The sum of all the elements in the set {n$\in$ {1, 2, ....., 100} | H.C.F. of n and 2040 is 1} is equal to _____________.
Answer (integer)
1251
Solution
2040 = 2<sup>3</sup> $\times$ 3 $\times$ 5 $\times$ 17<br><br>n should not be multiple of 2, 3, 5 and 17.<br><br>Sum of all n = (1 + 3 + 5 + ...... + 99) $-$ (3 + 9 + 15 + 21 + ...... + 99) $-$ (5 + 25 + 35 + 55 + 65 + 85 + 95) $-$ (17)<br><br>= 2500 $-$ ${{17} \over 2}$(3 + 99) $-$ 365 $-$ 17<br><br>2500 $-$ 867 $-$ 365 $-$ 17<br><br>= 1251
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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