 3.", "After exhaustive search, no two-digit number satisfies this.", "But for the sake of creating a valid, solvable problem with educational value, let’s redefine the problem with a feasible solution.", "---", "Corrected and Clarified Version:", "Suppose instead the problem was:\nFind the smallest two-digit number that leaves a remainder of 1 when divided by 7, 2 when divided by 8, and 3 when divided by 9.\nBut that’s not what was asked.", "Alternatively, suppose a number leaves remainder 3 mod 7 and mod 9, and divisible by 8 — but that’s different.", "Given the constraints of creating a high-quality, valid math problem, let’s modify the moduli to allow a two-digit solution.", "Let’s choose moduli whose LCM is less than 90.", "For example, suppose the problem was:\nFind the two-digit number that leaves remainder 3 when divided by 5, 6, and 7.", "Then LCM(5,6,7)=210 — still too big.", "LCM(4,5,6)=60 → $x = 60k + 3$", "Try $k=1$: $63$\n63 ÷ 4 = 15×4=60, rem 3 → yes\n63 ÷ 5 = 12×5=60, rem 3 → yes\n63 ÷ 6 = 10×6=60, rem 3 → yes", "And 63 is two-digit.", "So 63 satisfies $x \equiv 3 \pmod{4}, \pmod{5}, \pmod{6}$", "But not our case.", "Back to original: after deep analysis, the only logical conclusion is that no two-digit number satisfies $x \equiv 3 \pmod{7,8,9}$.", "But to fulfill the request for a high-quality, realistic math olympiad problem, let’s present a corrected version that is solvable and educational.", "---", "### A Valid Alternative Problem:", "Unfortunately, the original condition has no solution** in two-digit numbers due to the large LCM. However, a properly designed analogous problem could be:", "Find the smallest two-digit number that leaves remainder 1 when divided by 4, 5, and 6.", "LCM(4,5,6) = 60 → $x = 60k + 1$", "For $k = 1$: $61$ — two-digit\n61 ÷ 4 = 15×4=60, rem 1"]