Find the smallest positive integer which when divided by $12, 15, 20,$ and $24$ leaves a remainder of $7$ in each case, but is completely divisible by $13$. ________ (answer in integer).
Answer Key:
247
Step-by-Step Solution & Explanation
**Method 1: LCM and Modular Congruence**
The required number has the form:
$N = \text{LCM}(12, 15, 20, 24) \times k + 7$
$\text{LCM}(12, 15, 20, 24) = 120$
$N = 120k + 7$
We require $N \equiv 0 \pmod{13}$:
$120k + 7 \equiv (13 \times 9 + 3)k + 7 \equiv 3k + 7 \equiv 0 \pmod{13}$
$3k \equiv -7 \equiv 6 \pmod{13}$
Since $\gcd(3, 13) = 1$, divide both sides by 3:
$k \equiv 2 \pmod{13}$
For the smallest positive integer, choose $k = 2$:
$N = 120(2) + 7 = 240 + 7 = 247$
Verification: $247 = 13 \times 19$ (divisible by 13), and $247 = 120 \times 2 + 7$ (leaves remainder 7 when divided by 12, 15, 20, 24).