Questions on Word Problems: Numbers, consecutive odd/even, digits answered by real tutors!

    Is it true that for every k-digit positive integer n, 
    there is a power of 2 whose first k digits coincide with the number n.

The statement is true. [1]  

For every k-digit positive integer n, there is a power of two whose first digits coincide with the number n. 
This is a consequence of the fact that log_{10}(2) is an irrational number, which implies that the fractional parts 
of m*log_{10}(2) are uniformly distributed in the interval [0, 1) (by the Kronecker's Approximation Theorem). [1, 2, 3]  


Understanding the condition 

For a positive integer P to be the leading digits of a power of two, say 2^m, it must satisfy the inequality: 
   P times 10^d <= 2^m < (P+1) times 10^d 

for some integer d >= 0. This means 2^m has the same leading digits as P, followed by d more digits. 

Answer: 

The statement is true because any finite sequence of digits can be the leading digits of a power of two. 
This relies on the irrationality of log_{10}(2) and the uniform distribution of the fractional parts of its multiples. [1, 5]  

References

[1] https://www.quora.com/Prove-that-for-any-natural-number-n-there-esists-a-power-of-2-lets-say-2-k-such-that-2-k-in-its-decimal-expression-is-1-followed-by-n-zeros-and-eventually-other-digits-For-example-for-n-2-2-196-works-How-to-prove-it
[2] https://www.antonellaperucca.net/didactics/Powers-of-2.pdf
[3] https://math.stackexchange.com/questions/1370645/using-kroneckers-theorem-can-we-prove-theres-some-power-of-two-yielding-a-numb
[4] https://math.stackexchange.com/questions/328655/proving-prime-p-divides-binompk-for-k-in-1-ldots-p-1
[5] https://www.reddit.com/r/askmath/comments/18o86pr/is_it_true_that_for_any_positive_integer_n_there/


The link to this  Google  AI  response is

https://www.google.com/search?q=Is+this+statement+true%3F+For+every+k-digit+positive+integer+n%2C+there+is+a+power+of+two+whose+first+n+digits+coincide+with+the+number+n.&rlz=1C1CHBF_enUS1071US1071&oq=Is+this+statement+true%3F+For+every+k-digit+positive+integer+n%2C+there+is+a+power+of+two++whose+first+n+digits+coincide+with+the+number+n.&gs_lcrp=EgZjaHJvbWUyBggAEEUYOTIHCAEQIRiPAjIHCAIQIRiPAjIHCAMQIRiPAtIBCTE3OTZqMGoxNagCCLACAfEFNet2ZYDKsLI&sourceid=chrome&ie=UTF-8

of Dec.4, 2025.

    Find the degree of number 2 which starts 129.

     = 1.29807E+33.

If one number is x, then the other number is  .


Hence, the sum of the numbers is  (x + ).


Thus the function they ask for is  sum(x) = x + .    ANSWER

Let x be a number.


Write inequality as you read the problem

    2x + 5 > -17.


Simplify step by step

    2x > -17 - 5,

    2x > -22,

     x > ,  or,  which is the same,  x > -11.


ANSWER.  The solution to the given inequality is the set of real numbers  x > -11.

Let x be the number of students per van, and

let y be the number of students per bus.



Write equations as you read the problem

    10x +  8y = 394,    (1)

     5x + 10y = 365.    (2)


Solve the system by the Elimination method.


For it, multiply equation (2) by 2 (both sides).  Keep equation (1) as is

    10x +  8y = 394,     (1')

    10x + 20y = 730.     (2')


From equation (2') subtract equation (1').  You will get

          20y - 8y = 730 - 394,

              12y  =    336

                y  =    436/12 = 28.


Now from equation (1) find x, substituting x = 28 there

    10x + 8*28 = 394,

    10x = 394 - 8*28 = 170,

      x              = 170/10 = 17.


ANSWER.  17 students in each van and 28 students in each bus.