let-S-1-2-2018-compute-S-mod-2-S-mod8-S-mod2018-

Question Number 34159 by candre last updated on 01/May/18

l e t S = 1 + 2 + \dots + 2018

c o m p u t e S (\mod 2) + S (\mod 8) + S (\mod 2018)

Commented by Rasheed.Sindhi last updated on 05/May/18

S = (1 + 3 + 5 + \dots + 2017) + (2 + 4 + \dots + 2018)

Sum of even numbers is even number

∴ 2 + 4 + \dots + 2018 even number

Sum of odd numbers, when number of

numbers is odd, is odd .

∴ 1 + 3 + 5 + \dots + 2017 is odd number

∴ S = odd + even = odd number

∴ S (\mod 2) = 1

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S (\mod 8) ?

- - - - - -

S = (1 + 2 + 3 + \dots + 2017) + 2018

= (1 + 2017) + (2 + 2016) + \dots + 1009 + 2018

= 2018 + 2018 + \dots + 1009 + 2018

= (multiple of 2018) + 1009

∴ S (\mod 2018) = 1009

Continue

Answered by Rasheed.Sindhi last updated on 05/May/18

S = 1 + 2 + \dots + 2018; an AP : a = 1, d = 1, n = 2018

S = \frac{n}{2} [2 a + (n - 1) d]

S = \frac{2018}{2} [2 (1) + (2018 - 1) (1)]

= 1009 [2018 + 1]

= 2018 \times 1009 + 1009

∴ S (\mod 2018) = 1009 \dots \dots \dots A

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S = 1009 \times 2019

= (1008 + 1) (2018 + 1)

= 1008 \times 2018 + 1008 + 2018 + 1

= 1008 \times 2019 + 2018 + 1

= 2 (504 \times 2019 + 1009) + 1

∴ S (\mod 2) = 1 \dots \dots \dots \dots \dots . B

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S = 1009 \times 2019

= (1009) (2000 + 16 + 3)

= 2000 \times 1009 + 16 \times 1009 + 3 \times 1009

= 2000 \times 1009 + 16 \times 1009 + 3000 + 27

= 2000 \times 1009 + 16 \times 1009 + 3000 + 24 + 3

2000, 16, 3000 & 24 are divisible by 8

[Complete thousands are always divisible

by 8]

∴ S (\mod 8) = 3 \dots \dots \dots \dots \dots C

From A, B & C

S (\mod 2) + S (\mod 8) + S (\mod 2018)

= 1 + 3 + 1009 = 1013

Commented by Rasheed.Sindhi last updated on 09/May/18

Mr . candre please confirm the answer .