1998 AHSME Problems/Problem 30
Problem
For each positive integer , let
Let denote the smallest positive integer for which the rightmost nonzero digit of is odd. The rightmost nonzero digit of is
Solution
We have .
The value can be written as , where is not divisible by 2 and 5. The number of trailing zeroes is . The last non-zero digit is the last digit of .
Clearly, the last non-zero digit is even iff iff .
Thus we are looking for the smallest such that the power of that divides is at least equal to the power of that divides .
The number is a product of consecutive integers. Out of these, are divisible by . Out of those , at least are divisible by , and out of those , one is divisible by . Therefore for all .
On the other hand, exactly of our ten integers are divisible by , and at most one of them can be divisible by a higher power of . As we need , one of the integers from to must be divisible by . Therefore .
We can now take numbers starting with , and write each of them in the form . We are looking for 10 consecutive rows where the sum of s is at least equal to the sum of s.
number x y z 78116 2 0 19529 78117 0 0 78117 78118 1 0 39059 78119: 0 0 78119 78120: 3 1 1953 78121: 0 0 78121 78122: 1 0 39061 78123: 0 0 78123 78124: 2 0 19531 78125: 0 7 1 78126: 1 0 39063
At this point we can stop, as we just found out that is of the form . Therefore the we seek is .
Now all we need to do is to compute the last non-zero digit. As the powers of and that divide are equal, the last non-zero digit is simply the product of the last digits of the ten s. This is .
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
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