A L 217 Report post Posted January 31, 2014 Is there a condition it does not fulfil? Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted January 31, 2014 997= 27+970, 57+940, 87+910.. And so on. Quote Share this post Link to post Share on other sites
A L 217 Report post Posted January 31, 2014 Ah. Back to the lab again. (Lab = wasting pens) Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted January 31, 2014 ....okay..... Quote Share this post Link to post Share on other sites
A L 217 Report post Posted January 31, 2014 On 1/31/2014 at 3:46 PM, Nara-chan said: ....okay..... I'm actually a very slow person. Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted January 31, 2014 Stop spamming Quote Share this post Link to post Share on other sites
A L 217 Report post Posted January 31, 2014 On 1/31/2014 at 3:54 PM, Nara-chan said: Stop spamming Okay. I was just letting you know not to expect another answer anytime soon. :V Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted January 31, 2014 OK Quote Share this post Link to post Share on other sites
tengaku squared 291 Report post Posted February 1, 2014 Reveal hidden contents My answer is 17. It is neither a multiple of 3 or 10, nor is it a sum of a multiple of 3 and a multiple of 10. To eliminate all larger numbers, note the numbers that ARE multiples of 3 or 10, or sums of a multiple of 3 / multiple of 10, first from 10 to 20: 10 12 13 15 16 18 19 20 If you take the last three numbers, 18, 19, and 20, and add 3 to each of them, you get 21, 22, and 23. The first is a multiple of 3, the second can be represented by 10 + 12, and the third can be represented by 20 + 3. Keep adding 3 and you get 24, 25, 26. Multiple of 3, represented by 10 + 15, and represented by 20 + 6. One last time for 27, 28, 29. Once again, multiple of 3, represented by 10 + 18, and represented by 20 + 9. Adding 3 to a multiple of three simply makes another multiple of 3, adding 3 to (10 + 3x) gets (10 + 3 (x+1)), which is still a sum of a multiple of 10 and a multiple of 3, and adding 3 to (20 + 3x) gets (20 + 3 (x+1)). From this, it can be inferred that all numbers larger than 17 can be represented by either 3x, 10 + 3x, or 20 + 3x. Therefore, all numbers larger than 17 are disqualified. 3 Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 1, 2014 ^Correct Reveal hidden contents But my solution involved subtracting 10 from multiple of 3, 3 up to 27, since any number ending in 0-9 results from these numbers plus a multiple of 10. Biggest=3x9-10=17. But it's still that. x/y= 4/5+ 4/50+ 4/500+ 4/5000..... x+y=? Quote Share this post Link to post Share on other sites
PhiBrainChild 27 Report post Posted February 1, 2014 Reveal hidden contents Just saw this thread 'n' the most recent problem, so I wanted to give it a quick try. Is the answer 17? Here's the explanation: 4/5=0.8 4/50=0.08 4/500=0.008 . . . So, the final result will be 0.888... From that, we can say that x<y I've just started searching for the first two numbers that would meet these two conditions, 'n' I ended up with 8 'n' 9. 1 Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 1, 2014 ^Wow you're good. Yup. that's correct! Why don't ya post a problem? Quote Share this post Link to post Share on other sites
PhiBrainChild 27 Report post Posted February 1, 2014 I don't really have any problems to post, except for some random problems that I've solved while on the University. But that'll just kill your will to live. So, if anyone has an interestin' problem, feel free to post it in my stead. Quote Share this post Link to post Share on other sites
scudzilla 2 Report post Posted February 1, 2014 Simple problem: x is the smallest nonnegative number that satisfies the ff: 1) x-1, when divided by a number y wherin 2<=y<=10, the remainder is y-1; and 2) x+1, when divided by a number y wherin 2<=y<=10, the remainder is always 1 What is x? Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 1, 2014 Reveal hidden contents If x+1 has a remainder of 1, x has a remainder of 0. So x is divisible by y. Becuase it's the least possible number, y must be 2. And so x is either 2 or 1 (factors of 2) If x is one, then x+1=2, which does not fulfill the conditions. If x is 2, x-1=1, which has a remainder of 2-1=1 when divided by 2, and x+1=3, which has a remainder of 1 when divided by 3/ My answer: 2 But if y is less than or equal to 2 and also greater than or equal to 10, y does not exist. Quote Share this post Link to post Share on other sites
scudzilla 2 Report post Posted February 1, 2014 Oops sorry. My mistake. I meant to type 2<=y<=10 Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 1, 2014 But was my answer correct? Quote Share this post Link to post Share on other sites
scudzilla 2 Report post Posted February 1, 2014 Nope. It has to satisfy all values of y in that range. If y is 10, then the remainder of 1 (x-1) over 10 is 1, while 3 (x+1) over 10 is 3. Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 1, 2014 Ah, I see. Reveal hidden contents So x is a multiple of the GCF of 2-10, which is 2520. So x is at least 2520. I think. 1 Quote Share this post Link to post Share on other sites
scudzilla 2 Report post Posted February 1, 2014 Yup, Correct. Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 2, 2014 New question! What is the next number in.. 5, 6 1/4, 8 1/3, 12 1/2....? Quote Share this post Link to post Share on other sites
astraculpa 22 Report post Posted February 2, 2014 I'm here... And Reveal hidden contents 19 for absolutely no reason Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 2, 2014 Wrong Quote Share this post Link to post Share on other sites
astraculpa 22 Report post Posted February 2, 2014 Reveal hidden contents I am just like my twin Reveal hidden contents He'll kill me for saying this Quote Share this post Link to post Share on other sites
Nara-chan 38 Report post Posted February 2, 2014 ^Kill him first. And stop spamming Quote Share this post Link to post Share on other sites
