Find the smallest number which, when divided by 8, 12 and 20, leaves remainders 3, 7 and 15 respectively.

Class 10Real NumbersNumber Systems

Step-by-Step Solution

Each divisor falls short by the SAME amount: 8 − 3 = 5, 12 − 7 = 5, 20 − 15 = 5. So the number is LCM(8, 12, 20) − 5. With

8 = 2^{3}, 12 = 2^{2} \cdot 3, 20 = 2^{2}

·5, the LCM is

2^{3}

·3·5 = 120, giving 120 − 5 = 115. Check:

115 = 8 \times 14 + 3 = 12 \times 9 + 7 = 20 \times 5 + 15

. Adding 5 instead of subtracting gives 125.

Answer: D) 115

**Theorem**: A rational number

\frac{p}{q}

(in lowest terms) has a terminating decimal expansion if and only if

q

has the form **2ᵐ × 5ⁿ** (m, n ≥ 0).

**Examples**:
-

\frac{3}{8} : q = 8 = 2^{3}

→ terminates

. (\frac{3}{8} = 0.375.) - \frac{7}{40} : q = 40 = 2^{3} \times 5

→ terminates

. (\frac{7}{40} = 0.175.) - \frac{1}{3} : q = 3

→ does NOT terminate

. (\frac{1}{3} = 0.333

… recurring.)
-

\frac{4}{15} : q = 15 = 3 \times 5

→ does NOT terminate (because of the 3).

**Why**: any decimal expansion is a fraction with denominator 10ⁿ = 2ⁿ × 5ⁿ. To convert

\frac{p}{q}

to decimal form, you need

q

to ONLY have 2s and 5s as primes — otherwise you can't multiply numerator/denominator into a power of 10.

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