From 2019 to digits
Is it possible to obtain the digits from 0 to 9 starting from 2019 and using its digits in the same order, together with the usual operations +, *, -, /, concatenation of digits, and the less usual operators ^, !, sqrt(), int()? For example, 1 = 20-19. Unary minus is allowed too.
I manage to use only basic operations and elevation to a power for all digits except 4 and 5, but maybe somebody will do better!
arithmetic
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Is it possible to obtain the digits from 0 to 9 starting from 2019 and using its digits in the same order, together with the usual operations +, *, -, /, concatenation of digits, and the less usual operators ^, !, sqrt(), int()? For example, 1 = 20-19. Unary minus is allowed too.
I manage to use only basic operations and elevation to a power for all digits except 4 and 5, but maybe somebody will do better!
arithmetic
add a comment |
Is it possible to obtain the digits from 0 to 9 starting from 2019 and using its digits in the same order, together with the usual operations +, *, -, /, concatenation of digits, and the less usual operators ^, !, sqrt(), int()? For example, 1 = 20-19. Unary minus is allowed too.
I manage to use only basic operations and elevation to a power for all digits except 4 and 5, but maybe somebody will do better!
arithmetic
Is it possible to obtain the digits from 0 to 9 starting from 2019 and using its digits in the same order, together with the usual operations +, *, -, /, concatenation of digits, and the less usual operators ^, !, sqrt(), int()? For example, 1 = 20-19. Unary minus is allowed too.
I manage to use only basic operations and elevation to a power for all digits except 4 and 5, but maybe somebody will do better!
arithmetic
arithmetic
asked 6 hours ago
mau
9941233
9941233
add a comment |
add a comment |
1 Answer
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$0 = 2 cdot 0 cdot 1 cdot 9$
$1 = 20 - 19$
$2 = 2^0 + 1^9$
$3 = 2+0+1^9$
$4 = lfloor sqrt{20} rfloor + lfloor 1/9 rfloor$
$5 = 2 + 0 cdot 1 + sqrt{9}$
$6 = -(2+0+1) + 9$
$7 = -(2+0 cdot 1) + 9$
$8 = -(2 cdot 0 + 1) + 9$
$9 = -(2 cdot 0 cdot 1) + 9$
Ok for the first three, but I think you're not supposed to rearrange the digits
– deep thought
5 hours ago
1
@deepthought Fixed.
– Display name
5 hours ago
1
Nice! 4 may also be $2cdot 0 + 1 + sqrt 9$
– mau
5 hours ago
1
and $lfloorsqrt{201}rfloor-9=5$
– JonMark Perry
5 hours ago
1
Fun fact: $-lfloor -x rfloor = lceil x rceil.$
– Display name
4 hours ago
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$0 = 2 cdot 0 cdot 1 cdot 9$
$1 = 20 - 19$
$2 = 2^0 + 1^9$
$3 = 2+0+1^9$
$4 = lfloor sqrt{20} rfloor + lfloor 1/9 rfloor$
$5 = 2 + 0 cdot 1 + sqrt{9}$
$6 = -(2+0+1) + 9$
$7 = -(2+0 cdot 1) + 9$
$8 = -(2 cdot 0 + 1) + 9$
$9 = -(2 cdot 0 cdot 1) + 9$
Ok for the first three, but I think you're not supposed to rearrange the digits
– deep thought
5 hours ago
1
@deepthought Fixed.
– Display name
5 hours ago
1
Nice! 4 may also be $2cdot 0 + 1 + sqrt 9$
– mau
5 hours ago
1
and $lfloorsqrt{201}rfloor-9=5$
– JonMark Perry
5 hours ago
1
Fun fact: $-lfloor -x rfloor = lceil x rceil.$
– Display name
4 hours ago
|
show 1 more comment
$0 = 2 cdot 0 cdot 1 cdot 9$
$1 = 20 - 19$
$2 = 2^0 + 1^9$
$3 = 2+0+1^9$
$4 = lfloor sqrt{20} rfloor + lfloor 1/9 rfloor$
$5 = 2 + 0 cdot 1 + sqrt{9}$
$6 = -(2+0+1) + 9$
$7 = -(2+0 cdot 1) + 9$
$8 = -(2 cdot 0 + 1) + 9$
$9 = -(2 cdot 0 cdot 1) + 9$
Ok for the first three, but I think you're not supposed to rearrange the digits
– deep thought
5 hours ago
1
@deepthought Fixed.
