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USA Mathematical Talent SearchSolutions to Problem 2/2/17

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2/2/17. Write the number1

2 32as the sum of terms of the form 2q, where q is rational. (For example, 21 + 21/3 + 28/5 is asum of this form.) Prove that your sum equals 1/(

2 32).

Credit This problem was proposed by Sidney Kravitz.

Comments This problem is effectively an exercise in rationalizing the denominator, butthe twist is the presence of both the square root and the cube root. A knowledge of basicalgebraic identities can take care of both. James Sundstrom arrives at the answer usingtwo steps, Justin Hsu shows how to the same calculation in one step, and Jeffrey Manningprovides a clever solution using geometric series. Solutions edited by Naoki Sato.

Solution 1 by: James Sundstrom (11/NJ)

The identities (a b)(a + b) = a2 b2 and (a b)(a2 + ab + b2) = a3 b3 suggest thefollowing approach for rationalizing the denominator. For example, setting a =

2 and

b = 32 gives

(2 3

2)(2 +

32) = 2 3

4,

so1

2 32 =2 + 3

2

(2 32)(2 + 32) =

2 + 3

2

2 34 .

Then setting a = 2 and b = 34 gives

(2 34)(4 + 2

34 + 2

32) = 8 4 = 4,

so2 + 3

2

2 34 =(2 + 3

2)(4 + 2 3

4 + 2 3

2)

(2 34)(4 + 2 34 + 2 32)=

42 + 4 3

2 + 2 3

42 + 4 + 2 3

22 + 2 3

4

4

=2 +

32 +

342

2+ 1 +

322

2+

34

2

= 212 + 2

13 + 2

23

(212

) (21

)+ 20 + 2

13

(212

) (21

)+ 2

23

(21

)= 2

12 + 2

13 + 2

16 + 20 + 2

16 + 2

13 .

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USA Mathematical Talent SearchSolutions to Problem 2/2/17

www.usamts.org

Solution 2 by: Justin Hsu (11/CA)

We have the identity

(a b)(a5 + a4b+ a3b2 + a2b3 + ab4 + b5) = a6 b6.

Now, we let a =2 and b = 3

2, and multiply the numerator and denominator by a5+a4b+

a3b2 + a2b3 + ab4 + b5. This gives us

12 32

42 + 4 3

2 + 2 3

42 + 4 + 2 3

22 + 2 3

4

42 + 4 3

2 + 2 3

42 + 4 + 2 3

22 + 2 3

4

=42 + 4 3

2 + 2 3

42 + 4 + 2 3

22 + 2 3

4

4

= 212 + 2

13 + 2

16 + 20 + 2

16 + 2

13 ,

giving us the required sum of rational powers of 2.

Solution 3 by: Jeffrey Manning (10/CA)

Notice that if x 6= 0, then the sum x3 + x2 + x+ 1+ x1 + x2 is a geometric series withcommon ratio x1 and initial term x3. If x 6= 1, then we can use the formula for a geometricseries to get

x3 + x2 + x+ 1 + x1 + x2 =x3(1 x6)1 x1 =

x3 x31 x1 =

(x3 x3)x3(1 x1)x3 =

x6 1x3 x2 .

Applying this formula with x = 21/6 gives

12 32 =

21 121/2 21/3

=(21/6)6 1

(21/6)3 (21/6)2= (21/6)3 + (21/6)2 + 21/6 + 1 + (21/6)1 + (21/6)2

= 21/2 + 21/3 + 21/6 + 20 + 21/6 + 21/3.