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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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  • 4
    $\begingroup$ Don't underestimate the brains of this community... ;) $\endgroup$ – Mordechai Mar 20 '17 at 3:15
  • $\begingroup$ Haha! Nice one Mordechai! I'm new so... $\endgroup$ – user35295 Mar 20 '17 at 3:54
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    $\begingroup$ Can we safely assume that leading digits a,d,g are nonzero? (Yes I know not assuming this just creates more solution space to be explored.) $\endgroup$ – smci Mar 20 '17 at 10:42
  • $\begingroup$ I don't think so. I think that this would just be like throwing darts in darkness at a target on the roof of the CN tower. $\endgroup$ – user35295 Mar 20 '17 at 13:47
  • $\begingroup$ Allan, is this question one you made up or did you get it from somewhere else? If from somewhere else, please add proper attribution to the question. Thanks! $\endgroup$ – Gareth McCaughan Oct 21 '17 at 15:19
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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 $c=6$.

$c=6$

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 $13$.

$13^6$

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.

$12^6=2,985,984$

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.

$c=5$

The valid range is $13^5$ to $23^5$. We've ruled out numbers ending in $0,1,5$, so this leaves $\{13,14,16,17,18,19,23\}$.

$13^5=371,298$

Since $13$ is prime, the two numbers must be $13^2=169$ and $13^3=2197$. However, there are too many repeated digits.

$14^5$

$14$ only has $2$ and $7$ as factors. The only 3 digit numbers containing these factors are;

  • $2^4 \times 7 = 112$ - duplicate $1$s
  • $2^4 \times 7^2 = 784$ - duplicate $4$
  • $2^5 \times 7 = 224$ - duplicate $2$s
  • $2^3 \times 7^2 = 392$ - good so far, but the remaining factors $2^2 \times 7^3 = 1372$ creates duplicate $1,2, 3$
  • $2^2 \times 7^2 = 196$ - duplicate $1$
  • $2 \times 7^3 = 686$ - duplicate $6$
  • $7^3 = 343$ - duplicate $3$

$16^5$

Since $16$ is a power of $2$, we have 20 factors to spread around. The only valid 3 digit numbers are:

  • $2^7=128$
  • $2^8=256$
  • $2^9=512$

$17^5$

Again, this is prime. $17^2=289$ works, but $17^3=4913$ introduces a duplicate $1$ and $9$.

$18^5$

This consists of $18^5=2^5 \times 3^{10}$. The following are the valid factors:

  • $2^5 \times 3^2 = 288$
  • $2^5 \times 3^3 = 864$
  • $2^4 \times 3^2 = 144$
  • $2^4 \times 3^3 = 432$ - good so far, but $2 \times 3^7 =4374$
  • $2^3 \times 3^3 = 216$
  • $2^3 \times 3^4 = 648$
  • $2^2 \times 3^3 = 108$
  • $2^2 \times 3^4 = 324$ - good so far, but $2^3 \times 3^6=5832$
  • $2^2 \times 3^5 = 972$ - good so far, but $2^3 \times 3^5=1944$
  • $2 \times 3^4 = 162$
  • $2 \times 3^5 = 486$
  • $3^5 = 243$ - good so far, but $2^5 \times 3^5=7776$
  • $3^6 = 729$ - good so far, but $2^5 \times 3^4=2592$

$19^5$

A prime. $19^2=361$ has a duplicate $1$. Out.

$23^5$

A prime again. $23^2=529$ has a duplicate $2$, so it is out.

$c=4$

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 $\{26, 27, 28, 32, 36, 38, 39\}$.

$26^4$

Since $26^4=13^4\times 2^4$, we don't have that many options for 3 digit numbers, all of which create duplicates:

  • $13^2 = 169$
  • $13^2 \times 2 = 338$
  • $13^2 \times 2^2 = 676$
  • $13 \times 2^3 = 104$
  • $13 \times 2^4 = 208$

$27^4$

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.

