A computer is programmed to multiply consecutive even integers 2*4*6*8*…*n until the product is divisible by 2013, What is the value of n?
Choices:
A) 22
B) 38
C) 62
D) 122
E) 672
Same logic but with application of factorials.   2*4*6*8*...*n = 2(1*2*3*4*...n/2) = 2*(n/2)!   Now we have: 2*(n/2)!/2013=2*(n/2)!/(3×11×61)   For the expression 2*(n/2)! To be divisible by 2013 min value of n/2 should be 61.   So n/2=61 =&gt; n=122   Answer 122.
<result>D</result>
A computer is programmed to multiply consecutive even integers 2*4*6*8*…*n until the product is divisible by 2013, What is the value of n?