but, in my school, the hardest is math+social studeis.
if people that r really smart, they can solve problems so easily. in my class, only 9 students including me is 9 total got to his accelerated math class. one of the problem that we got from him is soo easy, it took us 2 min to figure it out. but for our accelerated teacher, he ONLY took 1 min or less to figure it out. if u people think ur so smart, try it: How many positive threedigit integers are divisible by both 11 and 5? p.s. there r only 1 answer to this problem. 



Eh 16 numbers. Here I'll even list them for you. Of course I might be wrong, but the patterns there.
220, 330, 440, 550, 660, 770, 880, 990, 165, 275, 385, 495, 605, 715, 825, 935 In all seriousness, I wasn't here to offend you. I was just pointing out that although you may be good at math for a 13 yr old. There are people here who have taken levels far beyond that. 



According to your question you gave us the answer. I think you should reword it a little differently.


Just came back... The nostalgia!


^^ isnt it 17? i think ur missing 110. wen u divide (100099)/55 = 16.3, u add 16 to the 1st 3 digit number... =p




^ Yepp missed the first one when I was recounting. I knew I messed up somewhere.




all right then.. if u all want a real challenge, try solving this... i solved it wen i was 12.. my math teachers couldnt even solve this without using calculus... but all u really need is geometry and algebra...
find the area of the shaded region. it is the intersection of the four quartercircles with radius 10. pm me if u think u have the answer, or if u think u need a hint... give it to ur math teachers too n see if they can solve it! 



sure. r those only the measurements that is in there?
o, and also, wushukid, ur answer to my problem is rite. it is 17. how much minutes did u take? 



wushukid wrote: all right then.. if u all want a real challenge, try solving this... i solved it wen i was 12.. my math teachers couldnt even solve this without using calculus... but all u really need is geometry and algebra... find the area of the shaded region. it is the intersection of the four quartercircles with radius 10. pm me if u think u have the answer, or if u think u need a hint... give it to ur math teachers too n see if they can solve it! Omg this feels like the SATs all over again. At least this is remotly challenging... Imma try and solve it w/o calculus 



fahrenheitSHE1314 wrote: sure. r those only the measurements that is in there? o, and also, wushukid, ur answer to my problem is rite. it is 17. how much minutes did u take? yup.. those are the only measurements... and i already saw jamehze's answer b4 trying to solve it so it took like less than a minute to spot the mistake... 



Why is everyone going off topic here?
If you want a thread that talks about outsmarting one another with different math problems, might as well make one. As far as this thread goes, it's about playing a game of numbers. Allow me to restart it: 2#'s whose sum is 38 product is 312 

My thoughts on Anime @ www.animananime.com


12 & 26
2#s whose sum is 72 product is 855 

void n devoid similar yet different contradictory dats wat.....


I don't understand~~~~




^ The person above gives you a sum to solve.
The answer to the question that qwevfr gave would be 15 and 57 15 and 57 gives a sum of 72 when added together and when you product [times] them, you get 855. Sorry if that was too patronising an explanation. 2 numbers whose sum is 58 product is 672 



I\'m back


^50 and 9
Start with the number of planets in the solar system, multiply by the number of Australian States, then add my age (as it is today) and finally subtract the number of yards in a mile. 

I\'m back
