+49 1.How many distinct subsets of the set S=1,8,9,39,52,91 have odd sums? (As an example, 1,8,52 is one such subset, because 1+8+52 equals 61, which is odd.)
2.How many distinct subsets of the set S=1,8,9,39,52,91 have three-digit sums? (As an example, 39,91 is one such subset, because 39+91 equals 130, which has three digits.)
3.Consider the set S=1,2,3,4,5,6,7,8,9,12,13,14,15,16,17,18,19,23,24,. . . . ,123456789\}, which consists of all positive integers whose digits strictly increase from left to right. This set is finite. What is the median of the set?
4.How many paths of minimum length are there from A to b in the grid below?
5.Consider the sequence 1,3,4,9,10,12,13,... which consists of every positive integer that can be expressed as a sum of distinct powers of 3. What is the 75th term of this sequence?
THANK YOU SO MUCH
+128408 4.How many paths of minimum length are there from A to b in the grid below?
Notice that if we consider a move downward to the right as "Southeast" = "SE" and a move upward to the right as "Northeast" = "NE," we have the following possible set of moves from A to B :
( SE,SE, SE, NE, SE, NE, NE, NE )
So...the total number of possible minimum paths is given by choosing any four of eight positions in the set for "SE" (or, alternatively, choosing any four of eight positions in the set for "NE" ) =
C(8, 4) = 70 possible minimum paths
+198 I also need help on the first two..... Also, isnt 4 and 5 from AoPS homework?
+128408 5.Consider the sequence 1,3,4,9,10,12,13,... which consists of every positive integer that can be expressed as a sum of distinct powers of 3. What is the 75th term of this sequence?
Note the pattern that emerges :
3^0 = 1 1
3^1 = 3 3^0 + 3^1 = 4 1 1
3^2 = 9 3^2 + 3^0 = 10 3^2 + 3^3 = 12 3^2 + 3^1 + 3^0 = 13 1 2 1
Note that the number of terms - 7- is just the sum of the first 2 rows of Pascal's Triangle [ the first entry is "Row 0" ]
Note that the next terms are
3^3 = 27
3^3 + 3^0 = 28
3^3 + 3^1 = 30
3^3 + 3^1 + 3^0 = 31
3^3 + 3^2 = 36
3^3 + 3^2 + 3^0 = 37
3^3 + 3^2 + 3^1 = 39
3^3 + 3^2 + 3^1 + 3^0 = 40
There are 8 terms which is the sum of the elements of the 3 row of Pascal's triangle = 1 3 3 1
And each row adds 2^n more terms [ where n is the row number ]
So...after the 4th row we have 15 + 2^4 = 15 + 16 = 31 terms
After the 5th row we have 31 + 2^5 = 31 + 32 = 63 terms
The first term in the 6th row will represent 64 term ....this will be = 3^6 = 729
Notice that the next few terms are
3^6 + 3^0 = 730
3^6 + 3^1 = 732
3^6 + 3^1 + 3^0 = 733
3^6 + 3^2 = 738 ....
And note that the terms in each case appended after the first term just follow the pattern of the first 15 terms
So...the 75th term will be the sum of this first term plus the 11th term in the above series [ 31]
So....the 75th term is 729 + 31 = 760
