Question

$find \: the \: smallest \:5 \: digit \: number \: which \: is \: exactly \: divisible \: by \: 27 \: 19 \: and \: 48$
How to do this question
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Answer 1

Answer:

12= 2 × 2 × 3

24 = 2 × 2 × 2 × 3

48= 2 x 2 x 2×2×3

60 = 2 × 2 × 3 × 5

96 = 2 × 2 ×2×2×2×3

⇒ LCM of 12, 24, 48, 60 and 96=2×2×2×2×2 × 3 × 5 = 480

Smallest 5 digit number = 10000

On dividing 10000 by 480 we get 400 as a remainder. [10000 = 480 x 20 + 400]

... Smallest 5 digits number divisible by 12, 24, 48, 60 and 96 = 10000 - 400 + 480 = 10080

Answer 2

Answer:

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Let N be the desired 5-digit number.

Since the prime factors of 27, 19 and 48 are 3^3, 19 and 3 * 2^4 respectively,

it follows that N must be a multiple of 3^3, 19 and 2^4 which is 8208.

The smallest multiple of 8208 which is a 5 digit number is 8208 * 2 = 16416