I have an amazing puzzle that should be a little difficult even for you awesome puzzlers.
$(ab)^c = def × ghij$
where each letter stands for a number from 0 to 9 (inclusive). Each digit must be unique and may not be repeated.
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Sign up to join this communityI have an amazing puzzle that should be a little difficult even for you awesome puzzlers.
$(ab)^c = def × ghij$
where each letter stands for a number from 0 to 9 (inclusive). Each digit must be unique and may not be repeated.
A more rigorous proof showing all possible answers.
First, note that the product of a 3 digit number and a 4 digit number is somewhere between $102 \times 3,456 = 352,512$ and $987 \times 6,543 = 6,457,941$.
Since $98^2 \lt 352,512 \lt 98^3$ and $10^6 \lt 6,457,941 \lt 10^7$, we can infer that $3 \le c \le 6$.
Also, we know that $a,d,g \ne 0$.
Lastly, we can see that $b=0,1,5 \implies j=0,1,5$ so $b,j \ne \{0,1,5\}$ as well.
Since we have only 4 values for $c$, lets start with $c=6$.
The only numbers that fall in the right range are $13$ or lower since $14^6 = 7,529,536 \gt 6,457,941$, which was our maximum 3-digit by 4-digit product.
Since $10$ and $11$ are ruled out, we only need to check $12$ and $13$.
This is the easiest one to rule out. Since $13$ is prime, the the 3-digit and 4-digit factors must be powers of $13$. However, $13^2=169$ and $13^3=2197$ are the only ones that fit - both of which contain a 1, ruling them out.
To get a last digit of $4$, the only option that doesn't involve, $1, 2, 6$ is $f,j \in \{3,8\}$.
But the only factors of $12$ are $2$ and $3$. That means that both the 3 and 4 digit factors must also have only $2$ and $3$ as their factors. Any number with a $2$ as a factor must be even, so the number ending in $3$ must only contain $3$s. Turns out, the only 3 or 4 digit powers of $3$ ending in $3$ is $3^5=243$. But this contains a $2$, so it cannot be one of the factors.
The valid range is $13^5$ to $23^5$. We've ruled out numbers ending in $0,1,5$, so this leaves $\{13,14,16,17,18,19,23\}$.
Since $13$ is prime, the two numbers must be $13^2=169$ and $13^3=2197$. However, there are too many repeated digits.
$14$ only has $2$ and $7$ as factors. The only 3 digit numbers containing these factors are;
Since $16$ is a power of $2$, we have 20 factors to spread around. The only valid 3 digit numbers are:
Again, this is prime. $17^2=289$ works, but $17^3=4913$ introduces a duplicate $1$ and $9$.
This consists of $18^5=2^5 \times 3^{10}$. The following are the valid factors:
A prime. $19^2=361$ has a duplicate $1$. Out.
A prime again. $23^2=529$ has a duplicate $2$, so it is out.
The range of numbers is $25^4$ to $50^4$. But trailing $0,1,5$ are out, as are primes because the only way to divide 4th power of primes is 1 and 3, which will not yield a 3 digit number. Also out are doubled numbers and obviously, all numbers containing a 4. Thus, we are left with $\{26, 27, 28, 32, 36, 38, 39\}$.
Since $26^4=13^4\times 2^4$, we don't have that many options for 3 digit numbers, all of which create duplicates:
This is simply $3^{12}$. The only 3 digit number is $3^5=243$ which has a duplicate $2$ and $4$. $3^6$ is also a 3 digit number, but the remaining factor would be the same, not a 4 digit number like required.
This is $28^4=7^4 \times 2^8$. Our 3 digit possibilities are:
$32$ is just a power of $2$. The only 3 digit powers of $2$ all contain a $2$.
This one is a bit complicated because $36^4=2^8 \times 3^8$ yields lots of factors. Also, 3 digit numbers under 168 will result in a 5 digit number for the remaining factors.
Since $38^4=19^4\times 2^4$, there aren't that many combinations to check. 3 digit numbers under 209 will result in a 5 digit factor.
Since $39^4 = 13^4 \times 3^4$, again, we don't have many combinations.
$71$ is the lowest number which when taken to the power of $3$ is within the valid range. But it is prime, so you cannot make a 3 digit and a 4 digit number using only 3 $71$s as factors.
In fact, this is true of all primes in this range. So, we can rule out $73, 79, 83, 89, 97$, as well as all numbers that contain a $3$ or double digits - $93, 77, 88$. We already eliminated $0, 1$ and $5$ as the last digit, so we are left with the following possibilities:
$\{72, 74, 76, 78, 82, 84, 86, 87, 92, 94, 96, 98\}$
$72^3 = 3^6 \times 2^9$. Looking at $c=2, 36^4$ we can see many of the options are already ruled out because they have duplicate digits. Also, any 3 digit number higher than than 373 will result in a second three digit number. The result is we only need to check one result further.
Very close, but a duplicate $3$ sets us back.
$74^3=37^3\times 2^3$. Factors are minimal.
$76^3 = 19^3 \times 2^6$
$78^3 = 2^3 \times 3^3 \times 13^3$
$82^3 = 41^3 \times 2^3$
$84^3 = 2^6 \times 3^3 \times 7^3$. Any 3 digit number greater than 592 will not result in a 4 digit number.
As you can see, a solution has been found!!
$$84^3=576 \times 1029$$
$86^3 = 43^3 \times 2^3$
$87^3=29^3 \times 3^3$
$92^3 = 23^3 \times 2^6$
$94^3 = 47^3 \times 2^3$
$96^3 = 3^3 \times 2^15$
$98^3 = 7^6 \times 2^3$
Therefore, there is a single solution.
$$84^3=576 \times 1029$$
This should work:
$84^3=576\times1029$
Method (if you could call it one):
Honestly I found it through trial and error. I knew the number had to be between $111\times1111=123321$ and $999\times9999=9989001$, so, starting with $98^3$ (the highest power of $98$ between $123321$ and $9989001$), I went down through the double digit numbers, looking at the factors of $(ab)^3$ using Wolfram Alpha and finding those that consisted of all distinct digits.