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LoneSurvivor
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What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + ... + 12 × 02 Oct 2019, 07:48
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

36% (01:47) correct 64% (02:18) wrong based on 22 sessions

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What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + ... + 12 × 12! is divided by 13?
(a) 1
(b) 12
(c) 11
(d) 0
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B
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abuc0112
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What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + ... + 12 × 10 Jul 2020, 13:27
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Can someone please solve this question

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firas92
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What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + ... + 12 × 10 Jul 2020, 13:40
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abuc0112
Can someone please solve this question

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Discussed here https://gmatclub.com/forum/what-is-the- ... l#p2534129
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What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + ... + 12 × 17 Aug 2020, 19:42
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Just start solving in pairs of two, then 3 and so on...
observe a pattern of, 3!-1, 4!-1....

1! + 2*2! = 5 = 3!-1
1! + 2*2! + 3*3! = 23 = 4!-1

Generalised formula of the series

1*1!+2*2!+3*3!......a*a! is

= (a+1)!-1

So, Total sum = 13!-1/13
Reminder = 12

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