Triple Double Set

Number Theory Level 4

There are triplets of positive integers satisfying $2^{x} + 2^{y} + 2^{z} = 2336$

Evaluate $\displaystyle \sum_{i = 1}^{n} (x_{i} + y_{i} + z_{i} )$

Dedicated to Brian Charlesworth
Brian Charlesworth's challenge - generalize the problem( for odds , evens and primes) , you will be appriciated

The answer is 144.

2 solutions

Pranjal Jain
Nov 19, 2014

Assume x<y<z WLOG,

$2336=2^{x}+2^{y}+2^{z}=2^{x}(1+2^{y-x}+2^{z-x})$

Note that the term $1+2^{y-x}+2^{z-x}$ is odd. So $2^{x}$ will contain max power of 2 in 2336. $2336=2^{5}×73$

So $\boxed{x=5}$

$73=1+2^{y-x}+2^{z-x} \Rightarrow 72=2^{y-x}+2^{z-x}=2^{y-x}(1+2^{z-y})$

Same as before, $72=2^{3}(1+2^{3}) \Rightarrow \boxed{y=8}$ and $\boxed{z=11}$

So the possible values of x, y, z are (5,8,11). Permuting x, y, z, there will be 3!=6 arrangements. So answer would be $6×(5+8+11)=\boxed{144}$

Nice solution, Pranjal. Note, however, that we don't know at the start that $x,y,z$ are all distinct. If the equation had been something like

$2^{x} + 2^{y} + 2^{z} = 2304$

then the solution triples would be all permutations of $7,7,11$ and $8,10,10$ .

@Megh Choksi Thanks for the dedication. :)

With regards to a generalization, we won't be able to generate odd numbers this way if we don't allow for $x,y,z$ to also be, potentially, $0$ . Then all non-negative numbers up to $2^{n} - 1$ can be represented as an $n$ -bit binary number representing a summation of powers of $2$ . So, for example, $2336$ would be $2^{5} + 2^{8} + 2^{11} \Longrightarrow 000001001001$ .

There will then be $\binom{12}{3} = \frac{12!}{9!*3!} = 220$ numbers from $0$ to $4095$ inclusive that can be represented at the sum of $3$ distinct non-negative powers of $2$ . If the powers don't need to be distinct then there are more numbers that can be represented. If two of the powers can be the same, e.g., $2^{8} + 2^{8} + 2^{11} = 2560$ , then there are $12*11 = 132$ more numbers that can be represented, and if all three powers are the same then $12$ more numbers can be represented. This gives us $364$ numbers from $0$ to $4095$ inclusive that can be represented as the sum of $3$ not necessarily distinct powers of $2$ .

(Note that when the powers are distinct then this a unique representation, i.e., there are no other sums of distinct powers of $2$ that will add to this number.)

Similarly, $\binom{12}{n}$ numbers in this range can be represented as the sum of $n$ distinct powers of $2$ .

Now $2336$ can be represented as a sum of powers of $2$ in many other ways, but there is only one way in which it can be expressed as the sum of three distinct (ascending) powers of $2$ , namely $2^{5} + 2^{8} + 2^{11}$ . If the powers do not need to be distinct then we could have as many as $2336$ powers of $2$ , all of them being the power $2^{0}$ . I'll have to think about how to count the number of ways in which multiple powers of $2$ add up to $2336$ .

Brian Charlesworth - 6 years, 6 months ago

Thanks for pointing it out!! I think since 2336 is not divisible by 3, all x, y, and z cannot be equal!! For two of them to be equal (x=y), solution for $2^{x+1}+2^{z}=2336$ could be found by same method as I did!

Pranjal Jain - 6 years, 6 months ago

Yes, that's good reasoning. :)

Brian Charlesworth - 6 years, 6 months ago

@Brian Charlesworth – Thanks! Why are you not posting some new cool Level 5 geometry questions like "A tight fit" and others?

Pranjal Jain - 6 years, 6 months ago

Sauditya Yo Yo
Jan 8, 2015

$2336_{10}$ = $10010010000_{2}$ as the binary representation of a number is unique and binary representation of 2336 has exactly 3 ones therefore the only way of writing:- 2336 as sum of three powers of two is = $2^{11}$ + $2^{8}$ + $2^{5}$ therefore (x, y, z)=(11, 8, 5) or any permutation => answer is 144

Triple Double Set

The answer is 144.

2 solutions

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