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 Post subject: Magic sequence.
PostPosted: Tue Jan 14, 2014 1:59 pm 
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Magic sequence.

TASK. The sequence of numbers from 1 up to 9 is given. With the help of substitution between numbers of a sequence of arithmetic marks incorporated in the usual accounting calculator, such how "+", "-", "", ":", "", "(" and ")" to express any year with 1-st up to 2100 (up to the end XXI of century).

THE DECISION. The decision of a task is resulted selectively for 101 meanings of years to show an opportunity to express ABSOLUTELY any year offered in a task. In a course of accounts for some years some answers are found. The most original decisions are selected with colour.

1 = 1–2-3–4+5–6–7+8+9 = 1▪2▪3▪(4 + 5):6–7+8-9
2 = 1+2▪3▪(4+5):6–7+8-9 = (1+2+3+4):5-6+7+8-9
3 = (-1-234):5+67-8-9 = (-1+23)▪4+5+6-7-89
4 = -12-34+56-7-8+9 = 1▪2+(3+45):6-7-8+9
5 = (-12:3):4+5▪6-7-8-9 = 1-2+3-4+5-6+7-8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
10 = 123-45-67+8-9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
20 = 1▪2+3+45+6▪7-8▪9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
30 = 1▪2▪34-5▪6-7+8-9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
40 = 12-3-4+5+6+7+8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
50 = (1+2+345):6-7+8-9 = 1+2▪34+56-78+√9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
99 = 1-2▪3+(4+5+6)▪7+8-9
100 = (1+23)▪4+56:(7▪8)+√9 = -1+23+(4+5)▪6+7+8+9 = 12+34+5▪6+7+8+9 =
= (-1+2)▪(34+5▪(6+7))-8+9 = -12+3+4-5▪6+(7+8)▪9 = (1+2+3+4+5)▪6-7+8+9 =
= 1-23+(4+5+6)▪7+8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
200 = 1▪2-3+45+67+89
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
300 = 1▪2+345+6▪7-89
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
400 = -1+23▪4▪5-6▪7-8-9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
500 = (12+34)▪(5+6)-7-8+9
501 = 123▪4-56+7▪8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
666 = -1+(2+3+45)▪(6+7)+8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
999 = (123-4-5-√(-6+7+8))▪9
1000 = 123▪(4+5)-(6+7)▪8-√9 = (-1+23-4)▪56-7+8-9
1001 = 1-2-3-(√4+5+6)▪78-9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1100 = 1▪23▪45+6+7▪8+√9 = (1+2+3-4)▪(567-8-9)
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1200 = 1234-5▪6+7-8-√9 = 1▪234▪5+6+7+8+9 = 1▪23+4▪5+(6+7)▪89
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1300 = 1234-5▪6+7+89 = 1+2-3+4+(5+6+7)▪8▪9 = 123+4▪5+(6+7)▪89
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1400 = -1-2+3▪456+7▪(8-√9)
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1492 = 1▪2▪34▪(5▪6-7)-8▪9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1500 = -1-2+3▪456+(7+8)▪9 = -1▪2+345+(6+7)▪89 = -12+(3+4+56)▪(7+8+9)
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1812 = 1▪2345-67▪8+√9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1900 = 12-3+45▪6▪7-8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1910 = 12+34▪56-7-8+9 = -1+2+3+4▪(567-89) = 123▪(4+5+6)+7▪8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1920 = 123▪(4+5+6)+78-√9 = (12▪3▪4+5)▪(6+7)-8-9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1930 = 1▪23+45▪6▪7+8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1940 = -1▪2+(34-5)▪67+8-9
1941 = 123▪(4+5+6)+7+89 = 1-2+(34-5)▪67+8-9
1942 = (12▪3▪4+5)▪(6+7)+8-√9
