Browse · MATH Print → jmc prealgebra intermediate Problem A number x factors as 107⋅1095, and 107 and 109 are primes. What is the exponent of 109 in the prime factorization of x11? Solution — click to reveal We have x11=(107⋅1095)11=10711(1095)11=1071110955, so our answer is 55. Final answer 55 ← Previous problem Next problem →