Respuesta :
Answer:
Which is the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively? 204.
Step-by-step explanation:
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Answer:
204 is the smallest number.
Step-by-step explanation:
28 - 8 = 20 and 32 - 12 = 20
The required number will be 20 less than the LCM of 28 and 32.
Prime factorization of 28,
→ 2 × 2 × 7
Prime factorization of 32,
→ 2 × 2 × 2 × 2 × 2
LCM(28, 32) = 2 x 2 x 2 x 2 x 2 x 7 = 224
Hence, the required smallest number,
→ 224 - 20 = 204