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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, 07:10
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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, 07:33
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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, 07:37
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, 08:23
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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, 05:43
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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  [#permalink]

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New post 30 Sep 2018, 05:45
Used a different approach.

So we have a series of n=99 numbers.
first + last together sum up to 1^2 + 99^2 = 100^2 , second and second last the same etc. All the combined 100^2 will have a zero at last digit.
Since the list contains 99 numbers, the middle number would be the 50, and 50^2 clearly has a zero as the last digit.

Thus, zero will be the last digit.

Pls correct me if I'm wrong.
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Re: find the last digit of the number 1^2+2^2+.....99^2 &nbs [#permalink] 30 Sep 2018, 05:45
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