what is the greatest number that will divide 2930 & 3250 and will leave reminders 7 and 11?  its 79 ok.\n\nbut can someone explain me how to get it.
135215
Phew, hard question.\n\nSTEP 1: minus the remainders from the numbers. you would get 2923 and 3239 respectively.\n\nSTEP 2: this is the hard part. take both the numbers individually and take their factors.\n\nfor 2923 you wold get 37*79, and for 3239 you would get 41*79. you get this by painstaikingly checking with each prime number whether or not the number is divisible by it or not.\n\n37 is the 12th prime number (excluding "1") and 41 is the 13th (again excluding "1"). This means you have to do 12+13=25 calculations. then divide the main number you got from STEP 1 with 37 and 41 respectively, and you will get 79 in both case.\n\nSTEP 3: Notice that 79 is common in both the cases. This is the greatest common number.\n\nIf there were more than one common number, you would just multiply all the common numbers to get the answer.\n\nHAVE FUN
391781
Use this little jingle that my Uncle Ed taught me:\n\n5 times six is thirty-six!\n8 time eleven is a twenty-two!\n4 times pi is a fist in the eye!\n3 times 3 is ze-e-e-e-ro!\n\nThere is a second verse, but I can't remember it.\n\nEmail me if you need help with division.\n\nJohn Z
497740
This question has been asked so many times!?!?!\n\nGood challenge question!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nSomeone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
408290
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
104991
paul67337 wrote:\n"228 divided by 12 is 19.\n19 is a common factor in the 3 numbers. go from there and figure out the rest by trial and error."\n\nThen just multiply 3, 4 and 5 by 19 and get 54, 76 and 95.
368830
Looks like it's 50 numbers. For example , let the numbers be\n51 thru 100 inclusive. Since even the lowest one can't divide\ninto the highest, none are divisible by the other.  Now lets' suppose there is another number less than 51 in this set. Then\nthat number times 2 will be some number less than or equal to\n100 so by contradiction, there is no other number less than 51\nin the set.
144992
my claim is that the number is infinite.  here is the reason, it depends on who you classify as a mathematician and what date you want to start at.  Also i claim that you can move forward in time, which i will claim is infinite, and find more mathematicians bore in march.
659308
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
420160
Let us call the first 2 "2 sub 1", the second "2 sub 2" ... same thing for the 6's.\n\nLet us say that "2 sub 1" is distinct from "2 sub 2", and so on.\n\nWe have 8 choices for the first number, seven choices for the second number, etc. This means that there are 8*7*6*5*4*3*2*1 choices, or 8!. \n\nNow if you look at any number, take all of the 2's out. You see that you can rearrange the 2's in 3! different ways (including the original arrangement) using the procedure in the paragraph above. If we take out the 2's rearrange them, and put them back in we have created the a new arrangement. So for each arrangement of numbers, we have counted it 3! times. So we divide by 3!. The same thing for the sixes, we divide by 2!.\n\nThe result it (8!)/((3!)(2!)) = 3360 choices.
244068
add the numbers\n15\n19\n42\n19\nnext number equals 15.....     and each has a 2 and a 7 in it\n\ntry 2742 sounds like a good answer for neopets lenny conundrum answer possibility but not fully sure..... lucidlyinsane
215356
In the number: 495,703 the value of 9 is 9 X 10,000 = 90,000.\n\nActually IF I recall correctly (been a long time since 3rd grade math) the value is 10,000.\n\nI think this lesson is probably about learning the value placement in the positioning of numbers so that understanding big numbers becomes easier. But the wording of this question to a 3rd grader (AND some other people too) IS not easy to comprehend, so hopefully, I am telling this answer correctly.\n\nSounds like it's time for a parent/teacher conference.
