Victor Amelkin

Hello Erin, Could you please delete my account? [Confirmation] Do I really want to delete my account? Yes, I do. Thanks in advance.

-- 1 --------------------------------------------------------------------- k = 6n + 5, where n is an integer 2k = 2(6n + 5) = 12n + 10 = 6(2n + 1) + 4 = 6c + 4, where c is an integer => remainder = 4 -- 2 --------------------------------------------------------------------- The only prime factor of 3^n (where n is a positive integer) is 3; of 5^n – is 5 3 B -- 3 --------------------------------------------------------------------- Of the following which is most nearly equal to 2/3? a) 3/4 b) 5/6 c) 7/9 d) 11/15 e) 15/21 A) 3/4 - 2/3 = 9/12 - 8/12 = 1/12 B) 5/6 - 2/3 = 5/6 - 4/6 = 1/6 C) 7/9 - 2/3 = 7/9 - 6/9 = 1/9 D) 11/15 - 2/3 = 11/15 - 10/15 = 1/15 E) 15/21 - 2/3 = 15/21 - 14/21 = 1/21 => 15/21 deviates from 2/3 less than other numbers. -- 4 --------------------------------------------------------------------- 1) 1st day of the month is Monday: 4th Tuesday = 2 (mon->tue) + 3 * 7 (3 full weeks) = 23 2) 1st day of the month is Tuesday: 4th Tuesday = 1 (tue->tue) + 3 * 7 (3 full weeks) = 22 3) 1st day of the month is Wednesday: 4th Tuesday = 7 (wed->tue) + 3 * 7 (3 full weeks) = 28 4) 1st day of the month is Thursday: 4th Tuesday = 6 (thu->tue) + 3 * 7 (3 full weeks) = 27 5) 1st day of the month is Friday: 4th Tuesday = 5 (fri->tue) + 3 * 7 (3 full weeks) = 26 6) 1st day of the month is Saturday: 4th Tuesday = 4 (sat->tue) + 3 * 7 (3 full weeks) = 25 7) 1st day of the month is Sunday: 4th Tuesday = 3 (sun->tue) + 3 * 7 (3 full weeks) = 24 => 4th Tuesday may be on 22th - 28th (7 different dates) -- 5 --------------------------------------------------------------------- $10,000 -> 1) 1st portion: n$ (invested at 4% per year) 2) 2nd portion: (10,000 - n)$ (invested at 6% per year) Annual interest = 4/100 * n + 6/100 * (10,000 - n) = 520 Multiplying by 100 both parts of the latter equation: 4n + 6(10,000 - n) = 52,000 4n + 60,000 - 6n = 52,000 2n = 8,000 n = 4,000 L: 4,000$ R: 10,000$ - 4,000$ = 6,000$ L B

Oatmeal, if you experience problems with counting, it may be easier for you to visualize such a problems with pairs of number: [ATTACH=CONFIG]5819[/ATTACH] http://www.www.urch.com/forums/images/misc/pencil.png

I believe the problem is simpler that you're trying to make it. -- The probability that the test is successfully passed: p = m / n m - number of ways to successfully pass the test n - number of ways to do anyhow on the test 1) There are ten questions each of which may be solved either correctly or incorrectly. Hence, n = 2 * 2 * ... * 2 = 2^10 2) To pass the test successfully, we are allowed to make 0, 1, or 2 errors. Hence, m = C{10,0} + C{10,1} + C{10,2} => p = m / n = (C{10,0} + C{10,1} + C{10,2}) / 2^10 = 7 / 128

Until the contrary is specified, the order of books should not matter.

> Though the solution looked easy... It is easy only if you have an off-the-shelf formula for the standard deviation. Otherwise, you'll have to spend 20min deriving it. -- By the way, the formula provided above may be a bit extended to cover an n-item chunk of any arithmetical progression. a[i + 1] = a + d Series (n items or an arithmetical progression a beginning from the item a): a, a[i + 1], ..., a[i + (n-1)] std_dev = d * sqrt(1/12 * (n^2 - 1)); which is d times greater than the std_dev for a natural series, which, in turn, is quite expected because a chunk of an arithmetical progression may be derived from a natural series by multiplying elements of the later on d (which will increase std_dev d times) and shifting all elements by a (which does not change std_dev). Example: 3, 5, 7 - memebers of an arithmetical progression with d = 2 (1) direct calculation: avg = 5 std_dev = sqrt(1/3 * ((3 - 5)^2 + (7 - 5)^2)) = sqrt(8/3) (2) workaround: std_dev = d * sqrt(1/12 * (n^2 - 1)) = 2 * sqrt(1/12 * (3^2 - 1)) = sqrt(8/3)

