Trick for Numbers Divisible by 3 or 9Date: 02/24/98 at 10:39:33 From: Leon Planken Subject: Division by 3 and by 9 A lot of people know the trick that to see if a number is divisible by 3 or by 9, the figures should be added, and if the result is divisible by 3 or by 9, the original number was too. I'm looking for the proof. Thanks in advance, Leon Planken Date: 02/24/98 at 10:54:07 From: Doctor Sam Subject: Re: Division by 3 and by 9 Hi Leon, The proof depends upon the observation that 9, 99, 999, 9999, ... etc. are all multiples of 3 and 9 and also 1 less than a power of 10. Any number can be written as a sum of powers of tens. I'll illustrate with a four digit number but the proof is easy to extend to an n-digit number. abcd = a(10^3) + b(10^2) + c(10) + d Rewrite each power: = a(999+1) + b(99+1) + c(9+1) + d Rearrange the terms: = a(999) + b(99) + c(9) + [a + b + c + d] The first three terms are divisible by 3 and by 9, so abcd a+b+c+d ---- = A whole number quotient + ------- 9 9 The division will be exact if the last division, the sum of the digits by 9, is exact. The proof actually shows even more. Not only is a number divisible by 3 or by 9 if the sum of its digits is, but the remainder of the division is the same as the remainder of the division of the sum of the digits. Example: 9876543/9 has the same remainder as (9+8+7+6+5+4+3)/9, which is 6. Okay? -Doctor Sam, The Math Forum Check out our web site http://mathforum.org/dr.math/ |
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