128. The integers 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth number is marked (that is, 1, 16, 31, etc.).

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1 28. The integers to 000 are written in order around a circle. Starting at, every fifteenth number is marked (that is,, 6, 3, etc.). This process is continued until a number is reached which has already been marked. How many unmarked numbers remain? Solution: One does not have to actually mark every number down to solve the problem. If one starts marking every th integer, one starts with, 6, 3, 6,..., one may note these are each more than a multiple of. That is, they are 0 +, +, 2 +, 3 +,.... The largest multiple of less than 000 is 66 = 990, so the last integer one would mark before going around the circle again would be 66 + = 99. At that point, one would have marked 67 numbers. The next group of numbers marked would be 6, 2, 36,,.... One sees these are each more than the corresponding numbers in the first group and may be written as 0 + 6, + 6, 2 + 6, 3 + 6,..., This last number in this group of 67 numbers is 66 + = 996. We have now marked another 67 numbers, for a total of 3 numbers. The next group of numbers would be, 26, 36,,.... These are once again each more than the corresponding numbers in the second group and may be written as 0 +, +, 2 +, 3 +,..., 6 +. The last number in this group of 66 (not 67 this time) numbers is 6 + = 986. We have now marked another 66 numbers, for a total of 200 numbers. The next number would be again, so that we d just be repeating the numbers already marked. Since we ve marked 200 numbers out of the 000 integers from to 000, there will be 800 numbers unmarked. Some students noted that all the numbers marked end in and 6. Here s an extra credit opportunity: Prove, without manually marking all the appropriate integers, the conjecture that all such integers and only such integers will be marked. Note that if one can prove that (it is true), it follows that 20% of the integers between and 000 are marked and 80% unmarked, so that 800 will be unmarked.

2 29. If the integers from to 222, 222, 222 are written down in succession, how many Os are written? Solution: This problem is an excellent example of how one can benefit from intelligent organization. We may break the set of integers into various intervals, based mainly on the number of digits, starting with the single digit integers, which contain 0 0s. The two digit numbers are the integers between 0 and 99, inclusively. For these, the 0 0 (unit) digit is 0 a tenth of the time, while the 0 (tens) digit is never 0. Since there are 90 such integers, the number of 0 digits is 0 90 = 0 (02 0 ) = (0 0 0 ). The three digit numbers are the integers between 00 and 999, inclusively. For these, the 0 0 (unit) digit is 0 a tenth of the time and the 0 (tens) digit is 0 a tenth of the time, while the 0 2 (hundreds) digit is never 0. Since there are 900 such integers, the number of 0 digits is = 2 0 ( ) = 2 (0 2 0 ). Similarly, the four digit numbers are the integers between 000 and 9999, inclusively. For these, the 0 0 (unit) digit is 0 a tenth of the time, the 0 (tens) digit is 0 a tenth of the time and the 0 2 (hundreds) digit is 0 a tenth of the time, while the 0 3 (thousands) digit is never 0. Since there are 9000 such integers, the number of 0 digits is = 3 0 (0 0 3 ) = 3 ( ). The same reasoning can be continued, obtaining the following results: Interval Number of 0s 0 99 (0 0 0 ) (0 2 0 ), 000 9, ( ) 0, , 999 (0 0 3 ) 00, , 999 (0 0 ), 000, 000 9, 999, (0 6 0 ) 0, 000, , 999, ( ) One must now start counting a little more carefully. For example, between 200,000,000 and 29,999,999, the rightmost seven digits are each 0 a tenth of the time, but the eighth digit from the right ( (the 0) 7 digit) is 0 half ( the time. ) Thus, the number of zeros in that interval is = = We can use similar reasoning in other intervals to obtain the following results.

3 Interval Number of 0s 200, 000, , 999, , 000, , 999, , 000, , 99, , 200, , 29, , 220, , 22, , 222, , 222, , 222, , 222, Finally, 222,222,220 contains a single 0. If we write all these numbers of zeros in a column as follows we note a lot of cancellation and the total can be calculated fairly easily: Looking at the terms and all the cancellation, we can see that the total number of zeros is = 87, 6, 32.

4 30. (Dividing Coconuts) Five sailors and a monkey are captured on a deserted island, where the sailors gather a pile of coconuts. They decide to share them equally the next morning. During the night one of the sailors gets up, divides the pile in five parts with one nut left over, which he gives to the monkey. He hides one part, puts the other nuts back in a pile and gets back to sleep. During the course of the night, each of the other sailors does the same thing in turn, each time giving one nut to the monkey. The next morning they get up and divide the remaining nuts in five equal portions, once again giving one nut to the monkey. What is the smallest possible number of nuts they could have started with? Solution: One could, of course, do this by trial and error but it would take an awfully long time. The use of a little algebra can save some time. Suppose we let x k represent the size of each of the five piles the k th sailor divided the coconuts into before giving one to the monkey, hiding one of the piles and then putting the rest back into a single pile. If we let x represent the original number of coconuts and let x 6 be the size of each of the five portions they wind up dividing among themselves in the morning, we have the following equations: x = x + x = x 2 + x 2 = x 3 + x 3 = x + x = x + x = x 6 + For each x k, k =, 2,...,, we have x k = xk+ +, so we get x = x + = x 2+ ( ) 2 ( ) 2 + = x ( ) ( ) ( ) + =... x = ( ) ( ) 6 ( ) ( ) ( ) x 6 + = x 6 + = (x 6 + ). We can write this as x = 6 x6 +. Obviously, x6 + must be some positive integer, so we must have x = 6 n for some positive integer n. The smallest positive integer is, so x can certainly be no smaller than 6 =, 62. If we check, 62, it works as follows. The first sailor divides,62 coconuts into five piles of 32 with one left over, which he gives to the monkey. He hides one pile of 32 coconuts, leaving 2,96 coconuts. The second sailor divides 2,96 coconuts into five piles of 2,99 with one left over,

5 which he gives to the monkey. He hides one pile of 2,99 coconuts, leaving 9,996 coconuts. The third sailor divides 9,996 coconuts into five piles of,999 with one left over, which he gives to the monkey. He hides one pile of,999 coconuts, leaving 7,996 coconuts. The fourth sailor divides 7,996 coconuts into five piles of,99 with one left over, which he gives to the monkey. He hides one pile of,99 coconuts, leaving 6,396 coconuts. The fifth sailor divides 6,396 coconuts into five piles of,279 with one left over, which he gives to the monkey. He hides one pile of,279 coconuts, leaving,6 coconuts. In the morning, the sailors divide the,6 coconuts into five piles of,023 with one left over, which they give to the monkey. We thus confirm that the smallest number of coconuts they could have started with is,62.