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how many numbers between 25 and 400 are divisible by 11? what is their sum?

Mathematics
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who can answer?
well, it starts with the number 33
can u pls show the equation sir???

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Other answers:

400-25 = 375 possible results divide that by ll 375/11 = 34.1 so ill assume there is like 34 numbers in that
and the sum sir?
there sum can be determined by the expression 33+11n from n=0 to 34 most likely
wait sir... 33+11n=????
answer is 7480
34 11 -- 34 34 ---- 374 11 ---- 385 11 ---- 396 .... found a few more
pfft, for got to add 33 :) 374 +33 ---- 407 so it looks like 34 is too large and 33 is perfect 0 to 33
this time i am right i double checked it
can i ask for the solution sir??? because the solution is greatly needed
33+11n i s a good expression yes since the first number divisible by 11 is 33
thanks sambhav
how to get the n sir???
\[\sum_{0}^{33}33+11n=\frac{33}{2}(33+374)\]
a=33 l=396 d=11 n=(396-33/11)+1=35 sum=(35)(2(33)+34(11))/2 =7480 use the formula i gave u earlier
http://www.wolframalpha.com/input/?i=table+%2833%2B11n%29%2C+n%3D0..33
thanks to to all of you friends :D
the sum is 7293
why is sambhav answer is different?
he most likely has a small error, or i do. but ive got the wolf to dbl chk me :) http://www.wolframalpha.com/input/?i=sum+%2833%2B11n%29%2C+n%3D0..33
hey @amistre64 ur formu;lla is correctbut ∑03333+11n=332(33+374) =6715.5
yeah, 374 should have been 396 :) but i accounted for that afterwards
you used another website for help i used my mind
at my age, the mind is not so good ;)
whats the correct answer monsieurs???
i have to look at my left foot just to remember what color sock i need to put on the right foot :)
∑03333+11n=(35/2)(33+396)=7507.5
whats ur age @amistre64
.... pretty old
sambhav and amistre what again is the answer???
7293 is what i get
pls wait for a minute pls sorry for the argument between us
its just ok no hard feelings
0 to 33 is 34 numbers entotal 34(33+396)/2 = 7293
its right tnx for all your help ive got 2 numbers to go can u cooperate me or help me sir???
between the both of us, we are sure to work out all the wrinkles ;)
@bienes_joshua the write answer is 7293
much thanks of all your help i just need to finish the 2 problems i will close this question and write a new one :D
starting an index at 1 makes life easier on the count but adds a little bit of complexity to the expression; therefore a modification that is equal is:\[\sum_{1}^{34}33+11(n-1)\]

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