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d is a power of 10
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b555
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#1
d is a power of 10
tornament of town 2001 --q5

Let a and d be positie integers. for any positive integer `n`, the number a+nd conntains a block of consecutive digits which constitute the number `n`. Prove that `d` is a power of 10.

PostPosted: Wed Oct 07, 2009 10:20 am  Back to top 
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pco
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#2
Re: d is a power of 10
tornament of town 2001 --q5

b555 wrote:
Let a and d be positie integers. for any positive integer `n`, the number a+nd conntains a block of consecutive digits which constitute the number `n`. Prove that `d` is a power of 10.


Let d=\overline{d_1xx...xxd_20...0} with d_1\neq 0,d_2\neq 0, q trailing zeroes and p+q total digits.
Let a=\overline{d_3xx...xx} with r total digits.

Let k=1000+\max(p+q,r)

Let N=a+10^kd =\overline{d_1xx...xxd_200...00d_3xx...xx}
The left part (\overline{d_1xx...xxd_2}) has p<k digits and so cant contain the representation of 10^k
The right part (\overline{d_3xx...xx}) has r<k digits and so cant contain the representation of 10^k
The middle part contains k+q-r zeroes

In order the middle part contains the representation of 10^k, we need d_2=1 and k+q-r\geq k \iff q\geq r

Let then M=a+(10^k+1)d=N+d =\overline{d_1xx...xxd_200...00xxx...xx}
The left part (\overline{d_1xx...xxd_2}) has p<k digits and so cant contain the representation of 10^k+1
The right part is a+d and so has p+q<k or p+q+1<k digits and so cant contain the representation of 10^k+1
The middle part contains k+q-(p+q)=k-p or k+q-(p+q+1)=k-p-1 zeroes.

In order the middle part contains the representation of 10^k+1, we need k-p\geq k-1 or k-p-1\geq k-1 and so p\leq 1

So p=1 and d_1=d_2=1 and so d=\overline{10...0}=10^q

Q.E.D.
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PostPosted: Thu Oct 08, 2009 11:01 am  Back to top 
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