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Divisibility of Nine-Digit Numbers


Date: 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/   
    
Associated Topics:
Elementary Division
Middle School Division

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