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What is the sum of all 3 digit numbers that leave a remainder of '2'

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What is the sum of all 3 digit numbers that leave a remainder of '2'  [#permalink]

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New post 14 Feb 2016, 04:34
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Re: What is the sum of all 3 digit numbers that leave a remainder of '2'  [#permalink]

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New post 14 Feb 2016, 11:20
3
3
First term in AP,a = 101 = 33*3 + 2
Last term in AP,l = 998= 332*3 + 2
Number of terms in sequence
= 332- 33+1
= 300
Sum of n terms in AP
= number of terms/2 *[a+l]
= 300/2 *[101 + 998]
=150 * 1099
=164850
Answer B
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Re: What is the sum of all 3 digit numbers that leave a remainder of '2'  [#permalink]

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New post 21 Mar 2016, 08:11
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Re: What is the sum of all 3 digit numbers that leave a remainder of '2'  [#permalink]

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New post 12 Jun 2016, 10:26
Bunuel wrote:
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

A. 897
B. 164,850
C. 164,749
D. 149,700
E. 156,720


Hi Bunnel,

I solved this question using the equations of Arithmetic progression. Is there an alternate way so that long multiplication can be avoided?

Thanks.
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Re: What is the sum of all 3 digit numbers that leave a remainder of '2'  [#permalink]

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New post 24 Jul 2016, 23:54
1
all three digit numbers 100~999

the first number that satisfies the description "leave a remainder of '2' when divided by 3" is 101, and the last one is 998.

total numbers that satisfy the description is:
n=[(998-101)/3]+1, and we have n=300

As the numbers are evenly spaced sequence, we could calculate the sum by:
S=n*mean
mean=(a1+an)/2, so mean=(101+998)/2=549.5

Finally S=300*549.5=146850

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Re: What is the sum of all 3 digit numbers that leave a remainder of '2'  [#permalink]

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New post 08 Apr 2018, 05:55
1
1
First term in AP,a = 101 = 33*3 + 2
Last term in AP,l = 998= 332*3 + 2
Number of terms in sequence
= 332- 33+1
= 300
Sum of n terms in AP
= number of terms/2 *[a+l]
= 300/2 *[101 + 998]
=150 * 1099
=164850
Answer B
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What is the sum of all 3 digit numbers that leave a remainder of '2'  [#permalink]

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New post 09 Apr 2018, 08:13
1
From 1 to 99 we have 33 numbers which leave a remainder of 2 when divided by 3 (You call it as 1/3rd of the numbers) .
33 numbers leave a remainder of 0.
33 numbers leave a remainder of 1.
33 numbers leave a remainder of 2.
If we take three digit number, then try to find sum of 100 to 999 and divide by 3 (1/3rd).

(900/2)*(100+999) = 164850
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What is the sum of all 3 digit numbers that leave a remainder of '2' &nbs [#permalink] 09 Apr 2018, 08:13
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What is the sum of all 3 digit numbers that leave a remainder of '2'

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