find the sum of all three digit natural numbers, which are divisible by 3 and not divisible by 6.
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(1)st condition : Divisible by 3 will be multiples of 3 = 3,6,9,12, 15,18,21,24,27,30,33………
(2)nd condition: Not divisible by 6,
Divisible by 6 will be multiples of 6 = 6,12,18,24,30,36,42,48, 54 60, 66…….
So, we conclude from the above that, all odd multiples of 3 are divisible by 3 but not divisible by 6.
And all even multiples of 3 are divisible by 3 as well as 6
So REQUIRED NUMBERS are to be odd multiples of 3 …
(3)rd condition: numbers should have 3 digits
Since the smallest 3 digit number is 100.
100÷ 3, quotient = 33, remainder not equal to 0.
So, we start with 35th multiple of 3 ( skip 34th multiple, as we explained above even multiple of 3 not to be cosidered)
Now, take 35th multiple onwards
3 x 35 = 105
3 x 37 = 111
3 x 39 = 117
……………..
3 x 333 = 999
So the series is..
105,111,117 ……………. 999
This is an AP , first term a = 105, common difference d = 6
Tn = a + (n-1)d
=> 105 + ( n-1) 6 = 999
=> 105 + 6n -6 = 999
=> 6n = 999 - 99
=> n = 900/6
=> n = 150………(1)
Now the sum of these 3 digit natural numbers
= Sn = n/2 ( a + l)
=> S = 150/2 ( 105 + 999)
=> S = 75 x 1104
=> S = 82800 ●●●●●●●●ANS
(2)nd condition: Not divisible by 6,
Divisible by 6 will be multiples of 6 = 6,12,18,24,30,36,42,48, 54 60, 66…….
So, we conclude from the above that, all odd multiples of 3 are divisible by 3 but not divisible by 6.
And all even multiples of 3 are divisible by 3 as well as 6
So REQUIRED NUMBERS are to be odd multiples of 3 …
(3)rd condition: numbers should have 3 digits
Since the smallest 3 digit number is 100.
100÷ 3, quotient = 33, remainder not equal to 0.
So, we start with 35th multiple of 3 ( skip 34th multiple, as we explained above even multiple of 3 not to be cosidered)
Now, take 35th multiple onwards
3 x 35 = 105
3 x 37 = 111
3 x 39 = 117
……………..
3 x 333 = 999
So the series is..
105,111,117 ……………. 999
This is an AP , first term a = 105, common difference d = 6
Tn = a + (n-1)d
=> 105 + ( n-1) 6 = 999
=> 105 + 6n -6 = 999
=> 6n = 999 - 99
=> n = 900/6
=> n = 150………(1)
Now the sum of these 3 digit natural numbers
= Sn = n/2 ( a + l)
=> S = 150/2 ( 105 + 999)
=> S = 75 x 1104
=> S = 82800 ●●●●●●●●ANS
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2
Here's the thing bro:
To solve your problem we need to add all the numbers from 100 to 999 and then subtract the numbers that can be divided exactly by 33.
The sum of the numbers is the sum of the first 900900 terms of a progression with a=100a=100 and d=1d=1, that is 9002(200+899)=4945509002(200+899)=494550
Now we need to calculate the sum of the first ⌈999−1003⌉=300⌈999−1003⌉=300 terms of the sequence with a=102a=102and d=2d=2, that is 3002(2(102)+3(299))=1651503002(2(102)+3(299))=165150, the result is then 494550−165150=329400
the number's which are not divisible by 2 and 3 are not divisible by 6 so if the numbers are divisible by 3 then they will be divisible by 6 because the is a common point 2 between these two number's
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