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# Sum

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261
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What is the sum of all the five-digit positive integers in which each of the digits 1, 2, 3, 5 and 8 appear exactly once?

Jul 8, 2022

#1
+289
0

There are 120 ways to arrange these 5-digit positive integers. That means that the digits will be added 24 times in each digit of the sum. Therefore, our sum would be equal to:
5(24(10000 + 20000 + 30000 + 50000+80000) + 24(1000 + 2000 + 3000 + 5000 + 8000) + ... + 24(1 + 2 + 3 + 5 + 8)))

We know that this is equal to:

120(11111 + 22222 + 33333 + 55555 + 88888)

= 25333080

Jul 8, 2022
#2
0

sum(12358, 12385, 12538, 12583, 12835, 12853, 13258, 13285, 13528, 13582, 13825, 13852, 15238, 15283, 15328, 15382, 15823, 15832, 18235, 18253, 18325, 18352, 18523, 18532, 21358, 21385, 21538, 21583, 21835, 21853, 23158, 23185, 23518, 23581, 23815, 23851, 25138, 25183, 25318, 25381, 25813, 25831, 28135, 28153, 28315, 28351, 28513, 28531, 31258, 31285, 31528, 31582, 31825, 31852, 32158, 32185, 32518, 32581, 32815, 32851, 35128, 35182, 35218, 35281, 35812, 35821, 38125, 38152, 38215, 38251, 38512, 38521, 51238, 51283, 51328, 51382, 51823, 51832, 52138, 52183, 52318, 52381, 52813, 52831, 53128, 53182, 53218, 53281, 53812, 53821, 58123, 58132, 58213, 58231, 58312, 58321, 81235, 81253, 81325, 81352, 81523, 81532, 82135, 82153, 82315, 82351, 82513, 82531, 83125, 83152, 83215, 83251, 83512, 83521, 85123, 85132, 85213, 85231, 85312, 85321)==Total permutations ==120==Total sum==5,066,616

Jul 8, 2022
#3
+289
-1

Sorry with my previous answer, I accidentally multiplied by 5 fo rsome reason.

There are 120 ways to arrange these 5-digit positive integers. That means that the digits will be added 24 times in each digit of the sum. Therefore, our sum would be equal to:
24(10000 + 20000 + 30000 + 50000+80000) + 24(1000 + 2000 + 3000 + 5000 + 8000) + ... + 24(1 + 2 + 3 + 5 + 8))

We know that this is equal to:

24(11111 + 22222 + 33333 + 55555 + 88888)

= 5,066,616

Jul 8, 2022
edited by Voldemort  Jul 8, 2022
edited by Voldemort  Jul 8, 2022