Let a 1 , a 2 , a 3 , a 4 , … be a sequence of all numbers which can be expressed as sums of different non-negative integer powers of 4 (which is the form of 4 k 1 + 4 k 2 + 4 k 3 + … + 4 k p , k m = k n , m , n = 1 , 2 , 3 , … , p , m = n where k is a non-negative integer) arranging from the smallest to the largest.
The first few terms are as follows, a 1 = 4 0 = 1 a 2 = 4 1 = 4 a 3 = 4 0 + 4 1 = 5 a 4 = 4 2 = 1 6 a 5 = 4 0 + 4 2 = 1 7 …
Find the value of a 6 4 .
Note: The k i need not to be the same for different a j .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Note that a i expressed in base-4 is the same as i expressed in binary (base-2). Problem solved.
We notice that 6 4 = 2 6 ⟹ a 6 4 = 4 6 = 4 0 9 6
Let a n = b 0 ˙ 4 0 + b 1 ˙ 4 1 + b 2 ˙ 4 2 + . . . + b p ˙ 4 p , where b i = 0 or 1 are digits of n in base 2 ((n) {(10)} = \overline {b p ...b 2 b 1 b_0 _ {(2)}}.
For example:
n = 1 ( 1 0 ) = 1 ( 2 ) ⇒ a 1 = b 0 ˙ 4 0 = 1 ˙ 1 = 1
n = 2 ( 1 0 ) = 1 0 ( 2 ) ⇒ a 2 = b 0 ˙ 4 0 + b 1 ˙ 4 1 = 0 ˙ 1 + 1 ˙ 4 = 4
n = 3 ( 1 0 ) = 1 1 ( 2 ) ⇒ a 3 = 1 ˙ 4 0 + 1 ˙ 4 1 = 5
n = 4 ( 1 0 ) = 1 0 0 ( 2 ) ⇒ a 4 = 0 + 0 + 1 ˙ 4 2 = 1 6
n = 5 ( 1 0 ) = 1 0 1 ( 2 ) ⇒ a 5 = 1 + 0 + 1 6 = 1 7
Therefore, n = 6 4 ( 1 0 ) = 1 0 0 0 0 0 0 ( 2 ) ⇒ a 6 4 = 4 6 = 4 0 9 6
Problem Loading...
Note Loading...
Set Loading...
Let A be a set of numbers which can be expressed as 2 m where m is a non-negative integer and B be a set of numbers which can be expressed as 4 n where n is a non-negative integer. Then, A = { 2 0 , 2 1 , 2 2 , 2 3 , … } B = { 4 0 , 4 1 , 4 2 , 4 3 , … } We can establish a correspondence relationship between the two sets: 2 0 ↔ 4 0 2 1 ↔ 4 1 2 2 ↔ 4 2 … So, a 1 = 4 0 ↔ 2 0 = 1 a 2 = 4 1 ↔ 2 1 = 2 a 3 = 4 0 + 4 1 ↔ 2 0 + 2 1 = 3 a 4 = 4 2 ↔ 2 2 = 4 a 5 = 4 0 + 4 2 ↔ 2 0 + 2 2 = 5 … Notice that the corresponding values are always the same as the term number, hence 6 4 = 2 6 ↔ 4 6 = 4 0 9 6 = a 6 4 So, the value of a 6 4 is 4096.