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Determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Given: Numbers 110000, 100000, 8, 15 and 21.
To find: Here we have to find the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Solution:
LCM is the least common multiple of any three numbers and to find the number nearest to 110000 we have to check if 110000 is divisible by the LCM of 8, 15 and 21.
Now, calculating LCM of 8, 15 and 21 using prime factorization method:
Writing the numbers as a product of their prime factors:
Prime factorization of 8:
- $2\ \times\ 2\ \times\ 2\ =\ 2^3$
Prime factorization of 15:
- $3\ \times\ 5\ =\ 3^1\ \times\ 5^1$
Prime factorization of 21:
- $3\ \times\ 7\ =\ 3^1\ \times\ 7^1$
Multiplying highest power of each prime number:
- $2^3\ \times\ 3^1\ \times\ 5^1\ \times\ 7^1\ =\ 840$
LCM(8, 15, 21) $=$ 840
Now,
$110000\ =\ 840\ \times\ 130\ +\ 800$
Here remainder is 800. So,
Required number $=$ 110000 $-$ 800 $=$ 109200
So, the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21 is 109200.
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