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A062400
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Smallest multiple of n with property that digits are even and each digit is two more (mod 10) than the previous digit; or 0 if no such multiple exists.
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1
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2, 2, 6, 4, 80, 6, 46802, 8, 468, 80, 68024, 24, 468, 46802, 4680, 80, 68, 468, 6802, 80, 80246802468, 68024, 46, 24, 0, 468, 680246802, 80246802468, 680246802468, 4680, 24680246802468, 0, 680246802468024, 68, 802468024680, 468, 24680246802468, 6802
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OFFSET
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1,1
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COMMENTS
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25 is impossible; its multiples end either with the digits 00 or 50.
Multiples of 16 except 16 and 80 are impossible. Of the 625 multiples of 16 mod 10000, none are 246, 2468, 4680, 6802, or 8024. That leaves only 80 as a possible value for multiples of 16. It appears that the multiples of 16 and 25 are the only numbers for which a(n) = 0 - Franklin T. Adams-Watters, Nov 03 2009
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LINKS
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EXAMPLE
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a(7) = 7*6686 = 46802 and this number has increasing larger even digits (mod 10). a(12) = 24 = 12*2 has increasing even digits.
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MATHEMATICA
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f[n_] := Block[{x = 0, a = IntegerDigits[n], i = 1}, l =Length[a]; While < l, If[ Mod[ a[[i]] + 2, 10] != a + 1]], x = 1]; i++ ]; Return[x]]; Dock = n; While[ Union[ even[ IntegerDigits[k]]] != {True} || Fmk] == 1, k += n]; Print[k], {n, 1, 20}]
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PROG
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(PARI) evenincr(n)=local(d, r); d=n%4*2+2; n\=4; r=0; for(k=0, n, r=r*10+(d+2*k)%10); r
a(n)=if(n%25==0 || (n%16==0&80%n!=0), 0, k=0; while(evenincr(k)%n!=0, k++); evenincr(k)) /* This program will loop if the conjecture above is incorrect. */
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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