Problem:
The two-digit integers from 19 to 92 are written consecutively to form the large integer
N=19202122…909192.
If 3k is the highest power of 3 that is a factor of N, then k=
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. more than 3
Solution:
Since 0+1+2+⋯+9=45 and
N=1910⋅2+452021⋯2910⋅3+453031⋯39⋯10⋅8+458081⋯89909192
the sum of the digits of N is
S=(1+9)+(10⋅2+45)+(10⋅3+45)+⋯+(10⋅8+45)+(3⋅9+3)=36⋅10+7⋅45+27+3=9(40+35+3)+3
Thus S has a factor of 3 but not 9, so the highest power of 3 which is a divisor of N is 31 and k=1.
Note. One could also compute S=705 and discover that it is divisible by 3 and not by 9.
OR
Note that 3 [or9] will divide N if and only if it divides the sum of 19,20,…,92. (Why?) Since
19+20+⋯+92=74⋅219+92=37⋅111=372⋅3
it follows that k=1.