1943 = 12▪3+45▪6▪7+8+9 = 1234-5+6▪7▪(8+9) = (12▪3+4)▪(56-7)-8-9
1944 = (1+234+5)▪6+7▪8▪9 = (1-2+34)▪56+7+89 = 123▪(√(4)▪5+6)-7-8-9
1945 = 1+2+(34-5)▪67+8-9 = 1+234▪(-5+6+7)+8▪9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1950 = 1-2-3+4+5▪6▪(7▪8+9)
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1960 = (-1+2)▪(34-5)▪67+8+9 = -1+2-3+45▪6▪7+8▪9 = -1+234▪(-5+6+7)+89
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1970 = (1+234)▪5+6+789
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1980 = 1+2-3+4▪(567-8▪9)
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
1987 = 1-2+345▪6+7-89
1988 = -1-2+34▪56+78+9
1989 = -1▪2+34▪56+78+9 = 1+2+34▪56-7+89
1990 = -1▪2+3-45+678▪√9
1991 = 1+2+345▪6+7-89
1992 = (1▪2▪3▪4▪5▪6-7▪8)▪√9 = -1+2+34▪56+78+9 = 123-(4+5-6) ▪7▪89 = (1▪2+34)▪56-7-8-9 =
= (-12+345)▪6-7-8+9
1993 = 1▪2+34▪56+78+9 = -12+345▪6-7▪8-9 = -12+345▪6+7-8▪9 = 1234+(5+6)▪(78-9)
1994 = 1+2+34▪56+78+9
1995 = 12+34▪56+7+8▪9 = 12+345▪6-78-9
1996 = -1+2-34-5+678▪√9
1997 = -1-2+34▪56+7+89 = 12▪34▪5-6▪7+8-9
1998 = 12▪34▪5-6▪7▪(-8+9)
1999 = 12▪34▪5+6-7▪8+9 = 12▪34▪5-6▪7-8+9
2000 = 1▪2345-6▪7▪8-9 = 12+345▪6+7-89
2001 = 1▪23▪(4▪5-6+7+8)▪√9 = -12▪3+(4▪5+6)▪78+9
2002 = -1-2+345▪6+7-8▪9
2003 = 12+345▪6-7-8▪9 = 12+34▪56+78+9 = 1+2+34▪56+7+89 = -1▪2+345▪6+7-8▪9 =
= (1+234-5)▪6+7▪89 = 12+3▪456+7▪89
2004 = (-1+2-3-4)▪5+678▪√9 = 1-2+345▪6+7-8-9
2005 = -(1+2)▪3-4▪5+678▪√9
2006 = -1+2+345▪6-7▪8-9
2007 = 1▪2+345▪6-7▪8-9
2008 = (-1▪2+34)▪5▪(6+7)-8▪9 = -1-2-3-4▪5+678▪√9 = 1+2+345▪6-7▪8-9 = 1+2+345▪6+7-8▪9
2009 = -1▪2+345▪6-7▪8-√9 = (1▪2-3-4)▪5+678▪√9 = -123+(4▪5+6)▪(-7+89)
2010 = 12▪34▪5-6-7-8-9 = 1-2+345▪6-7▪8-√9 = 12▪34▪5+6▪7-8-9 = (12+3)▪4+5▪6▪(7▪8+9) =
= -1+(2+3+45)▪6▪7-89 = 12▪34+(5+6+7)▪89
2011 = -12+345▪6-7▪8+9
2012 = 1234-5-6+789 = 12+34▪56+7+89 = 1▪2345-6▪7▪8+√9
2013 = 12+345▪6-78+9 = 1▪2+345▪6-7▪8-√9
2014 = (-1-2+34)▪5▪(6+7)+8-9 = -1-2+3-4▪5+678▪√9
2015 = -1▪2+345▪6-7▪8+√9 = 12▪34▪5-6▪7+8+9 = (1▪2▪3▪4+5)▪67+8▪9 = 1-23+(4▪5+6)▪78+9
2016 = (-1-2+34)▪5▪(6+7)-8+9
2017 = 12-34+5+678▪√9 = 12+345▪6-7▪8-9 = 12+345▪6+7-8▪9
2018 = 12▪34▪5+67-89 = 1▪2▪34+5▪6▪(7▪8+9) = 1▪2345-6▪7▪8+9
2019 = -1-2-3-4-5+678▪√9 = 1+2▪34+5▪6▪(7▪8+9)
2020 = -1▪2-3-4-5+678▪√9 = (123+4)▪(5+6)+7▪89 = -12+(34-5)▪67+89
2021 = 1-2-3-4-5+678▪√9
2022 = 1+(-2+3+45)▪6▪7+89 = 1234+5-6+789 = 12▪34▪5+6-7-8-9
2023 = (-1+2)▪345▪6-7▪8+9
2024 = 1234-5+6+789 = -12-34+5▪6▪(78-9)
2025 = 1+2-3-4-5+678▪√9 = 1▪2+345▪6-7▪8+9 = (1+2+34)▪56-7▪8+9
2026 = 12▪34▪5-6-7+8-9 = (1▪2+34)▪56-7+8+9
2027 = 12+(34-5)▪67+8▪9
2028 = (1+2)▪3+(4▪5+6)▪78-9 = ((1+2)▪34-5)▪(6+7+8)-9
2029 = -1-2+(34-5)▪67+89 = 12+(3+45)▪6▪7-8+9
2030 = -1-23+4▪5+678▪√9
2031 = 1-2+(34-5)▪67+89
2032 = ((1+2+3)▪4+5)▪67+89
2033 = -1-2-34+5▪6▪(78-9) = 1234+(5+6▪7)▪(8+9)
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
2040 = 1▪2+3-4+5+678▪√9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
2050 = 1+2▪(345+678)+√9 = 123▪(√(4)▪5+6)-7+89
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
2060 = 1+2+3+4▪5+678▪√9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
2070 = (123-4)▪(5+6+7)-8▪9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
2080 = 1▪2+345▪6+7-8+9 = (1+2+34)▪56+7-8+9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
2090 = 12▪3+4▪5+678▪√9
▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪
2095 = (123+45)▪(6+7)-89
2096 = (1+2+34)▪56+7+8+9
2097 = 12▪3▪45+6▪78+9
2098 = 1-234+5▪6▪78-9
2099 = (1+2+3+4)▪5▪6▪7+8-9
2100 = 1+(2+3+45)▪6▪7+8-9

