Question

12. Find the smallest number which on being added 5 to it, is exactly divisible
by 9, 21, 25 and 45.​

Answer 1

the the number is 4725 which is exactly divisible by all these numbers.

Answer 2

Answer:

The smallest number which on being added 5 is exactly divisible by the given numbers is 1570.

Step-by-step explanation:

Given to find the smallest number when added 5, is exactly divisible

by 9, 21, 25 and 45.

The smallest number that is exactly divisible by the given numbers must be the least common multiple(LCM) of the given numbers.

The prime factorization of 9:  $3\times 3$ or $3^2$.

The prime factorization of 21:  $3\times 7$ or $3^1\times7^1$.

The prime factorization of 25:  $5\times 5$ or $5^2$.

The prime factorization of 45:  $3\times 3\times5$ or $3^2\times 5^1$.

Thus the LCM of 9, 21, 25, 45 is

$3^2\times 7\times5^2=1575$

Given the number being added 5, then it is divisible by the given numbers.

So, $1575-5=1570$

Therefore, 1570 is the smallest number which on being added 5 is exactly divisible by the given numbers.

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