– Display name
5 hours ago
1
Nice! 4 may also be $2cdot 0 + 1 + sqrt 9$
– mau
5 hours ago
1
and $lfloorsqrt{201}rfloor-9=5$
– JonMark Perry
5 hours ago
1
Fun fact: $-lfloor -x rfloor = lceil x rceil.$
– Display name
4 hours ago
|
show 1 more comment
$0 = 2 cdot 0 cdot 1 cdot 9$
$1 = 20 - 19$
$2 = 2^0 + 1^9$
$3 = 2+0+1^9$
$4 = lfloor sqrt{20} rfloor + lfloor 1/9 rfloor$
$5 = 2 + 0 cdot 1 + sqrt{9}$
$6 = -(2+0+1) + 9$
$7 = -(2+0 cdot 1) + 9$
$8 = -(2 cdot 0 + 1) + 9$
$9 = -(2 cdot 0 cdot 1) + 9$
$0 = 2 cdot 0 cdot 1 cdot 9$
$1 = 20 - 19$
$2 = 2^0 + 1^9$
$3 = 2+0+1^9$
$4 = lfloor sqrt{20} rfloor + lfloor 1/9 rfloor$
$5 = 2 + 0 cdot 1 + sqrt{9}$
$6 = -(2+0+1) + 9$
$7 = -(2+0 cdot 1) + 9$
$8 = -(2 cdot 0 + 1) + 9$
$9 = -(2 cdot 0 cdot 1) + 9$
edited 5 hours ago
answered 5 hours ago
Display name
714216
714216
Ok for the first three, but I think you're not supposed to rearrange the digits
– deep thought
5 hours ago
1
@deepthought Fixed.
– Display name
5 hours ago
1
Nice! 4 may also be $2cdot 0 + 1 + sqrt 9$
– mau
5 hours ago
1
and $lfloorsqrt{201}rfloor-9=5$
– JonMark Perry
5 hours ago
1
Fun fact: $-lfloor -x rfloor = lceil x rceil.$
– Display name
4 hours ago
|
show 1 more comment
Ok for the first three, but I think you're not supposed to rearrange the digits
– deep thought
5 hours ago
1
@deepthought Fixed.
– Display name
5 hours ago
1
Nice! 4 may also be $2cdot 0 + 1 + sqrt 9$
– mau
5 hours ago
1
and $lfloorsqrt{201}rfloor-9=5$
– JonMark Perry
5 hours ago
1
Fun fact: $-lfloor -x rfloor = lceil x rceil.$
– Display name
4 hours ago
Ok for the first three, but I think you're not supposed to rearrange the digits
– deep thought
5 hours ago
Ok for the first three, but I think you're not supposed to rearrange the digits
– deep thought
5 hours ago
1
1
@deepthought Fixed.
– Display name
5 hours ago
@deepthought Fixed.
– Display name
5 hours ago
1
1
Nice! 4 may also be $2cdot 0 + 1 + sqrt 9$
– mau
5 hours ago
Nice! 4 may also be $2cdot 0 + 1 + sqrt 9$
– mau
5 hours ago
1
1
and $lfloorsqrt{201}rfloor-9=5$
– JonMark Perry
5 hours ago
and $lfloorsqrt{201}rfloor-9=5$
– JonMark Perry
5 hours ago
1
1
Fun fact: $-lfloor -x rfloor = lceil x rceil.$
– Display name
4 hours ago
Fun fact: $-lfloor -x rfloor = lceil x rceil.$
– Display name
4 hours ago
|
show 1 more comment
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