Enjoy! :)


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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 3:44 pm 
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What are the ▪ and : operators?

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 4:04 pm 
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Brandon, I think they are multiply and divide.

To pytlivyj_1, were these found by humans? Is it known which might be unique?

I guess a computer could search based on simply trying the 8 options in the 8 slots between the 9 numbers. (including no space in addition to your arithmetic ops.)

You could walk all 16,777,216 options, and place them in buckets based on the results they produce.
Hmmm.

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 4:19 pm 
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This reminds me of another similar sequence of equations...

Balance all the following equations using only the mathematical symbols you would find on a scientific calculator.

0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

e.g. 6 + (6 - 6) = 6 (no need to thank me!)

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 4:43 pm 
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JasonSmith wrote:
Brandon, I think they are multiply and divide.

To pytlivyj_1, were these found by humans? Is it known which might be unique?

I guess a computer could search based on simply trying the 8 options in the 8 slots between the 9 numbers. (including no space in addition to your arithmetic ops.)

You could walk all 16,777,216 options, and place them in buckets based on the results they produce.
Hmmm.


I've been thinking about how to go about finding these too. I think you could build a tree of options back-to-front. For example taking 9, the only option is 9 or sqrt(9). Taking 8 and (all options for 9), you have lots more options like 8 + sqrt(9), etc. The number of possibilities really balloons because of parenthesis and the sqrt() operator but since sqrt() pretty much needs perfect squares to be useful maybe it isn't as much as I think.

Even if you can only store a few million tree branches, you could probably get through 9, 8, ... N where N is pretty far down the list. After that you can brute-force CPU from 1 ... N. Seem like the sort of thing that has some elegant implementations for.

