Divisibility of Nine-Digit NumbersDate: 03/08/98 at 19:50:10 From: Jacob Hall Subject: math help Can you find a nine-digit number such that the first digit is divisible by one, the first two digits are divisible by 2, the first three digits are divisible by 3,..., the whole nine-digit number is divisible by nine? I have tried guess and check and have not gotten very far. Thanks for your time! Date: 03/09/98 at 17:02:02 From: Doctor Sam Subject: Re: math help Jacob, There are many possible answers. 123,252,568 is one of them. In order to come up with another, you need to be familiar with several divisibility rules for numbers. By 2: the last digit is divisible by 2. By 4: the last two digits are divisible by 4 (e.g., 24). By 8: the last three digits are divisible by 8 (e.g., 144). By 3: all the digits of the number add up to a multiple of 3. By 9: all the digits add up to a multiple of 9. By 6: the number is divisible by 2 and by 3. By 5: the number ends in 0 or 5. Okay. To solve your problem, start with ANY number (since any number is divisible by 1). This time, I'll start with a 7. First two numbers are divisible by 2. Pick either a 0,2,4,6 or 8 for the next digit. I'll pick 8. So far: 78. First three numbers are divisible by 3. The sum of these three digits should be a multiple of 3. Since 7+8 = 15, which is already a multiple of 3, I can pick 0 or 3 or 6 or 9. I'll pick 0. So far: 780. First four numbers are divisible by 4. The last two digits (I'll denote the new last digit with a ?), 0?, must be a multiple of 4. I can pick 0 or 4 or 8. I'll pick 4. So far: 7804. First five numbers are divisible by 5. The next digit must be a zero or a five. I'll pick 5: 78045. First six numbers are divisible by 6. The number must be even, so I can only pick a 0,2,4,6,8, and the six digits must add up to a multiple of 3: 7+8+0+4+5+? = 24+? ... I can use 0 or 6, since 24 and 30 are multiples of 3. I'll pick 6. So far: 780456. First 7 numbers are divisible by 7. I just divided 780456 by 7 to see what the remainder is. 780456/7 = 111493 with a remainder of 5. When I add the next digit (again, I'll call it ?) on the end and divide again, I'll get 5? for the last division. Since 7 goes into 56 exactly, I can pick 6. So far our number is 7804566. First 8 numbers are divisible by 8. The last three digits, 66?, must be divisible by 8. I can only pick 4. (Just start dividing 8 into 66? and you will see why.) So far: 78045664. Finally, the entire nine-digit number must be a multiple of 9. So all nine digits must add up to a multiple of 9. 7+8+0+4+5+6+6+4 = 40, so the last digit will make the sum 40+? ... only a 5 will make this a multiple of 9. So another solution is: 780456645. How's that? -Doctor Sam, The Math Forum http://mathforum.org/dr.math/ |
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