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Re: Can anyone answer this problem?
Posted:
Apr 10, 2002 1:27 AM
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Arturo Magidin wrote:
>
> In article <3CB37F24.36269B9@kidzmail.com.au>,
> Peter Johnson <goodlooking@kidzmail.com.au> wrote:
> >Hi,
> >
> >A question we have been asked in our maths class is: find an integer of
> >any size, such that, when then last digit becomes the first digit, the
> >result is double the original number.
> >
> >That is, if the initial number is:
> >
> >ABCDEF
> >
> >then find a number that is TWICE this number, and has the form
> >
> >FABCDE
> >
> >(ABCDEF * 2 = FABCDE)
> >
> >If anyone can give the initial number that would fit this problem, I
> >would be very greatful.
>
> You are trying to solve the following problem:
>
> Let x be a number with n digits. Then 2(10x+a) = 10^n*a + x, with
> 0<=a<=9.
>
> so 20x+2a = 10^n*a + x
> 19x = (10^n-2)a
>
> From this you should be able to deduce a number of things. For
> example, a must divide 19x, but cannot divide 19, so it must divide
> x. 19 must divide 10^n-2, which should give you possible values of n.
"A number of things" including the answer, which follows immediately
once you find n ( 17 works ) and evaluate s = (10^n-2)/19. Then
values of a = 2 thru 9 give solutions. ( a=1 requires 10*x+a to be
expressed with a leading 0 )
Lew Mammel, Jr.
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