Problem: Define naβ! for n and a positive to be
naβ!=n(nβa)(nβ2a)(nβ3a)β¦(nβka)
where k is the greatest integer for which n>ka. Then the quotient 7228β!/182β! is equal to
Answer Choices:
A. 45
B. 46
C. 48
D. 49
E. 412
Solution:
729β1=72Γ84Γ56Γ48Γ40Γ32Γ24Γ16Γ8=88Γ91
18z=18Γ16Γ14Γ12Γ10Γ8Γ6Γ4Γ2=2βΓ91
The quotsent 722β1/182β1=89/29=49