What is the Divisibility Rule for the Reflection of an Integer?

[cepheid]

"Reflection of an Integer"

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

 

[Werg22]

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

 

[cepheid]

so the difference becomes...

10^4 (a-e) + 10^3 (b-d) + 10^2 (c-c) + 10 (d-b) + (e-a)

= (10^4 - 1)(a-e) + (10^3 - 10)(b-d)

= 9999(a-e) + 990(b-d)

I guess that means the answer is divisible by 9. Did I do this right?

 

[Werg22]

Yes.

 

[CRGreathouse]

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

 

[Edgardo]

Hi Cepheid,

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

 

[cepheid]

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.

 

[ice109]

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.

this is in the math GRE?

 

[pwrstick]

this is in the math GRE?

Yes it is, I got it in the practice section right now as the third question, and if you don't know your first 10 questions are the most important (28 total, 45 mins to complete). Suffice to say this test is going to kick my *** at 12PM tomorrow... but hey, I'm a psych major.

 

[uart]

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.

 

[cepheid]

this is in the math GRE?

Yes it is

Somehow I doubt that such questions are asked in the GRE math *subject test*. This problem was from the math (quantitative reasoning) section of a GRE *general* test (which, I think, is what pwrstick was talking about).

 

[cepheid]

For any natural number the digit sum and the original number are equivalent modulo nine.

How do you know this? I mean, I convinced myself of it by figuring that the difference between the number and the sum of the digits can be expressed as:

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?

 

[adriank]

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

 

[uart]

How do you know this?

Just elaborating on what Adrian said.

10^n = 9*(\underbrace{111...1}_{\mbox{n ones}}) + 1 = 9 k_n +1.

So the quantity represented by the nth digit (in a decimal number) is 10^n d_n = 9k_n d_n + d_n. 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"

 

[cepheid]

Hey uart,

Thanks for the link and for adding something useful to a dredged up thread from a year ago.