If both 72 and 22 are the factors of the number (a x 43 x 55 x 172), then what is the smallest possible value of a?radio_button_unchecked49radio_button_unchecked343radio_button_unchecked64radio_button_unchecked25
Question
If both 72 and 22 are the factors of the number (a x 43 x 55 x 172), then what is the smallest possible value of a?radio_button_unchecked49radio_button_unchecked343radio_button_unchecked64radio_button_unchecked25
Solution
To find the smallest possible value of 'a', we need to ensure that 'a' contains all the prime factors of 72 and 22 that are not already in 43, 55, and 172.
First, let's find the prime factors of 72, 22, 43, 55, and 172.
- Prime factors of 72 are 2, 2, 2, 3, 3
- Prime factors of 22 are 2, 11
- Prime factors of 43 are 43
- Prime factors of 55 are 5, 11
- Prime factors of 172 are 2, 2, 43
From the above, we can see that 43, 55, and 172 already contain the prime factors 2, 11, and 43. However, the prime factor 3 from 72 is not present. Also, the number of 2's in 72 is three while in 43, 55, and 172 combined is only two. So, we need one more 2 in 'a'.
Therefore, the smallest possible value of 'a' should have the prime factors 2 and 3, 3. When multiplied together, these give 'a' = 2 * 3 * 3 = 18.
However, the options provided do not include 18. There might be a mistake in the question or the options provided.
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