252985
Hopefully, I'm not doing your homework for you, and you are just curious about how to find this answer. Follow along on a piece of paper so it is easier to understand.\n\nIn a 12 hour period of time, reguardless of what time you start the count, only 1:00 through 6:00 can possibly be used in order to stay within the "6" limit. Let's see how many numbers will add to 6 in 1:00. \n\n1:05, 1:50, 1:14, 1:41, 1:23, 1:32. (6 Total)\n\nNow for the rest of the hours:\n\n2:04, 2:40, 2:13, 2:31, 2:22. (5 Total)\n3:03, 3:30, 3:12, 3:21. (4 Total)\n\nSee a pattern developing?\n\n4:02, 4:20, 4:11. (3 Total)\n5:01, 5:10. (2 Total)\n6:00. (1 Total)\n\nNow add the totals to get 21 times that the digits add up to 6!\nCongratulations.
84975
There is a lot of answers\n\nSmallest: 32043^2 = 1 026 753 849\nLargest : 99066^2 = 9 814 072 356\n\n..A... ...B......\n------ -----------\n32043 1026753849 \n32286 1042385796 \n33144 1098524736 \n35172 1237069584 \n35337 1248703569 \n35757 1278563049 \n35853 1285437609 \n37176 1382054976 \n37905 1436789025 \n38772 1503267984 \n39147 1532487609 \n39336 1547320896 \n40545 1643897025 \n42744 1827049536 \n43902 1927385604 \n44016 1937408256 \n45567 2076351489 \n45624 2081549376 \n46587 2170348569 \n48852 2386517904 \n49314 2431870596 \n49353 2435718609 \n50706 2571098436 \n53976 2913408576 \n54918 3015986724 \n55446 3074258916 \n55524 3082914576 \n55581 3089247561 \n55626 3094251876 \n56532 3195867024 \n57321 3285697041 \n58413 3412078569 \n58455 3416987025 \n58554 3428570916 \n59403 3528716409 \n60984 3719048256 \n61575 3791480625 \n61866 3827401956 \n62679 3928657041 \n62961 3964087521 \n63051 3975428601 \n63129 3985270641 \n65634 4307821956 \n65637 4308215769 \n66105 4369871025 \n66276 4392508176 \n67677 4580176329 \n68763 4728350169 \n68781 4730825961 \n69513 4832057169 \n71433 5102673489 \n72621 5273809641 \n75759 5739426081 \n76047 5783146209 \n76182 5803697124 \n77346 5982403716 \n78072 6095237184 \n78453 6154873209 \n80361 6457890321 \n80445 6471398025 \n81222 6597013284 \n81945 6714983025 \n83919 7042398561 \n84648 7165283904 \n85353 7285134609 \n85743 7351862049 \n85803 7362154809 \n86073 7408561329 \n87639 7680594321 \n88623 7854036129 \n89079 7935068241 \n89145 7946831025 \n89355 7984316025 \n89523 8014367529 \n90144 8125940736 \n90153 8127563409 \n90198 8135679204 \n91248 8326197504 \n91605 8391476025 \n92214 8503421796 \n94695 8967143025 \n95154 9054283716 \n96702 9351276804 \n97779 9560732841 \n98055 9614783025 \n98802 9761835204 \n99066 9814072356
281955
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
347635
Roman please! Don't confuse the poor girl any more than she already is.  Which is  it, 23 or 24?\n  \nTat, Roman was on the right track with this one.  He had the right concept, but the wrong constant in the end.  Guess that's why the Caesars no longer rule.\n\nIf Greg now runs 35 miles today, then six weeks ago he started running 35 - 2(6) or 23 miles per week.  Letting m be the mileage run per week, and n the number of weeks he runs, then we come up with this equation for the mileage Greg runs each week he is on the program.:\n\nm = 23 + 2n.\n\nWe defy the normal convention of putting the constant at the end of the equation this time, because that was the mileage Greg started with, and everything else is added to it.
100235
1. Pick a number between 1 and 2023 \n   2\n2. Pick a number between 5 and 1338   \n   613\n3. Pick a number between 1 and 2108  \n   18\nThis is a religious forum :)
424556
Arrange your sequence of numbers from the smallest to largest...\n\nthen the middle number in your sequence is your median\n\nit's simple as that.\n\nfor example\n\nA.   6, 7, 10, 4, 9\n\nRearrange to\n\n4, 6, 7, 9, 10\n\nThe middle number in the sequence is 7 as you can see\n\nSimple :D\n\nhope it helps