Fahad, treat the square root as always non-negative: sqrt(a^2) = |a| Examples: # sqrt(0) = 0 # sqrt(2^2) = |2| = 2 # sqrt((-2)^2) = |-2| = 2 # sqrt((5 - 8)^2) = sqrt((-3)^2) = |-3| = 3 # sqrt((2 * x + 1)^2) = |2 * x + 1| = { 2 * x + 1, if x >= -1/2, - 2 * x - 1, if x } # sqrt((sin(x) + 1)^2) = |sin(x) + 1| = sin(x) + 1 (since (sin(x) + 1) is always non-negative)

Here is an example with a reduced number of books, images, and image slots: * 2 books; each has 2 image slots * 3 images: A, B, C All possible 2-image sets for one book are: A B A C B A B C C A C B (Their number is C{3, 2} * 2! = 6) Since there are two books, we are interested in the number of pairs of pairs. This number is 6 * 6 = 36 (each of provided 6 pairs may be combined with each of the same 6 pairs). 6 * 6 != 6 + 6 But maybe I'm missing something. Feel free to explicate your view on the problem.

avg = (1 + n) / 2 deviation^2 = (i - avg)^2 = (i - (1 + n)/2)^2 std_dev^2 = 1/n * sum(deviation^2, i=1..n) = 1/n * sum((i - (1 + n)/2)^2, i=1..n) = (summation of this series takes a while) = 1/12(n^2 - 1) std_dev^2 = 1/12(n^2 - 1) = 14 => n = 13 P.S. Such a question cannot appear on GRE (and, I believe, on any other standardized test). On the other hand, they say that there were questions like "the number of primes between 3000 and 3500" on GRE...

I. For one book: C{5, 3} = 10 II. For three books: C{5, 3}^3 = 1000 Whether I or II is asked for is unclear from the problem's formulation.

Could you please comment why it's March but not April? In March and April, Domestics are equal, but April's Import is less than the March's one. => Domestic/Import should be greater in April?! Am I missing something?

sueka, you're right, I indeed missed some non-prime factors. :-[

(s - t)^2 = 16, t = 2 (s - 2)^2 = 16 |s - 2| = 4 1) s >= 2: s - 2 = 4 => s = 6 2) s s = -2 => s and 6 are incomparable (D). sueka, there are no specific ETS rules on square roots extraction that would differ anyhow from traditional mathematics. sqrt(x^2) = |x|, where x is a real variable sqrt(a^2) = a, where a^2 is a non-negative real constant P.S. To be precise, a square root of a non-negative real constant may be negative by definition of a square root, but traditionally under "a square root of a non-negative real constant" we should understand "a principal square root", that is only a positive one. Thus, if it does not explicitely stated that a square root of a non-negative real constant may be negative, you should consider it to be positive.

C{n, m} does not take into accout order of elements. For instance, C{64, 2} count two-element sets (1, 2) and (2, 1) as one "combination" (the difference between "combinations" and "permutations" is exactly the lack or presence of order). Thus, you're counting only a half of all possible positions of two figures on the field. Instead, you should count the number of "permutations" A{n, m} = C{n, m} * m! = n! / (n - m)! which will produce 4032 instead of 2016 and lead you to the correct answer. As to card games, the order of cards you're holding is rarely significant (at least, I, an inveterate card player, have not seen such a game where the order of cards matters); what is significant is the "combination" you have. Hence, when dealing with cards, you're likely to use C{n, m}, without taking into account the order of cards.

N = 231 = 3 * 7 * 11 – 5 factors (1, 3, 7, 11, 231) 2N = 462 = 2 * 3 * 7 * 11 – 6 factors (1, 2, 3, 7, 11, 462) 6 = 5 + 1 != 5 * 2 2N = (1 + 1) * factor[1] * factor[2] * ... factor[M] = factor[1] * factor[2] * ... factor[M] + factor[1] * factor[2] * ... factor[M] You cannot simply count the elements in the last representation and proclaim all of them "factors" and their tally (including repetitions) "the number of factors", since there is an operation of addition involved, while a factorization involves only multiplications.