## Using all digits from 1 to 9 once w/ A^{B/C} + D^{E/F} + G^{H/I}

$A^{B/C} \; + \; D^{E/F} \; + \; G^{H/I}$

To determine the lowest/highest palindromes and prime numbers that can be obtained in this fashion.

Smallest and largest palindromes :

$1^{5/7} \; + \; 2^{6/3} \; + \; 9^{4/8} \; = \; 8$
$6^{9/3} \; + \; 7^{4/2} \; + \; 8^{5/1} \; = \; 33033$

next two palindromes:

$1^{5/6} \; + \; 7^{8/4} \; + \; 9^{3/2} \; = \; 77$
$1^{2/5} \; + \; 7^{6/3} \; + \; 9^{8/4} \; = \; 131$

Smallest and largest prime numbers:

$1^{5/7} \; + \; 2^{9/3} \; + \; 8^{4/6} \; = \; 13$
$5^{4/2} \; + \; 7^{8/1} \; + \; 9^{6/3} \; = \; 5764907$

Smallest and largest integers that can be obtained:

$1^{5/7} \; + \; 2^{6/3} \; + \; 9^{4/8} \; = \; 8 \; = \; 2^3$
$4^{7/2} \; + \; 5^{6/3} \; + \; 8^{9/1} \; = \; 134217881 \; = \; 7\times 19\times 1009157$

What is the highest square number that can be obtained?   highest cube?

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Pipo found that the highest square number that can be obtained is:

$8^{2/6} \; + \; 7^{9/3} \; + \; 4^{5/1} \; = \; 1369 \; = \; 37^2$
$8^{3/9} \; + \; 7^{6/2} \; + \; 4^{5/1} \; = \; 1369$

and the highest cube is:

$9^{7/2} \; + \; 1^{6/5} \; + \; 3^{8/4} \; = \; 2197 \; = \; 13^3$

Can you top that?

math grad - Interest: Number theory
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### 2 Responses to Using all digits from 1 to 9 once w/ A^{B/C} + D^{E/F} + G^{H/I}

1. pipo says:

According to me:
Highest cube:
9 ^(7 / 2) + 1 ^(6 / 5) + 3 ^(8 / 4) = 2197 (= 13^3)
Highest square:
8 ^(2 / 6) + 7 ^(9 /3) + 4 ^(5 / 1) = 1369 (= 37^2)
or 8 ^(3 / 9) + 7 ^(6 / 2) + 4 ^(5 / 1) = 1369

pipo

• benvitalis says:

Welcome back! I posted your results