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16 Mar 2020, 12:05
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
45% (02:42) correct
55%
(02:28)
wrong
based on 104
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Find the unit place of
2^1!+2!+3!+ .....+10! + 3^1!+2!+3!+ .....+10!+ 4^1!+2!+3!+ .....+10!+ 5^1!+2!+3!+ .....+10! +........+9^1!+2!+3!+ .....+10! ?
A. 4
B. 5
C. 0
D. 6
E. 9
Posted from my mobile device
2^1!+2!+3!+ .....+10! + 3^1!+2!+3!+ .....+10!+ 4^1!+2!+3!+ .....+10!+ 5^1!+2!+3!+ .....+10! +........+9^1!+2!+3!+ .....+10! ?
A. 4
B. 5
C. 0
D. 6
E. 9
Posted from my mobile device
16 Mar 2020, 12:39
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Raksat
Find the unit place of
2^1!+2!+3!+ .....+10! + 3^1!+2!+3!+ .....+10!+ 4^1!+2!+3!+ .....+10!+ 5^1!+2!+3!+ .....+10! +........+9^1!+2!+3!+ .....+10! ?
A. 4
B. 5
C. 0
D. 6
E. 9
Posted from my mobile device
2^1!+2!+3!+ .....+10! + 3^1!+2!+3!+ .....+10!+ 4^1!+2!+3!+ .....+10!+ 5^1!+2!+3!+ .....+10! +........+9^1!+2!+3!+ .....+10! ?
A. 4
B. 5
C. 0
D. 6
E. 9
Posted from my mobile device
The exponent for each term is 1! + 2! + 3! + ... + 10!
From 4! onward, each term is a multiple of 4
1! + 2! +3! = 1 + 2 + 6 = 9 = 4 * 2 + 1 = 'A multiple of 4' + 1
Thus, (1! + 2! + 3! + 4! + ... + 10!) is 'a multiple of 4' + 1
=> 1! + 2! + 3! + 4! + ... + 10! = 4k + 1 (where k is some positive integer)
Thus, we have: 2^(4k+1) + 3^(4k+1) + 4^(4k+1) + ... + 9^(4k+1)
Each of the digits (in the base) 2, 3, 4, 5, 6, 7, 8 and 9 have a cyclicity of 4 places (4 and 9 have a cyclicity of 2 places, but for all practical purposes can be thought of having a cyclicity of 4 places)
Thus, we have:
Units digit of 2^(4k+1) + 3^(4k+1) + 4^(4k+1) + ... + 9^(4k+1)
= Units digit of 2^1 + 3^1 + 4^1 + ... + 9^1
= Units digit of 2 + 3 + 4 + ... + 9
= Units digit of 44
= 4
Answer A
22 Dec 2024, 12:29
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