This is a very hard for this age group, but in fact all you need is to realize a series of relationships between numbers.
To fix ideas, we try to find some values \u200b\u200bfor the function in case we are asked (r = 5 s = 8), to predict what value F (2010), as requested in paragraph (1). Let's try
F (1). As F (1) = 1 + 5 or F (1) = 1 to 8, since it is a positive integer, the latter possibility has been discarded then F (1) = 6. Likewise, F (2) = 7, F (3) = 8 and so on, until we get F (9), in which we can choose between 9 + 5 and 9 to 8, because both are positive . However, if you think about it, we find that there is no alternative to choose from 9 to 8 = 1, since all numbers should be F (n) for some n, and 1 must be, therefore, of the form n - 8 or of the form n + 5. Of course, that 1 can not be of the form n + 5, so it must be F (9). Thus 1 = F (9), 2 = F (10), and so 5 = F (13).
words, the first 13 values \u200b\u200bbecome, successively, 6, 7, 8, 9, 10, 11, 12, 13, 1, 2, 3, 4 and 5. Now something similar happens again: F (14) can not be 14 - 8 = 6 because 6 is "used" (that is F (1)), then F (14) = 14 + 5 = 19. In summary, the following values \u200b\u200bare 19, 20, 21, 22, 23, 24. 25, 26, and then return to use the expression n - 8, because if we do not, the numbers 14 to 18 shall be without being the image of any number. So we continue with 14, 15, 16, 17, 18.
guess which clearly shows what the pattern is not it? Actually use the set of positive integers of 13 in 13 using the formula n + 5 in the first 8 and the formula n - 8 and the other five. Let's see what happens with the 2010. The
2010, if we try to divide by 13, gives 154, and a remainder of 8, ie 2010 is the eighth in the group that begins after 2002 = 154 * 13. After you apply the formula n + 5 and hence F (2010) = 2015.
Let's try now to answer the second part. In general, if r and s are not worth 5 and 8, if no other quantities to calculate F system is similar. Applies to sets of consecutive numbers r + s, so that the s first formula is applied to n + r, r and the following formula is applied n - s. It is very easy to extend the reasoning that we used before the general case.
To understand the second part, also applied to the case r = 5, s = 8, and then try to apply it in general. Since each set of 13 consecutive numbers become themselves the same way, simply look at the 13 prime numbers, because if you meet them, will be true for any set of numbers. Apply
F twice done, for example, that F (F (1)) = F (6) = 11. Obviously not enough. Applied 3 times, makes F (F (F (1))) = F (11) = 3. If we apply it four times, we will have 4 F (1) = F (3) = 8. If we apply again F 5 (1) = F (8) = 13. Thus, we will successively by 5, 10, 2, 7, 12, 4, 9, and finally arrive 1 in F 13 (1) = F (9) = 1. With a little work, we find that 13 is the lowest of times you have to apply this function to reach 13 F (n) = n.
Will the sum of r + s the answer? No, but before reaching it, see what happens when we apply repeatedly F. In fact, add subtract r s repeatedly, trying not to get out of the set of 1 to r + s, so we always results as ar - bs, where a and b are positive numbers. Combinations of integers are well known to fans of math problems, since you can always be achieved by choosing correctly and b, giving lower value as a result equal to the greatest common divisor of r and s.
That is, the number nearest to 1 on the pass will be 1 + d, where d is the greatest common divisor (in the case of 5, 8, passed by 2, since 1 = 5 * 5 - 3 * 8). Of course, we also 1 + 2d (3 in our case) and all terms of an arithmetic progression of difference d, ie, d hopping tours all numbers r + s. In the present case, since 5 and 8 have a GCD of 1, we have to give 13 jumps back to the original. Moreover, as we can not agree to the same value in two different dives, we will give exactly 13 jumps starting from any number.
Now if rys have another MCD, then it is impossible to achieve with subtraction of sums of r s values \u200b\u200bless than d = GCD (r, s), so just go (r + s) / d values, ie that the answer is actually the second paragraph is worth k (r + s) / MCD (r, s). For example, if r = 4 s = 6, k stands 5 (path 1 would be: 5, 9, 3, 7, 1).
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