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Question-222635

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Question Number 222635 by Mingma last updated on 02/Jul/25
Answered by Raphael254 last updated on 28/Jul/25
n + S(n) + S(S(n)) = 2007 n = 1983 1983 + (1+9+8+3) + S(1+9+8+3) = = 1983 + 21 + S(21) = = 2004 + 3 = 2007 ✓ What is necessary? The number has to be higher than 1973 (n > 1973); because 1969 will not be enough to reach 2007 (1969: 1969 + (1+9+6+9) + S(1+9+6+9) = 1969 + 25 + S(25) = 1994 + 7 = 2001); and neither 1973 (1973: 1973 + (1+9+7+3) + S(1+9+7+3) = 1973 + 20 + S(20) = 1993 + 2 = 1995) and lesser than 2002 (n < 2002); because 2002 will exceed 2007 (2002 + (2+0+0+2) + S(2+0+0+2) = 2002 + 4 + S(4) = 2006 + 4 = 2010 1973 < n < 2002 Now this is easier to solve: 1974: 1974 + (1+9+7+4) + S(1+9+7+4) = 1974 + 21 + S(21) = 1995 + 3 = 1998 (needs 9) 1976: 1976 + (1+9+7+6) + S(1+9+7+6) = 1976 + 23 + S(23) = 1999 + 5 = 2004 (needs 3) 1977: 1977 + (1+9+7+7) + S(1+9+7+7) = 1977 + 24 + S(24) = 2001 + 6 = 2007 ✓ See that when we went from 1974 to 1976 the sum went from 1998 to 2004 (plus 6); and from 1976 to 1977 (plus 3) We only need to think about that if the S(S(n)) is not changing the ten place′s number, so we can just sum 3x, where x∈N and means the quantity that we will add to n: 1978 and 1979 is not necessary to calculate because it will be higher than 2007. 1980: 1980 +(1+9+8+0) + S(1+9+8+0) = 1980 + 18 + S(18) = 1998 + 9 = 2007 ✓ See that if we sum two to n, 18 + 2 ⇒ S(20) = 2, and now the sum is lesser than before because of the change of S(S(n)); while had a sum of 4 because of 1980 + 2 and 18 + 2 that sums 4, by another side we have S(18) = 9 that went to S(20) = 2, causing the sum to decrease in 3 in total. To recuperate the sum of 2007, we need to sum 3 again, it means the next number n that will satisfies the equation is 1983; In fact: 1983: 1983 + (1+9+8+3) + S(1+9+8+3) = 1983 + 21 + S(21) = 2004 + 3 = 2007 ✓ 1984 to 1989 is not necessary to calculate; because will not occur a switch of number in the ten′s place: 1989: 1989 + (1+9+8+9) + S(1+9+8+9) = 1989 + 27 + S(27)= 2016 + 9 = 2025 Again: 1990: 1990 + (1+9+9+0) + S(1+9+9+0) = 1990 + 19 + S(19) = 2009 + 10 = 2019; in next, will sum 2 and decrease 8 (19 ⇒ 1+9 = 10 → 20 ⇒ 2+0 = 2); but will not be enough (total = 2013): 1991 = 1991 + (1+9+9+1) + S(1+9+9+1) = 1991 + 20 + S(20) = 2011 + 2 = 2013 The non−necessity of calculate will be extended until 1999. And so; two more numbers: 2000: 2000 + (2+0+0+0) + S(2+0+0+0) = 2000 + 2 + S(2) = 2002 + 2 = 2004; now you must know. The next number will be another n solution to the equation: 2001 = 2001 + (2+0+0+1) + S(2+0+0+1) = 2001 + 3 + S(3) = 2004 + 3 = 2007 ✓ Conclusion: N = set of elements (values of n) solution to the equation N = {1977, 1980, 1983, 2001} N(n) = number of elements of N N(n) = 4 Answer: D) 4