$28^4$

This is $28^4=7^4 \times 2^8$. Our 3 digit possibilities are:

  • $7^3 \times 2 = 686$
  • $7^3 = 343$
  • $7^2 \times 2^2 = 196$ - good so far, but $7^2 \times 2^6 = 3136$
  • $7^2 \times 2^3 = 392$
  • $7^2 \times 2^4 = 784$
  • $7 \times 2^4 = 112$
  • $7 \times 2^5 = 224$
  • $7 \times 2^6 = 448$
  • $7 \times 2^7 = 896$
  • $2^7, 2^8$ each contain a $2$

$32^4$

$32$ is just a power of $2$. The only 3 digit powers of $2$ all contain a $2$.

$36^4$

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.

  • $3^6 = 729$ but $3^2 \times 2^8 = 2304$
  • $3^5 = 243$
  • $3^5 \times 2 = 486$
  • $3^5 \times 2^2 = 972$ but $3^3 \times 2^6 = 1728$
  • $3^4 \times 2 = 162$
  • $3^4 \times 2^2 = 324$
  • $3^4 \times 2^3 = 648$
  • $3^3 \times 2^2 = 108$
  • $3^3 \times 2^3 = 216$
  • $3^3 \times 2^4 = 432$
  • $3^3 \times 2^5 = 864$
  • $3^2 \times 2^4 = 144$
  • $3^2 \times 2^5 = 288$
  • $3^2 \times 2^6 = 576$
  • $3 \times 2^6 = 192$ but $3^7 \times 2^2=8748$
  • $3 \times 2^7 = 384$
  • $3 \times 2^8 = 786$
  • $2^7=128$
  • $2^8=256$

$38^4$

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.

  • $19^2 \times 2 = 722$
  • $19^2 = 361$ - good so far, but $19^2 \times 2^4=5776$
  • $19 \times 2^3 = 152$
  • $19 \times 2^4 = 304$

$39^4$

Since $39^4 = 13^4 \times 3^4$, again, we don't have many combinations.

  • $13^2 = 169$
  • $13^2 \times 3 = 507$ but $13^2 \times 3^3 = 4563$
  • $13 \times 3^2 = 117$
  • $13 \times 3^3 = 351$

$c=3$

$71$ 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:

$\{72, 74, 76, 78, 82, 84, 86, 87, 92, 94, 96, 98\}$

$72^3$

$72^3 = 3^6 \times 2^9$. 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.

  • $3^3 \times 2^2 = 108$ but $3^3 \times 2^7 = 3456$

Very close, but a duplicate $3$ sets us back.

$74^3$

$74^3=37^3\times 2^3$. Factors are minimal.

  • $37 \times 2^2 = 148$
  • $37 \times 2^3 = 296$ but $37^2 = 1369$

$76^3$

$76^3 = 19^3 \times 2^6$

  • $19^2 \times 2 = 722$
  • $19^2 = 361$
  • $19 \times 2^3 = 152$ but $19^2 \times 2^3 = 2888$
  • $19 \times 2^4 = 304$

$78^3$

$78^3 = 2^3 \times 3^3 \times 13^3$

  • $13^2 = 169$ but $13 \times 2^3 \times 3^3 = 2808$
  • $13^2 \times 3 = 507$
  • $13^2 \times 2 = 338$
  • $13^2 \times 2^2 = 676$
  • $13 \times 3^3 = 351$
  • $13 \times 3^3 \times 2 = 702$
  • $13 \times 3^2 = 117$
  • $13 \times 3^2 \times 2 = 234$
  • $13 \times 3^2 \times 2^2 = 468$
  • $13 \times 3^2 \times 2^3 = 936$
  • $13 \times 3 \times 2^2 = 156$ but $13^2 \times 3^2 \times 2 = 3042$
  • $13 \times 2^3 = 108$

$82^3$

$82^3 = 41^3 \times 2^3$

  • $41 \times 2^2 = 164$ but $41^2 \times 2 = 3362$
  • $41 \times 2^3 = 328$

$84^3$

$84^3 = 2^6 \times 3^3 \times 7^3$. Any 3 digit number greater than 592 will not result in a 4 digit number.

