The two-digit integers from 19 to 92 are written consecutively to form the large integer N=19202122…909192. If 3^k is the highest power of 3 that is a factor of N, then k=

1992 AMC 12 Problem 17

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17.

The two-digit integers from

19

92

are written consecutively to form the large integer

N=19202122\ldots909192.

3^k

is the highest power of

3

that is a factor of

N,

then

k=

0

1

2

3

more than

3

Use the digit-sum tests for divisibility by

3

and

9

Modulo

9,

the digit sum of the concatenation equals

19+20+\cdots+92

Modulo

9,

a number is congruent to its digit sum, so

\begin{aligned} N&\equiv19+20+\cdots+92\\ &=\frac{74(19+92)}2\\ &=4107\equiv3\pmod9. \end{aligned}

Thus

N

is divisible by

3

but not by

9,

so the highest power of

3

dividing it is

3^1.

Thus the correct answer is B.