Problem Statement
The least number which when divided by 12, 15, 20, and 54 leaves remainder 8 in each case is:
Explanation
The required number is of the form LCM of the divisors plus the remainder. Since all divisions leave the same remainder 8, we find LCM of 12, 15, 20, and 54, then add 8.
Prime factorization: 12 equals 2 squared times 3. 15 equals 3 times 5. 20 equals 2 squared times 5. 54 equals 2 times 3 cubed. LCM equals highest power of each prime equals 2 squared times 3 cubed times 5 equals 4 times 27 times 5 equals 540. Required number equals LCM plus remainder equals 540 plus 8 equals 548. Verification: 548 divided by 12 equals 45 remainder 8. 548 divided by 15 equals 36 remainder 8. 548 divided by 20 equals 27 remainder 8. 548 divided by 54 equals 10 remainder 8. All correct.
Code Solution
SolutionRead Only
// Numbers: 12, 15, 20, 54 // Each leaves remainder 8 // Prime Factorization: // 12 = 2² × 3 // 15 = 3 × 5 // 20 = 2² × 5 // 54 = 2 × 3³ // LCM = 2² × 3³ × 5 // = 4 × 27 × 5 // = 540 // Required Number = LCM + Remainder // = 540 + 8 // = 548 // Verification: // 548 ÷ 12 = 45 rem 8 ✓ // 548 ÷ 15 = 36 rem 8 ✓ // 548 ÷ 20 = 27 rem 8 ✓ // 548 ÷ 54 = 10 rem 8 ✓
Practice Sets
This question appears in the following practice sets:
