Browse · MATH Print → jmc prealgebra intermediate Problem Simplify 9999111⋅33. Solution — click to reveal Note that 111 and 9999 have a common factor of 3. Also, 33 and 3333 have a common factor of 33. We get \cancelto33339999\cancelto37111⋅33=\cancelto101333337⋅\cancelto133=10137. Final answer \dfrac{37}{101} ← Previous problem Next problem →