Lucky and "Elucky" Numbers in Consecutive NumbersDate: 05/21/98 at 06:23:24 From: Amy Subject: numbers A "lucky number" is any number where the sum of the digits is divisible by 7. For example: 7, 25, 849. Can you show that 13 consecutive numbers always contain at least one lucky number? Date: 05/21/98 at 15:49:06 From: Doctor Schwa Subject: Re: numbers Amy, Great question! Here's a little hint: Any 13 consecutive numbers will contain a multiple of 10. Then it'll either contain at least 6 numbers after it, or at least 7 numbers before it ... And here's a harder related problem for you, about numbers where the sum of the digits is divisible by 11. Maybe to stick with your terminology, we should call them "elucky" numbers. 1) Find 38 consecutive numbers none of which is elucky. Check out the individual round of the 1998 Polya competition at: http://www.gunn.palo-alto.ca.us/teacher/jzucker for a solution to that one. 2) Show that 39 consecutive numbers always contain at least one elucky number. (I bet you can challenge your math teacher with that one!) -Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 05/21/98 at 16:33:43 From: Doctor Rob Subject: Re: numbers Think about consecutive numbers with all the same digits except the units digit. There are 10 of them, which we will call a decade. Now, within each decade, we will show that there can be at most 6 consecutive unlucky numbers. If the sum of all the upper digits of the numbers in a decade is S, then the sum of the digits of the numbers in the decade will be all the integers from S to S + 9. Of these 10 consecutive numbers, at least one will be lucky, and sometimes two. To show this, we divide S - 1 by 7, obtaining quotient Q and remainder R. Then: S - 1 = 7*Q + R, 0 <= R <= 6 and so: S <= S + (6 - R) = 7*(Q + 1) <= S + 6 will be one of these digit sums, and will be divisible by 7, so the number in that decade whose unit digit is 6 - R will be lucky. If R = 4, 5, or 6, then the number whose unit digit is 13 - R will also be lucky and in the same decade. Between 0 and 6 - R, there are fewer than 6 unlucky numbers. If R = 0, 1, 2, or 3, between 6 - R and 10 there are at most 6 unlucky numbers. If R = 4, 5, or 6, between 6 - R and 13 - R there are exactly 6 unlucky numbers, and between 13 - R and 10 there are fewer than 6 unlucky numbers. Thus in each decade there are at most 6 consecutive unlucky numbers. This proves our claim above. When you move from one decade to the next higher one, this pattern -- of the digit sum moving up one unit for each unit the number moves up -- fails. At this point, what happens depends on in how many 9's the last number of the lower decade ends. If it ends in just one 9, the digit sum changes by -9 + 1 = -8, because we are replacing a 9 with a 0, then adding 1 to the tens digit. If it ends in two 9's, the digit sum changes by 2*(-9) + 1 = -17, because we are replacing two 9's with 0's, then adding 1 to the hundreds digit. You can discern the pattern of what happens in general. If there are just k 9's at the end, the digit sum changes by 1 - 9*k. Now, in any two consecutive decades, there can be at most 12 unlucky numbers in a row: 6 at the top of the lower decade, preceded by a lucky number, and 6 at the bottom of the upper decade, succeeded by a lucky number. For this to happen, you have to have k = 3 and R = 3. S = 7*Q + 4, so S - 9 - 9 = 7*(Q-2), and the sum of all the digits other than the last three must be a multiple of 7. The smallest example is when Q = 2 and S = 18. The numbers from 994 to 1005 are 12 consecutive unlucky numbers, surrounded by 993 and 1006, both lucky. If you do not understand all of this, write back and ask a specific question about the part or parts you don't get. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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