  • $7^3 = 343$
  • $7^3 \times 2 = 686$
  • $7^2 \times 3^2 = 441$
  • $7^2 \times 3^2 \times 2 = 882$
  • $7^2 \times 3 = 147$
  • $7^2 \times 3 \times 2 = 294$
  • $7^2 \times 3 \times 2^2 = 588$
  • $7^2 \times 2^2 = 196$ but $7 \times 3^3 \times 2^4=3024$
  • $7^2 \times 2^3 = 392$
  • $7^2 \times 2^4 = 784$
  • $7 \times 3^3 = 189$
  • $7 \times 3^3 \times 2 = 378$
  • $7 \times 3^3 \times 2^2 = 756$
  • $7 \times 3^2 \times 2 = 126$ but $7^2 \times 3 \times 2^5=4704$
  • $7 \times 3^2 \times 2^2 = 252$
  • $7 \times 3^2 \times 2^3 = 504$
  • $7 \times 3 \times 2^3 = 168$
  • $7 \times 3 \times 2^4 = 336$
  • $7 \times 2^4 = 112$
  • $7 \times 2^5 = 224$
  • $7 \times 2^6 = 448$
  • $3^3 \times 2^2 = 108$
  • $3^3 \times 2^3 = 216$ but $7^3 \times 2^3 = 2744$
  • $3^3 \times 2^4 = 432$
  • $3^3 \times 2^5 = 864$
  • $3^2 \times 2^4 = 144$
  • $3^2 \times 2^5 = 288$
  • $3^2 \times 2^6 = 576$ and $7^3 \times 3 =1029$
  • $3 \times 2^6 = 192$ but $7^3 \times 3^2 = 3087$

As you can see, a solution has been found!!

$$84^3=576 \times 1029$$

$86^3$

$86^3 = 43^3 \times 2^3$

  • $43 \times 2^2 = 172$ but $43^2 \times 2 = 3698$
  • $43 \times 2^3 = 344$

$87^3$

$87^3=29^3 \times 3^3$

  • $29^2 = 841$
  • $29 \times 3^2 = 261$ but $29^2 \times 3 = 2523$
  • $29 \times 3^3 = 783$

$92^3$

$92^3 = 23^3 \times 2^6$

  • $23^2 = 529$
  • $23 \times 2^3 = 184$ but $23^2 \times 2^3 = 4232$
  • $23 \times 2^4 = 368$
  • $23 \times 2^3 = 736$

$94^3$

$94^3 = 47^3 \times 2^3$

  • $47 \times 2^2 = 188$ but $43^2 \times 2 = 3698$
  • $47 \times 2^3 = 376$

$96^3$

$96^3 = 3^3 \times 2^15$

  • $3^3 \times 2^2 = 108$ but $2^13=8192$
  • $3^3 \times 2^3 = 216$
  • $3^3 \times 2^4 = 432$
  • $3^3 \times 2^2 = 864$
  • $3^2 \times 2^4 = 144$
  • $3^2 \times 2^5 = 288$
  • $3^2 \times 2^6 = 576$
  • $3 \times 2^6 = 192$
  • $3 \times 2^7 = 384$
  • $3 \times 2^6 = 768$
  • $2^7=128$ but $3^3 \times 2^8=6912$
  • $2^8=256$
  • $2^9=512$ but $3^3 \times 2^6=1728$

$98^3$

$98^3 = 7^6 \times 2^3$

  • $7^3 = 343$
  • $7^3 \times 2 = 686$
  • $7^2 \times 2^2 = 196$
  • $7^2 \times 2^3 = 392$

Solution

Therefore, there is a single solution.

$$84^3=576 \times 1029$$

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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.

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  • $\begingroup$ Could you share a link please? $\endgroup$ – user35295 Mar 20 '17 at 3:55
  • $\begingroup$ @AllanCao Edited in, I assume you meant a link to what I did on Wolfram Alpha. $\endgroup$ – DooplissForce Mar 20 '17 at 9:59
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    $\begingroup$ Thanks so much! I tried a different method but it didn't work. $\endgroup$ – user35295 Mar 20 '17 at 13:42
  • $\begingroup$ @AllanCao No worries :) $\endgroup$ – DooplissForce Mar 20 '17 at 15:11

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