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

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Intern
Joined: 31 May 2018
Posts: 17
Location: United States
Concentration: Finance, Marketing
find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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20 Jul 2018, 08:10
4
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Difficulty:

45% (medium)

Question Stats:

62% (01:40) correct 38% (01:18) wrong based on 58 sessions

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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
Senior Manager
Joined: 22 Feb 2018
Posts: 326
find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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20 Jul 2018, 08:33
4
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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##### General Discussion
Intern
Joined: 03 Feb 2016
Posts: 11
Re: find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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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'
Ans E
Senior Manager
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Posts: 464
Location: Malaysia
GMAT 1: 700 Q50 V33
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Re: find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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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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Intern
Joined: 17 May 2018
Posts: 32
Re: find the last digit of the number 1^2+2^2+.....99^2  [#permalink]

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23 Jul 2018, 06:43
1
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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# find the last digit of the number 1^2+2^2+.....99^2

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