the resulting number is four times as large as the original number.  What is the smallest n?
123220
such a no. cannot exist but hey dude a great question lets look at it this way\nthe last digit of the given no. will be 6\nand the resulting no will be 4 times that no\nlast digit of 4*6 is 4 so the resulting nos last digit digit has to 4 \nthat would mean the first digit of the original no will  be 4\nso just as an eg let us consider the no. 46\n46*4 will be 184\nthat is the first digit will be 1(or 2 or 3 or 4 but never 6)\nthis will be followed by every no \nso u c such a no just canno exist
155498
100 which is the binary equivalent of 4 if you are considering those 2 numbers to be added binary numbers
428803
You have to look for the repeating pattern of units digits for each problem.\n\n1998^1 ends in 8\n1998^2 ends in 4  (8X8=64)\n1998^3 ends in 2\n1998^4 ends in 6\n1998^5 ends in 8 so it's starting to repeat.  It repeats every 4 so 1998^1999 is the same as 1998^3, ends in a 2.\n\nNow for 1999^1998.\n1999^1 ends in a 9\n1999^2 ends in a 1 9X9 = 81\n1999^3 ends in 9  It's already repeating\n\nSo 1999^1998 ends just like 1999^2 did, with a 1\n\nFinally subtract these two end digits since that's all you care about: 2 - 1 = 1\n\nSo it ends in a 1
51153
The next number is 30651.\nIt has a recursion, a_n=a_(n-1)^2+2*a_(n-2).
339228
smaller integer=n\nn-10=(n+2)/2\nn-10=1+n/2\nn/2=11\nn=22\nintegers are 22 & 24
20804
You need to round off the number to 4 digits and use the proper exponent to reflect the entire number:\n\n1.496 x 10^11  Your first answer is correct.
512971
s=9\ne=5\nn=6\nd=7\nm=1\no=0\nr=8\nn=6\ny=2\n\nSEND=9567\nMORE=1085\n--------\nMONEY=10652\n\nBut I've assumed that you can't use the same number to be represented by different letters.
466110
2520,\nwhich is the least common multiple of 2,3,4...,10\n2^3 * 3^2 * 5 * 7
649968
6!/4!2! = 15.\n\nIn general, the number of combinations of n objects taken k at a time is n!/k!(n-k)!.\n\nedit: Janet*, those combinations all include a.  Certainly there are combinations that don't include a, such as bcde, bcdf, and cdef.  There's your 15.
670277
4 is the correct answer
618301
it would be 22 choose 4 =22!/(4!18!) minus the ways in which Mr and Mrs Math can be together = 20 choose 2 = 20!/(2!18!)\n=(22.21.20.19)/(4.3.2)-...\n...(20.19)/(2)\n=((20.19)/2).((22.21/4.3)-1)\n=190.75/2\n=7125\nSorry had to two line one bit due to Yahoos dodgy cutting off of things
507159
1\n\njust do a little bit of trial you will see that the smaller number you use it nears 1.
414399
12,000,198 microscopical creatures\n\nit's 7 x 7 x 7 x 7 x 17 x 7 x 6 x 7 = 12,000,198\n\nBy the way, why is this in Words and Wordplay not Math?\nAm I missing something?
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.
2865
There is no smallest number.\n\nEvery number (real number, for example) can be halved.\n\nThe set of real numbers is closed under the operation of multiplication.  (See item (II)-(5) at the linked reference.)\n\nSo, since 0.5 is a real number. if X is a real number, X*0.5 is also a real number.
605833
Any small square of four numbers totals 20
62491
The problem asks for:\na natural number n wich: n = k*3+1, n=q*5+1, n=r*7+1, n=p*9+1, n=m*11. => n = cM(3,5,7,9)+1 and 11 divides n. (cM(3,5,7,9) = a common multiple of 3,5,7,9). \nYou take the less cM(3,5,7,9) of all, wich is 945, add 1, and you find out that 11 divides it.\nAnswer is:\n\n-- 946 --