Can anyone solve this?
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Solution :
-> The sum of all natural numbers between 200 to 600 =
( 200 + 201 + ... + 600 ) = (401)(800) /2 = 160400
-> Let A = Sum of Numbers divisible by '8' = ( 200 + 208 + ... + 600 )
= (51)(800)/2 = 20400
-> Let B = Sum of Numbers divisible by '12' = ( 204 + 216 + ... + 600 )
= (34)(804) /2 = 13668
--> Sum of numbers added twice = A∩B
= Sum of numbers divisible by '24' = 6936
=> Total Sum of numbers between 200 and 600 ( both exclusive ) divisible by '8' and '12' = A∪B = ( A + B ) - A∩B = ( 20400 + 13668 - 6936 ) = 27132
=> Total Sum of numbers between 200 and 600 ( both exclusive ) NOT divisible by '8' and '12' = 160400 - 27132 = 133268 √√
-> The sum of all natural numbers between 200 to 600 =
( 200 + 201 + ... + 600 ) = (401)(800) /2 = 160400
-> Let A = Sum of Numbers divisible by '8' = ( 200 + 208 + ... + 600 )
= (51)(800)/2 = 20400
-> Let B = Sum of Numbers divisible by '12' = ( 204 + 216 + ... + 600 )
= (34)(804) /2 = 13668
--> Sum of numbers added twice = A∩B
= Sum of numbers divisible by '24' = 6936
=> Total Sum of numbers between 200 and 600 ( both exclusive ) divisible by '8' and '12' = A∪B = ( A + B ) - A∩B = ( 20400 + 13668 - 6936 ) = 27132
=> Total Sum of numbers between 200 and 600 ( both exclusive ) NOT divisible by '8' and '12' = 160400 - 27132 = 133268 √√
Yuichiro13:
Understood ??
Sorry if there's a little calculation mistake here or there ^^"
What are the values of n in the second and third steps that you have taken?
Values of 'n' ?? I skipped the part to avoid TIME DELAYS and bigger soln . ^^"
Then how did you find 51 and 34? Explain please
'51' and '34' are the number of 'TERMS' divisible by that particular 'number { 8 / 12 / 24 } falling with the range of 200 - 600 !!
Ok. I understood. Thanks a ton!!
A ton.. a thousand billion ... a monk or a nun ... its just so fun :p
:p Anytime :) I do maths just forrrrr funnnnnnnnnnn Yayyyyyyyyyyyyy !!!!!!!
:))
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