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K = sum of the integers from 1 to 500 inclusive that are div
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09 Dec 2018, 13:31
09 Dec 2018, 13:31
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K = sum of the integers from 1 to 500 inclusive that are divisible by 5.
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
K |
25,000 |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: K = sum of the integers from 1 to 500 inclusive that are div
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09 Dec 2018, 23:09
09 Dec 2018, 23:09
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first integer divided by 5 is 5 itself
last integer divided by 5 up till 500 is 500 itself
There are in total\(\frac{500-5}{5}\) \(+ 1 = 100\) integers between 5 and 500 inclusive that are divisible by 5
Hence sum of these integers = \(\frac{5+500}{2} * 100 = 25250\)
last integer divided by 5 up till 500 is 500 itself
There are in total\(\frac{500-5}{5}\) \(+ 1 = 100\) integers between 5 and 500 inclusive that are divisible by 5
Hence sum of these integers = \(\frac{5+500}{2} * 100 = 25250\)
Re: K = sum of the integers from 1 to 500 inclusive that are div
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24 Sep 2021, 11:29
24 Sep 2021, 11:29
amorphous wrote:
first integer divided by 5 is 5 itself
last integer divided by 5 up till 500 is 500 itself
There are in total\(\frac{500-5}{5}\) \(+ 1 = 100\) integers between 5 and 500 inclusive that are divisible by 5
Hence sum of these integers = \(\frac{5+500}{2} * 100 = 25250\)
last integer divided by 5 up till 500 is 500 itself
There are in total\(\frac{500-5}{5}\) \(+ 1 = 100\) integers between 5 and 500 inclusive that are divisible by 5
Hence sum of these integers = \(\frac{5+500}{2} * 100 = 25250\)
there is any kind of trik to sole these type proble.