It's probably especially useful to find "null sequences" that either produce 1 or 0 so that you can truncate out parts of the list. For example ((3 * 4) - (5 + 6)) = 1 so you can always just skip 3-6 (multiplying by 1) if all you really need is 1, 2, 7, 8, 9.

There are many other neat tricks that could be done.

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 4:56 pm 
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oxymoronicuber wrote:
This reminds me of another similar sequence of equations...

Balance all the following equations using only the mathematical symbols you would find on a scientific calculator.

Easy:
Quote:
(cos(0) + cos(0) + cos(0))! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3 * 3 - 3 = 6
4 + 4 - sqrt(4) = 6
5 + (5/5) = 6
6 + (6 - 6) = 6
7 - (7/7) = 6
8 - (sqrt(sqrt(8 + 8))) = 6
(9 + 9) / sqrt(9) = 6

(in white text)

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 5:10 pm 
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Some alternatives:
Quote:
(0!+0!+0!)!=6
sqrt(4)+sqrt(4)+sqrt(4)=6
sqrt(8+(8/8))!=6
sqrt(9)*sqrt(9)-sqrt(9)=6

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 5:59 pm 
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Aright so here is a question:

Given an arbitrary set of unary operators and functions like sin(x) or x! or gamma(x), or floor(x), etc., and allowing arbitrary nesting, can you make ANY integer from ANY other integer?

I'm pretty sure you can.

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 6:09 pm 
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Brandon Enright wrote:
Aright so here is a question:

Given an arbitrary set of unary operators and functions like sin(x) or x! or gamma(x), or floor(x), etc., and allowing arbitrary nesting, can you make ANY integer from ANY other integer?

This gave me an idea: transcendental numbers like e and pi can't reach integers with normal functions...but using them, can you reach any transcendental number from any other transcendental number? I'm honestly not completely sure...

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 7:34 pm 
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How far can you go making integers using 1 to 9 like this without a break in between? (In other words, what's the smallest integer that cannot be constructed like this?)


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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 7:51 pm 
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benpuzzles wrote:
Brandon Enright wrote:
Aright so here is a question:

Given an arbitrary set of unary operators and functions like sin(x) or x! or gamma(x), or floor(x), etc., and allowing arbitrary nesting, can you make ANY integer from ANY other integer?

This gave me an idea: transcendental numbers like e and pi can't reach integers with normal functions...but using them, can you reach any transcendental number from any other transcendental number? I'm honestly not completely sure...


This is possible, but there are restrictions. Example:

Ln(2) is transcendental
Log(e) is transcendental
Ln(2)/Log(e)=Log(2)
Log(2) is transcendental

You could come up with something similar with trig functions like sin and cos, but you would not be able to mix sin and cos together with pi and Log's and come out with a single transcendental function or symbol that is that number (unless of course you make a new symbol to equal whatever it is you get).

EDIT: I just found this:
i^i = 0.207879576... (Here i is the imaginary number sqrt(-1). Isn't this a real beauty? How many people have actually considered rasing i to the i power? If a is algebraic and b is algebraic but irrational then a^b is transcendental. Since i is algebraic but irrational, the theorem applies. Note also: i^i is equal to e^(- pi / 2 ) and several other values. Consider i^i = e^(i log i ) = e^( i times i pi / 2 ) . Since log is multivalued, there are other possible values for i^i.

So there is an infinite number of ways you can get a transcendental number from other transcendental numbers, but there is still certain restrictions.


Last edited by Elrog on Tue Jan 14, 2014 8:02 pm, edited 1 time in total.

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 7:55 pm 
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benpuzzles wrote:
Brandon Enright wrote:
Aright so here is a question:

Given an arbitrary set of unary operators and functions like sin(x) or x! or gamma(x), or floor(x), etc., and allowing arbitrary nesting, can you make ANY integer from ANY other integer?

This gave me an idea: transcendental numbers like e and pi can't reach integers with normal functions...but using them, can you reach any transcendental number from any other transcendental number? I'm honestly not completely sure...


