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Valentin [98]
2 years ago
5

Find K such that the number 23K10 is divisible by 2, 3, and 5 2021

Mathematics
1 answer:
blsea [12.9K]2 years ago
5 0

The values of K that will make 23K10 to be divisible by 2, 3 and 5 are; 0, 3, 6 and 9

<h3>How to use divisibility rules?</h3>

We want to find the number K in 23K10 that will make it divisible by 2, 3 and 5.

Now, for the number to be divisible by 2, 3 and 5, then it means it must be divisible by the LCM which is; 2 * 3 * 5 = 30

Now, the divisibility rule for 30 is that it must be divisible by 3 and 10.

If 23K10 is divisible by 3, sum of digits will be multiple of 3.

Thus;

2 + 3 + K + 1 + 0 = K + 6 must be equal to 0, 3, 6, 9, 12, 15, 18....

Thus, K could either be 0, 3, 6 or 9

Read more about Divisibility Rules at; brainly.com/question/9462805

#SPJ1

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