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find the last digit of the number 1^2+2^2+.....99^2

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find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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New post 20 Jul 2018, 08:10
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find the last digit of the number \(1^2+2^2+..........+99^2\).

a) 1
b) 2
c) 3
d) 4
e) 0
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find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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New post 20 Jul 2018, 08:33
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1
shridhar786 wrote:
find the last digit of the number \(1^2+2^2+..........+99^2\).

a) 1
b) 2
c) 3
d) 4
e) 0


OA:E
Sum of the Squares of First \(n\) Positive Integers = \(\frac{{n(n+1)(2n+1)}}{{6}}\)
Putting n=99
Sum of the Squares of First \(99\) Positive Integers = \(\frac{{99(100)(198+1)}}{{6}}={33*50*199}\)
Last Digit of the number would be \(0\).
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Re: find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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New post 20 Jul 2018, 08:37
formula is 1/6N(N+1)(2N+1)
N=99
so sum=(33*199*100)/2
so last digit is '0'
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Re: find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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New post 20 Jul 2018, 09:23
shridhar786 wrote:
find the last digit of the number \(1^2+2^2+..........+99^2\).

a) 1
b) 2
c) 3
d) 4
e) 0


Sum of the Squares of Number -

n(n+1)(2n+1)/6. Here, n = 99.
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Re: find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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New post 23 Jul 2018, 06:43
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If for some reason you don't know the formula mentioned in the previous solutions, we could try to find it manually:

Last digit of the square of a number ending in 1: 1
Last digit of the square of a number ending in 2: 4
Last digit of the square of a number ending in 3: 9
Last digit of the square of a number ending in 4: 6
Last digit of the square of a number ending in 5: 5
Last digit of the square of a number ending in 6: 6
Last digit of the square of a number ending in 7: 9
Last digit of the square of a number ending in 8: 4
Last digit of the square of a number ending in 9: 1
Last digit of the square of a number ending in 0: 0

Adding these (try to group them to ease the calculation: 1 with 9, 4 with 6, etc.) we end up with a 5 for the sequence 1 to 10 (or 11 to 20, or 21 to 31... or 91 to 99). We'll need to add this 10 times or, what is the same, multiply it by 10, to find out the last digit of the sequence presented in the stem. The result is a zero as the last digit.
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Re: find the last digit of the number 1^2+2^2+.....99^2 &nbs [#permalink] 23 Jul 2018, 06:43
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