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Its prime factorisation is:
2 x 5 x 5 x 7 = 350 which can be written in index form (in other words, using exponents) as 2 x 52 x 7.
In addition it has composite factors, which are found by multiplying combinations of these prime factors, and so we obtain this list:
The 12 factors of 350 are 1, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175, and 350
- If you write 350 = 2 x 52 x 7 = 21 x 52 x 71 then 350 will have (1+1)(2+1)(1+1) = 12 factors, because any factor of 350 consists of a product of (0 or 1) 2's, (0, 1, or 2) 5's and (0 or 1) 7's. So these are all of 12 of them.
- You can write (1+2)(1+5+25)(1+7) and expand this out to see all the factors (added), or evaluate it to get the sum of the factors of 350 is 3 x 31 x 8 = 744.
The factor pairs of 350 are 1 x 350, 2 x 175, 5 x 70, 7 x 50, 10 x 35, and 14 x 25.
The proper factors of 350 are 1, 2, 5, 7, 10, 14, 25, 35, 50, 70, and 175, or
if the definition you are using excludes 1, they are 2, 5, 7, 10, 14, 25, 35, 50, 70, and 175.
The 3 distinct prime factors of 350 are 2, 5, and 7.
1, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175, 350
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What is the prime factorization tree of the prime factorization of 350?
350=2x5x5x7
Prime factorization of 350?
7x5x5x2=350
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