Extraordinary Social Security NumberDate: 12/05/2001 at 21:13:02 From: Elle Subject: Combinations Dear Sir or Ma'am, My name is Elle Woods, and I am a 20 year old college student. I have been given a math problem that I feel is possible to solve, but may take more time than I have been given. Could you help me with my math problem? Problem: Dee finds that she has an extraordinary social security number. Its nine digits contain all the digits from 1 to 9. They also form a number with the following characteristics: a. When read from left to right the first two digits form a number divisible by two. b. The first three digits form a number divisible by three. c. The first four digits a number divisible by four, and so on, until the complete number is divisible by 9. What is Dee's social security number? If I use the method I am currently using, I will not have enough time to solve the problem. Could you please help me? Date: 12/06/2001 at 16:02:31 From: Doctor Greenie Subject: Re: Combinations Hello, Elle - This page in the Dr. Math archives shows an answer where this problem is solved: Divisibility Word Problem http://mathforum.org/library/drmath/view/56742.html Actually, the problem on that page uses digits 0-9 to form a 10-digit number; the only difference from your problem is that the 10-digit number in the problem in the archives has final digit "0" (because the 10-digit number must be divisible by 10); the first 9 digits of the 10-digit number are the same as the 9-digit number in your problem. I remembered seeing this problem in the archives a couple of years ago; but I thought I would also have a go at it myself. Following is my solution; the result is of course the same, but the path to the solution, while it has many similarities to the solution in the referenced problem, is different. Let's start by using divisibility rules to list the restrictions imposed by the conditions of the problem.... The number formed by the first digit is divisible by 1 (there are no restrictions imposed by this condition since every 1-digit number is divisible by 1) The number formed by the first two digits is divisible by 2 (1) digit 2 must be even The number formed by the first three digits must be divisible by 3 (2) the sum of digits 1-3 must be divisible by 3 The number formed by the first four digits must be divisible by 4 (3) digit 4 must be even (4) the number formed by digits 3-4 must be divisible by 4 The number formed by the first five digits must be divisible by 5 (5) digit 5 must be 5 The number formed by the first six digits must be divisible by 6 (6) digit 6 must be even (7) the sum of digits 1-6 must be divisible by 3 The number formed by the first seven digits must be divisible by 7 (rules for divisibility by 7 are clumsy - let's not try to impose any restrictions on the digits here...) The number formed by the first eight digits must be divisible by 8 (8) digit 8 must be even (9) the number formed by digits 6-8 must be divisible by 8 The number formed by the first nine digits must be divisible by 9 (10) the sum of digits 1-9 must be divisible by 3 Now let's start combining these restrictions logically to try to determine some of the digits. From (1), (3), (6), and (8) above, digits 2, 4, 6, and 8 are even; so (11) digits 1, 3, 7, and 9 are odd (we already know digit 5 is odd) From (2), (7), and (10) above, the sum of digits 1-3, the sum of digits 1-6, and the sum of digits 1-9 are all divisible by 3; from this, we can conclude that (12) the sum of digits 4-6 is divisible by 3 (13) the sum of digits 7-9 is divisible by 3 From (1) and (11) above, digits 1-3 are odd, even, and odd; their sum is therefore even. From (2), this sum is divisible by 3. Therefore (14) the sum of digits 1-3 is an even multiple of 3 (6, 12, 18, or 24) From (3), (5), and (6) above, digits 4-6 are even, odd, and even; their sum is therefore odd. From (12), this sum is divisible by 3. Therefore (15) the sum of digits 4-6 is an odd multiple of 3 (9, 15, or 21) From (8) and (11) above, digits 7-9 are odd, even, and odd; their sum is therefore even. From (13), this sum is divisible by 3. Therefore (16) the sum of digits 7-9 is an even multiple of 3 (6, 12, 18, or 24) From (15), the sum of digits 4-6 is either 9, 15, or 21; and from (5) we know digit 5 is 5. From (3) and (6), digits 4 and 6 are different even numbers. These facts together give us only a small number of possible combinations for digits 4-6: (17) the only possibilities for digits 4-6 are 258, 852, 456, and 654 From (4) and (11), digits 3-4 form a 2-digit number with first digit odd that is a multiple of 4. With digit 3 odd, we then know (18) the only possibilities for digit 4 are 2 and 6 Combining (17) and (18), we now have (19) the only possibilities for digits 4-6 are 258 and 654 From (8), (9), and (11), digits 6-8 form a 3-digit number, divisible by 8, with middle digit odd. We can combine this with (19) to determine (20) if digits 4-6 are 258, then digit 8 must be 6, making digit 2 equal to 4 (21) if digits 4-6 are 654, then digit 8 must be 2, making digit 2 equal to 8 From (1), (11), and (14), digits 1-3 are odd, even, and odd, and the sum of those digits is an even multiple of 3. Here is a complete list of the possibilities for these digits: (sum = 6) 123 321 (sum = 12) 129 921 327 723 147 741 183 381 (sum = 18) 729 927 369 963 189 981 387 783 (sum = 24) 789 987 From (20), we know that if digits 4-6 are 258, then digit 8 must be 6, making digit 2 equal to 4. In the list of possible combinations for digits 1-3, there are only two combinations with digit 2 equal to 4 (147 and 741); so we now know that (22) if digits 4-6 are 258, then the possible solutions to the problem are 147 258 369 or 147 258 963 741 258 369 or 741 258 963 With these possible solutions, digits 6-8 are either 836 or 896; condition (9) eliminates the two possible solutions with digits 6-8 equal to 836. Then, the requirement that the number formed from digits 1-7 be divisible by 7 eliminates the two possible solutions with digits 6-8 equal to 896. So there are no solutions with digits 4-6 equal to 258. From (21), we know that if digits 4-6 are 654, then digit 8 must be 2, making digit 2 equal to 8. In the list of possible combinations for digits 1-3, there are eight combinations with digit 2 equal to 8; this tells us that (23) if digits 4-6 are 654, then the possible solutions to the problem are 183 654 729 or 183 654 927 381 654 729 or 381 654 927 189 654 327 or 189 654 723 981 654 327 or 981 654 723 387 654 129 or 387 654 921 783 654 129 or 783 654 921 789 654 123 or 789 654 321 987 654 123 or 987 654 321 With these possible solutions, digits 6-8 are either 412, 432, 472, or 492; condition (9) eliminates the possible solutions with digits 6-8 equal to either 412 or 492. The remaining possible solutions are 183 654 729 381 654 729 189 654 327 981 654 327 189 654 723 981 654 723 789 654 321 897 654 321 Only one of these satisfies the condition that the number formed by digits 1-7 be divisible by 7. The single solution to the problem is 381654729 This is a great problem. I hope my response and the referenced problem in the Dr. Math archives have helped give you an idea of possible ways to go about solving problems like this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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