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?
+289 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
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
+289 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
