The Fundamental Theorem of Arithmetic states that every positive integer can factored into primes uniquely, and in only one way (ignoring the order of multiplication). That is to say, for any positive whole number, \( n \), there exists only one possible prime factorization.

For example:

\[ 588 = 2^2 \times 3 \times 7^2 \]

and it is not possible to write 588 as some other product of prime factors (e.g., \( 588 \neq 2^2 \times 11 \times 13 \) ).

## What is the sum of the prime factors of 1190?

Since \( 1190 = 10 \times 119 = 2 \times 5 \times 119 \), the problem reduces to factoring 119. Factoring large integers is a very hard problem, but 119 is small enough that we can find the answer by inspection. \( 119 = 7 \times 17 \), so \( 1190 = 2 \times 5 \times 7 \times 17 \).

\( 2 + 5 + 7 + 17 = 31 \), which is our answer.

## What is the least common multiple of 14, 15, and 20?

It will be helpful to look at these numbers in their factored form.

\[ \begin{align} 14 &= 2 \times 7 \\ 15 &= 3 \times 5 \\ 20 &= 2^2 \times 5 \end{align} \]

Let \( M = 14a = 15b = 20c \) be the smallest number that satisfies this equality for positive integer values of \( a \), \( b \), and \( c \). Since \( \frac{M}{14}=a \), and \( a \) is an integer, \( M \) must contain both 2 and 7 as factors. Reasoning similarly for 15, and 20 makes clear that \( M = 2^2 \times 3 \times 5 \times 7 = 420 = (30)(14) = (28)(15) = (21)(20)\).

Finally, \( M = 420 \) must be the smallest such number because if we remove any of the stated factors, it will no longer be an integer multiple of one of: 14, 15, or 20.

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## Comments

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TopNewesti want an testing exercise

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You might try the Practice section of the site. There's a "prime factorization" skill early in the map.

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