Quoting Wikipedia:
Most sums, products, powers, etc. of the number π and the number e, e.g. π + e, π − e, πe, π/e, π^π, e^e, π^e, π^√2, e^π^2 are not known to be rational, algebraic irrational or transcendental. Notable exceptions are π + e^π, πe^π and e^π√n (for any positive integer n) which have been proven to be transcendental.

In general I think it's an open problem.

Given my problem of starting an a number and producing other numbers, starting at π should be fine could you could do cos(π) to get an integer and then go from there. starting at e is fine too because of ln e.

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 8:40 pm 
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Elrog wrote:
i^i = 0.207879576...

Reminds me of a song by Kajagoogoo:
Quote:
Cause you're too shy, shy, hush, hush, 0.207879576...?

Doesn't have the same ring to it.

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 8:47 pm 
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Brandon Enright wrote:
Given my problem of starting an a number and producing other numbers, starting at π should be fine could you could do cos(π) to get an integer and then go from there. starting at e is fine too because of ln e.

Well you could just do:

π-π = e-e = 0

π/π = e/e = 1

+/- (π/π + π/π + ... + π/π) = +/- (e/e + e/e + ... + e/e) = any integer

+/- (π/π + π/π + ... + π/π) / (π/π + π/π + ... + π/π) = +/- (e/e + e/e + ... + e/e) / (e/e + e/e + ... + e/e) = any fraction

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 Post subject: Re: Magic sequence.
PostPosted: Tue Jan 14, 2014 10:43 pm 
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KelvinS wrote:
Brandon Enright wrote:
Given my problem of starting an a number and producing other numbers, starting at π should be fine could you could do cos(π) to get an integer and then go from there. starting at e is fine too because of ln e.

Well you could just do:

π-π = e-e = 0

π/π = e/e = 1

+/- (π/π + π/π + ... + π/π) = +/- (e/e + e/e + ... + e/e) = any integer

+/- (π/π + π/π + ... + π/π) / (π/π + π/π + ... + π/π) = +/- (e/e + e/e + ... + e/e) / (e/e + e/e + ... + e/e) = any fraction

The rules are that you aren't allowed to introduce any new numbers or values. You used e and pi twice, so that doesn't count.
oxymoronicuber wrote:
Balance all the following equations using only the mathematical symbols you would find on a scientific calculator.

EDIT: Here's another one to add to the above challenge: 10 10 10 = 6
This one is EVIL!!! :twisted:

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 5:49 am 
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10+10+10=6 (in binary!) Took me 3 seconds. :D

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 6:52 am 
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Or, if you don't want to mix bases:
Quote:
(lg 10 + lg 10 + lg 10)!

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 8:48 am 
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KelvinS wrote:
10+10+10=6 (in binary!) Took me 3 seconds. :D

One side is in binary, while the other is in decimal, so...wrong. :lol:
Coaster1235 wrote:
Or, if you don't want to mix bases.

Correct! 2 alternatives:
Quote:
(log(10*10)+log(10))!=6
(log(10*10*10))!=6

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 11:51 am 
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So, in general, using the operators found on most scientific calculators, can a set of three equal integers be made equal to 6? If so, can the set of operators be reduced? If not, what additional operators are needed and what is the smallest integer that requires additional operators?

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 11:53 am 
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Jeffery Mewtamer wrote:
So, in general, using the operators found on most scientific calculators, can a set of three equal integers be made equal to 6? If so, can the set of operators be reduced? If not, what additional operators are needed and what is the smallest integer that requires additional operators?

6 XOR 6 XOR 6

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 1:19 pm 
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benpuzzles wrote:
KelvinS wrote:
10+10+10=6 (in binary!) Took me 3 seconds. :D

One side is in binary, while the other is in decimal, so...wrong. :lol:

Actually no, because I just apply the convert base operator to one side of the equation only, just as you apply the factorial operator to one side. So Hah! :mrgreen:

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 1:54 pm 
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KelvinS wrote:
apply the convert base operator to one side of the equation only

There is no such mathematical operator. Numbers are numbers and they (abstractly) exist without any base. Bases don't even have to be integers. For example http://en.wikipedia.org/wiki/Golden_ratio_base

A base only comes into play when interpreting a number that has been written down or deciding how to write down a number.

