Help with these Math questions

snipez90
Sat, 24 Nov 2007 20:00:15 +0000

E, 2 3 3 6 _ _ 20 satisfies conditions (only way for 3 to be mode)… least in the two blanks are 7 and 8, summing gets you 49, mean is 49/7 = 7, most in blanks could be 18 19, summing gets 71, 71/7 > 10, so all three
two ways to do this, first is to realize that n must end in 9 and go quickly divide each number with units digit 9 under 100 by 9 to see which gives you a remainder of 8, 89 works (9*9 = 81, so 89/9 has a remainder of 8)


second is more clever and quickest, add 1 to n and it will be divisible by 9 and 10, the lowest common multiple of 9 and 10 is 90 since they share no common factors. subtract 1 from n to get 89


9 possibilities, not hard by checking each case, 410 works, 411 does not, 412 does not. 5 is going to work obviously, 610 works, 611 does not, 612 does not => 5/9
the bottom left side that’s not given is 1 (Why?). split the figure right down the middle from top to bottom to get a square and rectangle. the area of the rectangle is 9/4 - 1 = 5/4. Since one side of this rectangle is 1, the only unknown side that’s left (to the right) is 5/4. Thus the perimeter is 5/4+5 = 25/4