These Two Cubes are The Only Ones That Are All Pure Prime..name themSubtraction square dance alphameticFind 5 digits, which have the most numbers of prime numbersConnected circles which sum of any two of them which are adjacent is a triangle numbersMy Social Security Card NumberHe said yes..she said no..they went back and forth..they agreed toWe are two immediate neighbors who forged our own powers to form concatenated relationship. Who are we?A B C in close knit relationship.. Who are they?Find the values of U, V, C based on the given relationship…useful for upcoming puzzlesUvc with Lots of Magic… Enjoy the Show!Find this Unique UVC Palindrome ( ignoring signs and decimal) from Given Fractional Relationship
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These Two Cubes are The Only Ones That Are All Pure Prime..name them
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These Two Cubes are The Only Ones That Are All Pure Prime..name them
Subtraction square dance alphameticFind 5 digits, which have the most numbers of prime numbersConnected circles which sum of any two of them which are adjacent is a triangle numbersMy Social Security Card NumberHe said yes..she said no..they went back and forth..they agreed toWe are two immediate neighbors who forged our own powers to form concatenated relationship. Who are we?A B C in close knit relationship.. Who are they?Find the values of U, V, C based on the given relationship…useful for upcoming puzzlesUvc with Lots of Magic… Enjoy the Show!Find this Unique UVC Palindrome ( ignoring signs and decimal) from Given Fractional Relationship
$begingroup$
Given:
CU and BBUE are concatenated Cubes with all prime digits B, C, U, E.
Which are these two Unique Cubes?
mathematics no-computers
asked 8 hours ago
UvcUvc
82313
$endgroup$
add a comment |
$begingroup$
Given:
CU and BBUE are concatenated Cubes with all prime digits B, C, U, E.
Which are these two Unique Cubes?
mathematics no-computers
asked 8 hours ago
UvcUvc
82313
$endgroup$
add a comment |
$begingroup$
Given:
CU and BBUE are concatenated Cubes with all prime digits B, C, U, E.
Which are these two Unique Cubes?
mathematics no-computers
asked 8 hours ago
UvcUvc
82313
$endgroup$
Given:
CU and BBUE are concatenated Cubes with all prime digits B, C, U, E.
Which are these two Unique Cubes?
mathematics no-computers
mathematics no-computers
asked 8 hours ago
UvcUvc
82313
asked 8 hours ago
UvcUvc
82313
asked 8 hours ago
UvcUvc
82313
asked 8 hours ago
UvcUvc
82313
asked 8 hours ago
UvcUvc
82313
82313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Solution
$$C=2,,,,,,, U=7,,,,,,,,, B=3,,,,,,,,, E=5$$
Reasoning
The only $2$-digit cubes are $27$ and $64$ which gives us $C$ and $U$ straight away. Then, $BB7E$ must be the cube of a $2$-digit number $x$ (since it is greater than $1000$). Let $x = 10a+b$. Then, we have $$(10a+b)^3 = 1000a^3 + 300a^2 b + 30ab^2 + b^3 = BB7E$$. This means that $b$ is either $7$ or $5$ (to make $E$ prime and different to $U$). If $b$ is $7$, then we have $$1470a + 343 equiv 70a + 43 equiv 73 mod 100.$$ The smallest solution to this equation is $a=9$ which is too large to cube to a $4$-digit number.
Otherwise, if $b=5$, we have $$250a + 125 equiv 75 mod 100$$ which means $a$ is odd. Since, $35^3 > 10000$, the only possibility is $a=1$.
Checking this, we have $$ 15^3 = 15 times 225 = 3375 $$
answered 8 hours ago
hexominohexomino
50.2k4149239
$endgroup$
1
$begingroup$
I figured it out through trial and error but didn't post it because I knew someone could do it mathematically.
$endgroup$
– Zimonze
8 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Solution
$$C=2,,,,,,, U=7,,,,,,,,, B=3,,,,,,,,, E=5$$
Reasoning
The only $2$-digit cubes are $27$ and $64$ which gives us $C$ and $U$ straight away. Then, $BB7E$ must be the cube of a $2$-digit number $x$ (since it is greater than $1000$). Let $x = 10a+b$. Then, we have $$(10a+b)^3 = 1000a^3 + 300a^2 b + 30ab^2 + b^3 = BB7E$$. This means that $b$ is either $7$ or $5$ (to make $E$ prime and different to $U$). If $b$ is $7$, then we have $$1470a + 343 equiv 70a + 43 equiv 73 mod 100.$$ The smallest solution to this equation is $a=9$ which is too large to cube to a $4$-digit number.
Otherwise, if $b=5$, we have $$250a + 125 equiv 75 mod 100$$ which means $a$ is odd. Since, $35^3 > 10000$, the only possibility is $a=1$.
