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vaibhav206
@vaibhav206
June 2022
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plzzz help me to solve
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ani99ket
Multiples of 3 are 3,6,9,12,15,18,21,24....
multiples of 4 are 4,8,12,16,20,24,...
so numbers which are divisible both by 3 and 4 are
12,24,36... which is an Ap of 12.
first 3 digit divisble by 12 is 108
and last digit divisble by 12 can be found out as
1000/12 = 250/3 = 83.something
thus 12 *83 is the last 3 digit divisible by 12.
so the sum is
108+....+12*83
taking 12 common
12(9+10+11+.....+83)
now sum of Ap is n*(t1 + tlast)/2
= 12*[(9+83)/2]*75
= 6*75*92
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Answers & Comments
multiples of 4 are 4,8,12,16,20,24,...
so numbers which are divisible both by 3 and 4 are
12,24,36... which is an Ap of 12.
first 3 digit divisble by 12 is 108
and last digit divisble by 12 can be found out as
1000/12 = 250/3 = 83.something
thus 12 *83 is the last 3 digit divisible by 12.
so the sum is
108+....+12*83
taking 12 common
12(9+10+11+.....+83)
now sum of Ap is n*(t1 + tlast)/2
= 12*[(9+83)/2]*75
= 6*75*92