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 Post subject: Magic sequence.
PostPosted: Wed Jan 15, 2014 2:00 pm 
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@ Brandon Enright, JasonSmith, oxymoronicuber, benpuzzles, Gus, Elrog, KelvinS, Coaster1235, Jeffery Mewtamer

No, there is no and more time no.
In a condition of a task is said, that it is impossible to use binary system, logarithms, trigonometry, erection in a degree and other difficult mathematical actions.
This task was published in a magazine together with crossword puzzles, sudoku and other logic puzzles in printed variant for the decision them manually (hobby, intellectual and logic games).
In the given task it is possible to use only simple ARITHMETIC actions: multiplication, division, addition, subtraction, extraction of a square root, mark a minus and brackets:

Image

The task is authorized to be solved only with the help of intelligence and without use of the computer programs and Internet. :D


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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 2:07 pm 
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Well talk about posting the rules *after* everyone has given their solution: We can't read your mind, you know! :lol:

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 2:10 pm 
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pytlivyj_1 wrote:
The task is authorized to be solved only with the help of intelligence and without use of the computer programs and Internet. :D

I'm an unauthorized problem solver 8-) .

I think we understood the rules based on your first post but just took the topic off onto several tangents and related problems.

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 2:57 pm 
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Brandon Enright wrote:
There is no such mathematical operator.

So that button I see and press on my calculator doesn't really exist? You mean it's just an illusion? Wow, that's spooky, it seems so REEEAL! :lol:

And yes, I know what you mean, I'm just being my usual awkward self. :wink:

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 3:28 pm 
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KelvinS wrote:
Brandon Enright wrote:
There is no such mathematical operator.
So that button I see and press on my calculator doesn't really exist? You mean it's just an illusion? Wow, that's spooky, it seems so REEEAL! :lol:
Is "9" a mathematical operator? How about "clear" :wink:

KelvinS wrote:
And yes, I know what you mean, I'm just being my usual awkward self. :wink:
Same here :mrgreen:

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 4:11 pm 
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Yes, both 9 and Clear are mathematical operators:

Clear adds the next number you press to 0, regardless of the preceding number
9 just puts the number 9 on the screen somewhere and then lets you press something else

And both send a tiny signal to the CPU, saying "I'm an operator". But you can't see it.

And I'm actually being serious: When you press any button on a calculator, you are just applying some mathematical (logical) operator to a bunch of 1's and 0's stored in memory. So Hah! :mrgreen:

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 5:19 pm 
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I must admit, when I first saw the n n n = 6 puzzle it was clearly stated that the solutions could not contain any other digits e.g. cube root, log to base 2 (although log implies log to base 10 but that was acceptable) and so on. Things like pi, e, i, ln, cos, atan etc. were allowed because you did not have to actually write any other digits.

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 5:41 pm 
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And just to pick more holes, *every* operator comprises arbitrary numbers, for example:

x {+} y = x {+(2*3!-4/2)/5-2+1*} y

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 Post subject: Re: Magic sequence.
PostPosted: Wed Jan 15, 2014 6:14 pm 
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Gus wrote:
I must admit, when I first saw the n n n = 6 puzzle it was clearly stated that the solutions could not contain any other digits e.g. cube root, log to base 2 (although log implies log to base 10 but that was acceptable) and so on. Things like pi, e, i, ln, cos, atan etc. were allowed because you did not have to actually write any other digits.

Earlier today I decided to make things harder and try to solve each problem without using any functions defined by a value. That means: no powers, roots, logarithms, greatest integers, etc. or extra numbers can be used. I managed to solve 8 of them, but the last 2 look to be impossible from what I can tell. Maybe someone else might have want to try?

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