"Reflection of an Integer"by cepheid Tags: None 

#1
Sep107, 12:38 AM

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I haven't encountered this before. I'm not sure how to approach it. At this point it's not even clear to me why the result should only be divisible by one number in *every* case.
The reflection of a positive integer is obtained by reversing its digits. For example, the reflection of 321 would be 123. The difference between a 5 digit integer and its reflection must be divisible by which of the following? A. 2 B. 4 C. 5 D. 6 E. 9 



#2
Sep107, 01:10 AM

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The number abcde is really [tex]10^{4}a + 10^{3}b + 10^{2}c + 10^{1}d + 10^{0}e [/tex]. You know what to do...




#3
Sep107, 02:17 AM

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so the difference becomes...
10^4 (ae) + 10^3 (bd) + 10^2 (cc) + 10 (db) + (ea) = (10^4  1)(ae) + (10^3  10)(bd) = 9999(ae) + 990(bd) I guess that means the answer is divisible by 9. Did I do this right? 



#4
Sep107, 02:28 AM

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"Reflection of an Integer"
Yes.




#5
Sep107, 03:34 PM

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That's it. Of course, with multiple choice looking for counterexamples might even be faster... heh.




#6
Sep207, 04:20 AM

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Hi Cepheid,
I find such problems interesting. Where did you find it? Do you have a website for it? 



#7
Sep207, 05:32 PM

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Edgardo,
This is a problem from a practice GRE exam. These exams are administered by ETS. However, I think that such problems are also common in math contests that are intended for junior high/high school students. Try a search for math contests on the net. 



#8
Sep207, 07:26 PM

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#9
Dec2008, 01:51 AM

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#10
Dec2008, 06:34 AM

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And note that this is true not only for 5 digit numbers, it's true for all natural numbers.
For any natural number the digit sum and the original number are equivalent modulo nine. Since the number and it's reflection both have the same digit sum it follows that there are equal modulo 9, hence their differnece must be a multiple of nine. 



#11
Dec2108, 12:52 AM

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#12
Dec2108, 12:59 AM

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Sum over all digits{ [(some power of ten)  1]digit } Therefore the difference between the actual number and its digit sum will always be a multiple of nine, hence they are equivalent modulo nine. Is there a simpler/more obvious way of explaining your statement, though? 



#13
Dec2108, 02:49 AM

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Any power of ten is congruent to 1 (mod 9).




#14
Dec2108, 07:27 AM

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[tex]10^n = 9*(\underbrace{111.....1}_{\mbox{n ones}}) + 1 = 9 k_n +1[/tex]. So the quantity represented by the nth digit (in a decimal number) is [itex]10^n d_n = 9k_n d_n + d_n[/itex]. That is, for each digit the quantity represented by that digit is congruent to the digit itself (mod 9). Cepheid, this property is used in a wide variety of "number tricks" and was used for centuries before computers and calculators as a quick easy "checksum" for testing the integrety of long hand calculations. See casting out nines : http://en.wikipedia.org/wiki/Casting_out_nines 


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