Find the number of positive integers < 1 0 0 0 that can be expressed as 2 k − 2 m , where k and m are non-negative integers.
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Note that if k = 1 0 , then 2 m > 2 4 so that 2 k − 2 m < 1 0 0 0 . Therefore m ≥ 5
If k ≤ 9 , then 0 ≤ m < k .
Since there are k different possibilities for m if 0 ≤ m < k , then the total number of possible ways for k ≤ 9 is 1 + 2 + ⋯ + 8 + 9 = 4 5 .
We add the possibilities where k = 1 0 . Since 5 ≤ m < 1 0 , there are 5 ways.
Therefore, the total number of ways is 4 5 + 5 = 5 0 .
You need to prove that the values with k ≤ 9 do not overlap.
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Does it have anything to do with the fact that for each 2 k − 2 m , you can factor out 2 m and have a unique 2 k − m − 1 for each m in [ 1 , 1 0 ] and each individual k ? After starting the prime factorizations for all of the values for 2 k − 2 m , you would have 2 m and some other number not divisible by 2 .
m and k are non negative. 2^k - 2^m will be positive integers if k>m. m is non negative so if: m=0, k={1,2,...,9}. there are 9 solutions. m=1, k={2,3,...,9}. there are 8 solutions. m=2, k={3,4,...,9}. there are 7 solutions. m=3, k={4,5,...,9}. there are 6 solutions. m=4, k={5,6,...,9}. there are 5 solutions. m=5, k={6,7,...,10}. there are 5 solutions. m=6, k={7,8,...,10}. there are 4 solutions. m=7, k={8,9,10}. there are 3 solutions. m=8, k={9,10}. there are 2 solutions. m=9, k={10}. there is 1 solution. if m>9, then k>10, 2^k - 2^m is larger than 1000. so there is no solutions for m>9. the answer is 9+8+7+6+5+5+4+3+2+1=50.
since the positive integer required is less than 1000,
maximum value of k = 10 and corresponding values of m range from 5 to 9 value of k = 9 and corresponding values of m range from 0 to 8 value of k = 8 and corresponding values of m range from 0 to 7 value of k = 7 and corresponding values of m range from 0 to 6 value of k =6 and corresponding values of m range from 0 to 5 value of k = 5 and corresponding values of m range from 0 to 4 value of k = 4 and corresponding values of m range from 0 to 3 value of k = 3 and corresponding values of m range from 0 to 2 value of k = 2 and corresponding values of m range from 0 to 1 value of k = 1 and corresponding values of m =0
totally we get 50 numbers :)
As 2^k - 2^m should be positive, then we can make cases and calculate the total number of integers.
For m=1, k can be {2,3,4,5,6,7,8,9}
m=2, k can be {3,4,5,6,7,8,9}
m=3, k be {4,5,6,7,8,9}
m=4, k be {5,6,7,8,9}
m=5, k be {6,7,8,9,10}
m=6, k be {7,8,9,10}
m=7, k be {8,9,10}
m=8, k be {9,10}
m=9, k be {10}
Adding all these cases
Therefore Total number of cases = 50
The largest power of 2 below 1000 is 9 as 2^9 = 512 and 2^{10} = 1024 Fixing k = 9, m = {8,7,6 ... 2,1,0} i.e. 9 solutions as 0 < 2^9 -2^m < 1000
similarly fixing k = {8,7,6,5,4,3,2,1} we get 8,7,6,5,4,3,2,1 solutions respectively adding no. of solutions given above = 45 Now let k =10 Then 2^10 - 2^m <1000 or 1024 > 2^m > 24 It follows 2^m =32, 64, 128, 256, 512 or m = {5,6,7,8,9} 5 solutions Total solutions = 45+5 = 50
In order for our number N 2 k − 2 m = N to be positive then k > m .
So if k=1 then m has to be 0 ⟶ 1 solution (notice the difference in the wording between non-negative and positive) if k=2 then m can either be 1 or 2.... ⟶ 2 solutions Therefore we have k solutions for every k value. if k=10 then 2 1 0 = 1 0 2 4 so m can't be 2 0 , 2 1 , 2 2 , 2 3 , 2 4 or otherwise N>1000
Since we have k solutions for every k value, and there are 9 values of k (plus the solutions for k=10 which are 5) ; then the number of solutions are \frac{9+1}{2}\dot 9}\=45} (sum of algebraic sequence) numer of solution from k=1 to k=9 \(\longrightarrow 45 + numer of solution from k=10 ⟶ 5 = 50 (45+5=50)
You need to prove that the values with k ≤ 9 do not overlap.
sorry for the mistake
In order for our number N 2 k − 2 m = N to be positive then k > m .
So if k=1 then m has to be 0 ⟶ 1 solution (notice the difference in the wording between non-negative and positive) if k=2 then m can either be 1 or 2.... ⟶ 2 solutions Therefore we have k solutions for every k value. if k=10 then 2 1 0 = 1 0 2 4 so m can't be 2 0 , 2 1 , 2 2 , 2 3 , 2 4 or otherwise N>1000
Since we have k solutions for every k value, and there are 9 values of k (plus the solutions for k=10 which are 5) ; then the number of solutions are 2 9 + 1 9 ˙ = 4 5 (sum of algebraic sequence) numer of solution from k=1 to k=9 ⟶ 45 + numer of solution from k=10 ⟶ 5 = 50 (45+5=50)
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