Posted 13 Aug 2014 This is a problem presented in local math olimpyad, N(12,34)=12131415.....31323334 Find a, b, c, d where ab+cd=99 and N(ab,cd) is dividable by abcd (4 digit number). 0 Share this post Link to post Share on other sites

0 Posted 13 Aug 2014 a b c d=4 9 5 0 ab=49, cd=50; 49+50 = 99 N(ab,cd) = N(49,50) = 4950 = abcd. 4950/4950 = 1 1 Share this post Link to post Share on other sites

0 Posted 14 Aug 2014 a b c d=4 9 5 0 ab=49, cd=50; 49+50 = 99 N(ab,cd) = N(49,50) = 4950 = abcd. 4950/4950 = 1 Yes that works . But are there other answers as well? (Should have thought about 49.50 before posting ) 0 Share this post Link to post Share on other sites

0 Posted 14 Aug 2014 I wrote a program to check. 49 + 50 = 99, 4950 | N(49,50) 39 + 60 = 99, 3960 | N(39,60) 24 + 75 = 99, 2475 | N(24,75) 23 + 76 = 99, 2376 | N(23,76) 19 + 80 = 99, 1980 | N(19,80) 15 + 84 = 99, 1584 | N(15,84) 14 + 85 = 99, 1485 | N(14,85) 9 + 90 = 99, 990 | N(9,90) 0 Share this post Link to post Share on other sites

0 Posted 14 Aug 2014 Testing a few targets other than 99, N(n,n+1) is always true for targets of 2n+1. 0 Share this post Link to post Share on other sites

0 Posted 15 Aug 2014 (edited) From 9 to 90.910111213...888990 divided by 990= 910111213...88899 divided by 99It's divisible by 9 if all digits sum to 9. The digits 1-9 sum to 45, and 4+5=9, so the "ones" digits of each set of numbers like 10-19 or 30-39 would be divisible by 9. As for the "tens" digits, there are ten numbers each in the range from 10-19, 20-29, 30-39, ... 80-89, (plus the starting 9 and ending 90 which wouldn't affect divisibility by 9). Pair up the tens digits: 10-19 & 80-89, 20-29 & 70-79 ... 40-49 & 50-59, and it's obvious that the tens digits will also be divisible by 9. So that whole number is divisible by 9.You also need to know that a number is divisible by 11 if and only if taking alternating sums and differences of its digits leads to a number that's divisible by 11. In this case that would be9 - 1 + 0 - 1 + 1 - 1 + 2 - 1 + 3 .... - 8 + 8 - 8 + 9 - 9 + 0It should be obvious that if you consider all the numbers from 10 to 89, there are 10 each of digits 1-8 in the tens spot (which will be added) and 8 each of the digits 1-9 in the ones spot (which will be subtracted), and the leading 9 and trailing 90 will cancel each other out. The digits in the tens spot will sum to 10x(1+2+...+7+8) = 10x(9x4) = 360, and the digits in the ones spot will sum to 8x(1+2+...+8+9) = 8x45 = 360. So it's divisible by 11.Edit: Out of curiosity, what sort of time and resources do the kids get to solve these problems? Edited 15 Aug 2014 by plasmid 0 Share this post Link to post Share on other sites

0 Posted 15 Aug 2014 (edited) From 9 to 90. 910111213...888990 divided by 990 = 910111213...88899 divided by 99 It's divisible by 9 if all digits sum to 9. The digits 1-9 sum to 45, and 4+5=9, so the "ones" digits of each set of numbers like 10-19 or 30-39 would be divisible by 9. As for the "tens" digits, there are ten numbers each in the range from 10-19, 20-29, 30-39, ... 80-89, (plus the starting 9 and ending 90 which wouldn't affect divisibility by 9). Pair up the tens digits: 10-19 & 80-89, 20-29 & 70-79 ... 40-49 & 50-59, and it's obvious that the tens digits will also be divisible by 9. So that whole number is divisible by 9. You also need to know that a number is divisible by 11 if and only if taking alternating sums and differences of its digits leads to a number that's divisible by 11. In this case that would be 9 - 1 + 0 - 1 + 1 - 1 + 2 - 1 + 3 .... - 8 + 8 - 8 + 9 - 9 + 0 It should be obvious that if you consider all the numbers from 10 to 89, there are 10 each of digits 1-8 in the tens spot (which will be added) and 8 each of the digits 1-9 in the ones spot (which will be subtracted), and the leading 9 and trailing 90 will cancel each other out. The digits in the tens spot will sum to 10x(1+2+...+7+8) = 10x(9x4) = 360, and the digits in the ones spot will sum to 8x(1+2+...+8+9) = 8x45 = 360. So it's divisible by 11. Edit: Out of curiosity, what sort of time and resources do the kids get to solve these problems? Usually 45 minutes per problem with pen, paper, ruler, compasses. Rarely they accept calculater and/or formula book. But state olimpyad is 90 minutes per problem, 3 problems, 2 day. I once sat 9 hours straight coz I was challengin in above grade too. Edited 15 Aug 2014 by Barcallica 0 Share this post Link to post Share on other sites

0 Posted 15 Aug 2014 (edited) I am adding this using my mobile and I don't see any spoiler button, so the mod may please put this in a spoiler. Here is a general method to solve for more possibilities Let ab = x Then cd = 99 -x Abcd = 100x + 99 -x = 99 (x+1) Now the condition is that N (x,99-x) should be divisible by 99 (x+1) All N are divisible by 9 and 11 - easy to prove. Now we just need to see which x+1 can divide N. Some numbers I can identify quickly are 1,98 divisible by 2 4,95 divisible by 5 7,92 divisible by 8 9,90 divisible by 10 19,80 divisible by 20 24,75 divisible by 25 39,60 divisible by 40 49,50 divisible by 50 There can be more possibilities. For example I don't remember the divisibility rules for 7,13, 27, 17, 19, 23 etc. So can not check for multiples of 3 or above numbers Edited 17 Aug 2014 by bonanova Added spoiler 0 Share this post Link to post Share on other sites

Posted

This is a problem presented in local math olimpyad,

N(12,34)=12131415.....31323334

Find a, b, c, d where ab+cd=99 and N(ab,cd) is dividable by abcd (4 digit number).

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