Question

please tell with whole solution please ​

Answer 1

To find:

The smallest square number divisible by the numbers 8, 9 and 10.

Solution:

To begin with, we must find the lowest common multiply of the numbers 8, 9 and 10, since the square number should be divisible by each.

$\Large \begin{array}{c|c} \underline{\sf {2}}&\underline{\sf {\; \; 8,9,10 \; \; \: }} \\ \underline{\sf {2}}&\underline{\sf {\; \; 4,9,5 \; \; \: }}\\ \underline{\sf {2}}&\underline{\sf {\; \; 2,9,5 \; \; \: }}\\ \underline{\sf {3}}&\underline{\sf {\; \; 1,9,5 \; \; \: }}\\ \underline{\sf {3}}&\underline{\sf {\; \; 1,3,5 \; \; \: }}\\ \underline{\sf {5}}&\underline{\sf {\; \; 1,1,5 \; \; \: }} \\ & {\sf \; 1,1,1 \; \; }\end{array}$

LCM of 8, 9 and 10 = 2 × 2 × 2 × 3 × 3 × 5

                                 = 360

We know that in the prime factorization of a square number, the factors must exist in pairs of two. As the factors 2 and 5 do not exist in pairs, 360 is not a square number. To make it a square number, we multiply it by 2 and 5 = 10. So,

• 360 × 10 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

Now each factor exists in a pair!

• 3600 = (2 × 2 × 3 × 5)²
• 3600 = 60²

Therefore, 3600 is the smallest square number divisible by each of the numbers 8, 9 and 10.