Q: What is the prime factorization of 1701 using exponents when repeated factors appear?

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441 = 32 x 72

23 x 32 x 11 = 792

22 x 52 x 13 = 1300

2^2 x 5^2 x 13

Suppose you have a prime factor with an exponent of n. Then that prime factor can appear 0, 1, 2, ... , n times in a compound factor: that is, there are (n+1) different exponents that it can take.

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22*72 = 196

2^2 = 4

245 = 51 x 72

441 = 32 x 72

22*52*43 = 4300

75 = 3 x 52

22 x 32 = 36

2 x 72 = 98

23 x 32 x 11 = 792

23 x 32 x 17 = 1224

936 = 23 x 32 x 13

22 x 52 x 13 = 1300