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A067734
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Number of ways writing n as a product of decimal digits of some other number which has no digits equal to 1.
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11
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0, 1, 1, 2, 1, 3, 1, 4, 2, 2, 0, 7, 0, 2, 2, 7, 0, 7, 0, 5, 2, 0, 0, 17, 1, 0, 3, 5, 0, 8, 0, 13, 0, 0, 2, 21, 0, 0, 0, 12, 0, 8, 0, 0, 5, 0, 0, 38, 1, 3, 0, 0, 0, 15, 0, 12, 0, 0, 0, 24, 0, 0, 5, 24, 0, 0, 0, 0, 0, 6, 0, 58, 0, 0, 3, 0, 0, 0, 0, 26, 5, 0, 0, 24, 0, 0, 0, 0, 0, 24, 0, 0, 0, 0, 0, 82, 0
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OFFSET
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1,4
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COMMENTS
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For n=36, this was given as an exercise for children of age 14 years.
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LINKS
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FORMULA
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EXAMPLE
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There are 21 other numbers with no digit 1 whose digit product equals 36: 49, 66, 94, 229, 236, 263, 292, 326, 334, 343, 362, 433, 623, 632, 922, 2233, 2323, 2332, 3223, 3232, 3322. If 1-digits were permitted then an infinite number of solutions would exist, e.g., 111114111113111113. If n has a prime divisor larger than 7, i.e., a prime divisor that is two or more digits in length, such as 11, then no solutions exist at all. The largest solution is a (decimal) number created by concatenating not-necessarily-distinct prime factors, such as 36 = 3*2*2*2. [edited by Jon E. Schoenfield, Jun 14 2017]
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MATHEMATICA
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id1[x_] := IntegerDigits[x]; id2[x_] := DeleteCases[id1[x], 1] f[x_] := Apply[Times, IntegerDigits[x]]; k=0; Do[s=f[n]; If[Equal[s, 36]&&!Greater[Length[id1[n]], Length[id2[n]]], k=k+1; Print[{k, n}]], {n, 1, 3400}]
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PROG
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(PARI) { A067734(n) = local(v, r, i2, i3); v=vector(4, i, valuation(n, prime(i))); if(n==1||n!=prod(i=1, 4, prime(i)^v[i]), return(0)); r=0; for(i6=0, min(v[1], v[2]), for(i8=0, (v[1]-i6)\3, for(i4=0, (v[1]-i6-3*i8)\2, i2=v[1]-i6-3*i8-2*i4; for(i9=0, (v[2]-i6)\2, i3=v[2]-i6-2*i9; r += (i2+i3+i4+v[3]+i6+v[4]+i8+i9)! / i2! / i3! / i4! / v[3]! / i6! / v[4]! / i8! / i9! )))); r } \\ Max Alekseyev, Sep 19 2009
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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