Distinct power numbers: Difference between revisions
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Let given all integer combinations of <math>a^b</math> for '''2 < a < 5''' and '''2 < b < 5''': |
Let given all integer combinations of <math>a^b</math> for '''2 <= a <= 5''' and '''2 <= b <= 5''': |
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<math>2^2=4 ,2^3=8, 2^4=16, 2^5=32</math> |
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<math>3^2=9,3^3=27,3^4=81,3^5=243</math> |
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<math>4^2=16,4^3=64,4^4=256,4^5=1024</math> |
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<math>5^2=25,5^3=125,5^4=6256,5^5=3125</math> |
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Place them in numerical order, with any repeats removed. |
Place them in numerical order, with any repeats removed. |
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Revision as of 04:25, 16 August 2021
Distinct power numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Task
Let given all integer combinations of for 2 <= a <= 5 and 2 <= b <= 5:
Place them in numerical order, with any repeats removed.
We get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
Ring
<lang ring> load "stdlib.ring"
see "working..." + nl see "Distinct powers are:" + nl row = 0 distPow = []
for n = 2 to 5
for m = 2 to 5
sum = pow(n,m)
add(distPow,sum)
next
next
distPow = sort(distPow)
for n = len(distPow) to 2 step -1
if distPow[n] = distPow[n-1]
del(distPow,n-1)
ok
next
for n = 1 to len(distPow)
row++
see "" + distPow[n] + " "
if row%5 = 0
see nl
ok
next
see "Found " + row + " numbers" + nl see "done..." + nl </lang>
- Output:
working... Distinct powers are: 4 8 9 16 25 27 32 64 81 125 243 256 625 1024 3125 Found 15 numbers done...