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Let's try to find out the numbers satisfying the cryptarithms below;

  • $ab=(a+b)^2$
  • $abc=(a+b+c)^3$
  • $abcd=(a+b+c+d)^4$

There are only 3 unique answers!

Note: $a$,$b$,$c$,$d$ are all digits and might be different than each other for each equation. $ab$, $abc$ and $abcd$ are positive integers and and all $a$ values are nonzero.

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  • 5
    $\begingroup$ So is it $ab => a * b$, or is it $ab => a * 10 + b$ $\endgroup$ – Brian J May 10 '17 at 15:43
  • $\begingroup$ Oh. I was trying to work it out as the product of a and b. That is obviously not right. Haha $\endgroup$ – Fogmeister May 11 '17 at 7:52
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As there aren't many possibilities for each of the equations, we can look at the options and deduce which one fits.

For the first equation, we have

$4^2=16$, $5^2=25$, $6^2=36$, $7^2=49$, $8^2=64$, $9^2=81$

of these options, only the last on fits, because $81=(8+1)^2$

For the second one, if we look at the cubes of numbers:

$5^3=125$, $6^3=216$, $7^3=343$, $8^3=512$, $9^3=729$

The only one to fit is $512=(5+1+2)^3$

For the third one, if we look at the fourth powers

$6^4=1296$, $7^4=2401$, $8^4=4096$, $9^4=6561$

The only one to fit is $2401=(2+4+0+1)^4$

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5
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$ab=(a+b)^2$

B can be

0, 1, 4, 5, 6, 9.
Taking step by step
for $b=0$ it's easy to see that no value for $a$ fits since a+b must end with a 0. And $a=0$ is not allowed
$b=1$. You can get this is $a+b$ ends with a 1 or a 9. Same as above 1 is not possible so a+b must end with a 9. Hence $a=8$. Testing out $81 = (8+1)^2$. It fits.
$b=4$. It means $ab$ can be 64 (the only 2 digit perfect square that ends with 4). But (6+4)^ = 100. So it does not fit.
$b=5$. $ab$ can be 25 (the only 2 digit perfect square that ends with 5). This doesn't work out.
$b=6$ means $ab$ can be $16, 36$. Doesn't fit.
$b = 9$ it means $ab=49$. Again doesn't fit.

Next:

$abc=(a+b+c)^3$

.

The 3 digit perfect cubes are 125, 316, 343, 512, 729.
$(1+2+5)^3 = 729$ Not good.
$(3+1+6)^3 = 1000$ Not good.
$(3+4+5)^3 = 1728$ Not good.
$(5+1+2)^3 = 512$ Found it.
$(7+2+9)^3 = 5832$. Not good

And last one:

$abcd=(a+b+c+d)^4$

.

The 4 digit perfect whatever 4th power is called:
1296: $(1+2+9+6)^4 = 18^4 > 6^4$. Way too much.
2401: $(2+4+0+1)^4 = 7^4 = 2401$. Found it.
4096: $(4+0+9+6)^4 = 19^4 > 8^4 = 40196$. Not good.
6561: $(6+5+6+1)^4 = 18^4 > 9^4 = 40196$. Not good.

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3
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For $abcd=(a+b+c+d)^4$

It is

$2401 = (2+4+0+1)^4$

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The very first one that comes to mind, as many will probably see, is making:

a, b, c, d all = to 0.

So that way:

* $00=(0+0)^2$
* $000=(0+0+0)^3$
* $0000=(0+0+0+0)^4$

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  • $\begingroup$ :) you got me! but fixed the question accordingly! $\endgroup$ – Oray May 10 '17 at 14:32
  • $\begingroup$ Bahh that's no fun now :P $\endgroup$ – n_plum May 10 '17 at 14:33

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