$${n}\:+\:{S}\left({n}\right)\:+\:{S}\left({S}\left({n}\right)\right)\:=\:\mathrm{2007} \\ $$$$ \\ $$$${n}\:=\:\mathrm{1983} \\ $$$$ \\ $$$$\mathrm{1983}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{3}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{3}\right)\:= \\ $$$$=\:\mathrm{1983}\:+\:\mathrm{21}\:+\:{S}\left(\mathrm{21}\right)\:= \\ $$$$=\:\mathrm{2004}\:+\:\mathrm{3}\:=\:\mathrm{2007}\:\checkmark \\ $$$$ \\ $$$${What}\:{is}\:{necessary}? \\ $$$$ \\ $$$${The}\:{number}\:{has}\:{to}\:{be}\:{higher}\:{than}\:\mathrm{1973}\:\left({n}\:>\:\mathrm{1973}\right);\:{because}\:\mathrm{1969}\:{will}\:{not}\:{be}\:{enough}\:{to}\:{reach}\:\mathrm{2007}\:\left(\mathrm{1969}:\:\mathrm{1969}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{6}+\mathrm{9}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{6}+\mathrm{9}\right)\:=\:\mathrm{1969}\:+\:\mathrm{25}\:+\:{S}\left(\mathrm{25}\right)\:=\:\mathrm{1994}\:+\:\mathrm{7}\:=\:\mathrm{2001}\right);\:{and}\:{neither}\:\mathrm{1973}\:\left(\mathrm{1973}:\:\mathrm{1973}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{3}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{3}\right)\:=\:\mathrm{1973}\:+\:\mathrm{20}\:+\:{S}\left(\mathrm{20}\right)\:=\:\mathrm{1993}\:+\:\mathrm{2}\:=\:\mathrm{1995}\right) \\ $$$${and}\:{lesser}\:{than}\:\mathrm{2002}\:\left({n}\:<\:\mathrm{2002}\right);\:{because}\:\mathrm{2002}\:{will}\:{exceed}\:\mathrm{2007}\:\left(\mathrm{2002}\:+\:\left(\mathrm{2}+\mathrm{0}+\mathrm{0}+\mathrm{2}\right)\:+\:{S}\left(\mathrm{2}+\mathrm{0}+\mathrm{0}+\mathrm{2}\right)\:=\:\mathrm{2002}\:+\:\mathrm{4}\:+\:{S}\left(\mathrm{4}\right)\:=\:\mathrm{2006}\:+\:\mathrm{4}\:=\:\mathrm{2010}\right. \\ $$$$\mathrm{1973}\:<\:{n}\:<\:\mathrm{2002} \\ $$$$ \\ $$$${Now}\:{this}\:{is}\:{easier}\:{to}\:{solve}: \\ $$$$ \\ $$$$\mathrm{1974}:\:\mathrm{1974}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{4}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{4}\right)\:=\:\mathrm{1974}\:+\:\mathrm{21}\:+\:{S}\left(\mathrm{21}\right)\:=\:\mathrm{1995}\:+\:\mathrm{3}\:=\:\mathrm{1998}\:\left({needs}\:\mathrm{9}\right) \\ $$$$\mathrm{1976}:\:\mathrm{1976}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{6}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{6}\right)\:=\:\mathrm{1976}\:+\:\mathrm{23}\:+\:{S}\left(\mathrm{23}\right)\:=\:\mathrm{1999}\:+\:\mathrm{5}\:=\:\mathrm{2004}\:\left({needs}\:\mathrm{3}\right) \\ $$$$\mathrm{1977}:\:\mathrm{1977}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{7}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{7}+\mathrm{7}\right)\:=\:\mathrm{1977}\:+\:\mathrm{24}\:+\:{S}\left(\mathrm{24}\right)\:=\:\mathrm{2001}\:+\:\mathrm{6}\:=\:\mathrm{2007}\:\checkmark \\ $$$$ \\ $$$${See}\:{that}\:{when}\:{we}\:{went}\:{from}\:\mathrm{1974}\:{to}\:\mathrm{1976}\:{the}\:{sum}\:{went}\:{from}\:\mathrm{1998}\:{to}\:\mathrm{2004}\:\left({plus}\:\mathrm{6}\right); \\ $$$${and}\:{from}\:\mathrm{1976}\:{to}\:\mathrm{1977}\:\left({plus}\:\mathrm{3}\right) \\ $$$$ \\ $$$${We}\:{only}\:{need}\:{to}\:{think}\:{about}\:{that}\:{if}\:{the}\:{S}\left({S}\left({n}\right)\right)\:{is}\:{not}\:{changing}\:{the}\:{ten}\:{place}'{s}\:{number},\:{so}\:{we}\:{can}\:{just}\:{sum}\:\mathrm{3}{x},\:{where}\:{x}\in{N}\:{and}\:{means}\:{the}\:{quantity}\:{that}\:{we}\:{will}\:{add}\:{to}\:\boldsymbol{{n}}: \\ $$$$ \\ $$$$\mathrm{1978}\:{and}\:\mathrm{1979}\:{is}\:{not}\:{necessary}\:{to}\:{calculate}\:{because}\:{it}\:{will}\:{be}\:{higher}\:{than}\:\mathrm{2007}. \\ $$$$ \\ $$$$\mathrm{1980}:\:\mathrm{1980}\:+\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{0}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{0}\right)\:=\:\mathrm{1980}\:+\:\mathrm{18}\:+\:{S}\left(\mathrm{18}\right)\:=\:\mathrm{1998}\:+\:\mathrm{9}\:=\:\mathrm{2007}\:\checkmark \\ $$$$ \\ $$$${See}\:{that}\:{if}\:{we}\:{sum}\:{two}\:{to}\:\boldsymbol{{n}},\:\mathrm{18}\:+\:\mathrm{2}\:\Rightarrow\:{S}\left(\mathrm{20}\right)\:=\:\mathrm{2},\:{and}\:{now}\:{the}\:{sum}\:{is}\:{lesser}\:{than}\:{before}\:{because}\:{of}\:{the}\:{change}\:{of}\:{S}\left({S}\left({n}\right)\right); \\ $$$${while}\:{had}\:{a}\:{sum}\:{of}\:\mathrm{4}\:{because}\:{of}\:\mathrm{1980}\:+\:\mathrm{2}\:{and}\:\mathrm{18}\:+\:\mathrm{2}\:{that}\:{sums}\:\mathrm{4}, \\ $$$${by}\:{another}\:{side}\:{we}\:{have}\:{S}\left(\mathrm{18}\right)\:=\:\mathrm{9}\:{that}\:{went}\:{to}\:{S}\left(\mathrm{20}\right)\:=\:\mathrm{2},\:{causing}\:{the}\:{sum}\:{to}\:{decrease}\:{in}\:\mathrm{3}\:{in}\:{total}. \\ $$$$ \\ $$$${To}\:{recuperate}\:{the}\:{sum}\:{of}\:\mathrm{2007},\:{we}\:{need}\:{to}\:{sum}\:\mathrm{3}\:{again},\:{it}\:{means}\:{the}\:{next}\:{number}\:\boldsymbol{{n}}\:{that}\:{will}\:{satisfies}\:{the}\:{equation}\:{is}\:\mathrm{1983}; \\ $$$$ \\ $$$${In}\:{fact}: \\ $$$$ \\ $$$$\mathrm{1983}:\:\mathrm{1983}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{3}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{3}\right)\:=\:\mathrm{1983}\:+\:\mathrm{21}\:+\:{S}\left(\mathrm{21}\right)\:=\:\mathrm{2004}\:+\:\mathrm{3}\:=\:\mathrm{2007}\:\checkmark \\ $$$$ \\ $$$$\mathrm{1984}\:{to}\:\mathrm{1989}\:{is}\:{not}\:{necessary}\:{to}\:{calculate};\:{because}\:{will}\:{not}\:{occur}\:{a}\:{switch}\:{of}\:{number}\:{in}\:{the}\:{ten}'{s}\:{place}: \\ $$$$ \\ $$$$\mathrm{1989}:\:\mathrm{1989}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{9}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{8}+\mathrm{9}\right)\:=\:\mathrm{1989}\:+\:\mathrm{27}\:+\:{S}\left(\mathrm{27}\right)=\:\mathrm{2016}\:+\:\mathrm{9}\:=\:\mathrm{2025} \\ $$$$ \\ $$$${Again}: \\ $$$$ \\ $$$$\mathrm{1990}:\:\mathrm{1990}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{9}+\mathrm{0}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{9}+\mathrm{0}\right)\:=\:\mathrm{1990}\:+\:\mathrm{19}\:+\:{S}\left(\mathrm{19}\right)\:=\:\mathrm{2009}\:+\:\mathrm{10}\:=\:\mathrm{2019};\:{in}\:{next},\:{will}\:{sum}\:\mathrm{2}\:{and}\:{decrease}\:\mathrm{8}\:\left(\mathrm{19}\:\Rightarrow\:\mathrm{1}+\mathrm{9}\:=\:\mathrm{10}\:\rightarrow\:\mathrm{20}\:\Rightarrow\:\mathrm{2}+\mathrm{0}\:=\:\mathrm{2}\right);\:{but}\:{will}\:{not}\:{be}\:{enough}\:\left({total}\:=\:\mathrm{2013}\right): \\ $$$$ \\ $$$$\mathrm{1991}\:=\:\mathrm{1991}\:+\:\left(\mathrm{1}+\mathrm{9}+\mathrm{9}+\mathrm{1}\right)\:+\:{S}\left(\mathrm{1}+\mathrm{9}+\mathrm{9}+\mathrm{1}\right)\:=\:\mathrm{1991}\:+\:\mathrm{20}\:+\:{S}\left(\mathrm{20}\right)\:=\:\mathrm{2011}\:+\:\mathrm{2}\:=\:\mathrm{2013} \\ $$$$ \\ $$$${The}\:{non}−{necessity}\:{of}\:{calculate}\:{will}\:{be}\:{extended}\:{until}\:\mathrm{1999}. \\ $$$$ \\ $$$${And}\:{so};\:{two}\:{more}\:{numbers}: \\ $$$$ \\ $$$$\mathrm{2000}:\:\mathrm{2000}\:+\:\left(\mathrm{2}+\mathrm{0}+\mathrm{0}+\mathrm{0}\right)\:+\:{S}\left(\mathrm{2}+\mathrm{0}+\mathrm{0}+\mathrm{0}\right)\:=\:\mathrm{2000}\:+\:\mathrm{2}\:+\:{S}\left(\mathrm{2}\right)\:=\:\mathrm{2002}\:+\:\mathrm{2}\:=\:\mathrm{2004};\:{now}\:{you}\:{must}\:{know}.\:{The}\:{next}\:{number}\:{will}\:{be}\:{another}\:\boldsymbol{{n}}\:{solution}\:{to}\:{the}\:{equation}: \\ $$$$ \\ $$$$\mathrm{2001}\:=\:\mathrm{2001}\:+\:\left(\mathrm{2}+\mathrm{0}+\mathrm{0}+\mathrm{1}\right)\:+\:{S}\left(\mathrm{2}+\mathrm{0}+\mathrm{0}+\mathrm{1}\right)\:=\:\mathrm{2001}\:+\:\mathrm{3}\:+\:{S}\left(\mathrm{3}\right)\:=\:\mathrm{2004}\:+\:\mathrm{3}\:=\:\mathrm{2007}\:\checkmark \\ $$$$ \\ $$$${Conclusion}: \\ $$$$ \\ $$$${N}\:=\:{set}\:{of}\:{elements}\:\left({values}\:{of}\:\boldsymbol{{n}}\right)\:{solution}\:{to}\:{the}\:{equation} \\ $$$${N}\:=\:\left\{\mathrm{1977},\:\mathrm{1980},\:\mathrm{1983},\:\mathrm{2001}\right\} \\ $$$$ \\ $$$$\boldsymbol{{N}}\left({n}\right)\:=\:{number}\:{of}\:{elements}\:{of}\:{N} \\ $$$${N}\left({n}\right)\:=\:\mathrm{4} \\ $$$$ \\ $$$$\left.{Answer}:\:{D}\right)\:\mathrm{4} \\ $$

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