Checking this, we have $$ 15^3 = 15 times 225 = 3375 $$
answered 8 hours ago
hexominohexomino
50.2k4149239
$endgroup$
1
$begingroup$
I figured it out through trial and error but didn't post it because I knew someone could do it mathematically.
$endgroup$
– Zimonze
8 hours ago
add a comment |
$begingroup$
Solution
$$C=2,,,,,,, U=7,,,,,,,,, B=3,,,,,,,,, E=5$$
Reasoning
The only $2$-digit cubes are $27$ and $64$ which gives us $C$ and $U$ straight away. Then, $BB7E$ must be the cube of a $2$-digit number $x$ (since it is greater than $1000$). Let $x = 10a+b$. Then, we have $$(10a+b)^3 = 1000a^3 + 300a^2 b + 30ab^2 + b^3 = BB7E$$. This means that $b$ is either $7$ or $5$ (to make $E$ prime and different to $U$). If $b$ is $7$, then we have $$1470a + 343 equiv 70a + 43 equiv 73 mod 100.$$ The smallest solution to this equation is $a=9$ which is too large to cube to a $4$-digit number.
Otherwise, if $b=5$, we have $$250a + 125 equiv 75 mod 100$$ which means $a$ is odd. Since, $35^3 > 10000$, the only possibility is $a=1$.
Checking this, we have $$ 15^3 = 15 times 225 = 3375 $$
answered 8 hours ago
hexominohexomino
50.2k4149239
$endgroup$
1
$begingroup$
I figured it out through trial and error but didn't post it because I knew someone could do it mathematically.
$endgroup$
– Zimonze
8 hours ago
add a comment |
$begingroup$
Solution
$$C=2,,,,,,, U=7,,,,,,,,, B=3,,,,,,,,, E=5$$
Reasoning
The only $2$-digit cubes are $27$ and $64$ which gives us $C$ and $U$ straight away. Then, $BB7E$ must be the cube of a $2$-digit number $x$ (since it is greater than $1000$). Let $x = 10a+b$. Then, we have $$(10a+b)^3 = 1000a^3 + 300a^2 b + 30ab^2 + b^3 = BB7E$$. This means that $b$ is either $7$ or $5$ (to make $E$ prime and different to $U$). If $b$ is $7$, then we have $$1470a + 343 equiv 70a + 43 equiv 73 mod 100.$$ The smallest solution to this equation is $a=9$ which is too large to cube to a $4$-digit number.
Otherwise, if $b=5$, we have $$250a + 125 equiv 75 mod 100$$ which means $a$ is odd. Since, $35^3 > 10000$, the only possibility is $a=1$.
Checking this, we have $$ 15^3 = 15 times 225 = 3375 $$
answered 8 hours ago
hexominohexomino
50.2k4149239
$endgroup$
Solution
$$C=2,,,,,,, U=7,,,,,,,,, B=3,,,,,,,,, E=5$$
Reasoning
The only $2$-digit cubes are $27$ and $64$ which gives us $C$ and $U$ straight away. Then, $BB7E$ must be the cube of a $2$-digit number $x$ (since it is greater than $1000$). Let $x = 10a+b$. Then, we have $$(10a+b)^3 = 1000a^3 + 300a^2 b + 30ab^2 + b^3 = BB7E$$. This means that $b$ is either $7$ or $5$ (to make $E$ prime and different to $U$). If $b$ is $7$, then we have $$1470a + 343 equiv 70a + 43 equiv 73 mod 100.$$ The smallest solution to this equation is $a=9$ which is too large to cube to a $4$-digit number.
Otherwise, if $b=5$, we have $$250a + 125 equiv 75 mod 100$$ which means $a$ is odd. Since, $35^3 > 10000$, the only possibility is $a=1$.
Checking this, we have $$ 15^3 = 15 times 225 = 3375 $$
answered 8 hours ago
hexominohexomino
50.2k4149239
edited 2 hours ago
answered 8 hours ago
hexominohexomino
50.2k4149239
answered 8 hours ago
hexominohexomino
50.2k4149239
answered 8 hours ago
hexominohexomino
50.2k4149239
50.2k4149239
1
$begingroup$
I figured it out through trial and error but didn't post it because I knew someone could do it mathematically.
$endgroup$
– Zimonze
8 hours ago
add a comment |
1
$begingroup$
I figured it out through trial and error but didn't post it because I knew someone could do it mathematically.
$endgroup$
– Zimonze
8 hours ago
1
1
$begingroup$
I figured it out through trial and error but didn't post it because I knew someone could do it mathematically.
$endgroup$
– Zimonze
8 hours ago
$begingroup$
I figured it out through trial and error but didn't post it because I knew someone could do it mathematically.
$endgroup$
– Zimonze
8 hours ago
add a comment |
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