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Find the sum of all odd three digit numbers

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Find the sum of all odd three digit numbers [#permalink] New post   31 Aug 2018, 08:04
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Find the sum of all odd three digit numbers that are divisible by 5.

1) 49500
2) 50000
3) 50250
4) 50500
5) 51250
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A
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Re: Find the sum of all odd three digit numbers [#permalink] New post   31 Aug 2018, 08:31
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Afc0892 wrote:
Find the sum of all odd three digit numbers that are divisible by 5.

1) 49500
2) 50000
3) 50250
4) 50500
5) 51250


Total 3 digit odd numbers = 9 (1 to 9)*10 (0 to 9)*1(odd digits 5) = 90

Smallest 3 digit odd number = 105
Biggest 3 digit odd number = 995
Sum of these two numbers = 1100

Second Smallest 3 digit odd number = 115
Second Biggest 3 digit odd number = 985
Sum of these two numbers = 1100

Third Smallest 3 digit odd number = 125
Third Biggest 3 digit odd number = 975
Sum of these two numbers = 1100

Total Numbers = 90

So total such Pairs of numbers = (90)/2
Each pair has sum = 1100
Sum of all pairs = (90/2)*1100 = 49500

Answer: Option A
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Re: Find the sum of all odd three digit numbers [#permalink] New post   31 Aug 2018, 11:32
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Afc0892 wrote:
Find the sum of all odd three digit numbers that are divisible by 5.

1) 49500
2) 50000
3) 50250
4) 50500
5) 51250


Numbers divisible by 5 will end with 0 or 5
So,Odd Numbers divisible by 5 will end with 5
First Odd three digit number number divisible by 5 is 105
second 115 , third 125 and so on ....
last odd three digit number divisible by 5 is 995

So the series
105 +115+125.........+995

number of terms ={ (995-105)/10} +1 = 89+1=90

Sum = n/2 [2a+(n-1)d] where n=90 ,a=105 & d=10
Sum = 49500

Ans:1
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Re: Find the sum of all odd three digit numbers [#permalink] New post   27 Dec 2023, 04:10
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Find the sum of all odd three digit numbers that are divisible by 5.

All odd three digit numbers divisible by 5 will have 5 as the units' digit:

First number = 105
Last Number = 995
Common difference, d = 10 (as alternate multiples of 5 will be odd numbers)
Number of terms, n = (Last term - First Term)/d + 1 = \(\frac{995 - 105}{10}\) + 1 = 89 + 1 = 90
Sum = n * (First Term + Last Term)/2 = 90 * \(\frac{(105 + 995)}{2}\) = 49,500

So, Answer will be A
Hope it helps!

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Re: Find the sum of all odd three digit numbers [#permalink] New post   18 Jul 2021, 07:39
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The first odd three-digit number divisible by 5 is 105.

The last odd three-digit number divisible by 5 is 995.

Number of terms: \(\frac{995 - 105 }{ 10} + 1 = 90\)

Sum = \(\frac{n}{2 }[a + l]\)

=> Sum = \(\frac{90}{2} [105 + 995]\)

=> Sum = 49,500

Answer A
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Re: Find the sum of all odd three digit numbers [#permalink]
18 Jul 2021, 07:39
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