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Algebra Level 2

What is the integral part of number ( 2 + 1 ) 6 (\sqrt2+1)^6 ?

Details and Assumptions :

  • As an explicit example, the integral part of 123.456 123.456 is 123 123 .


The answer is 197.

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Bhavya Jain by Bhavya Jain , 23, India

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5 solutions

Danish Mohammed
Apr 21, 2015

Let ( 2 + 1 ) 6 = I + F \displaystyle (\sqrt{2}+1)^6=I+F where I I is the integral part and 0 < F < 1 \displaystyle 0 < F<1 is the fractional part.

Consider the number f = 1 ( 2 + 1 ) 6 = ( 2 − 1 ) 6 \displaystyle f = \frac{1}{(\sqrt{2}+1)^6}=(\sqrt{2}-1)^6

Observe that 0 < f < 1 \displaystyle 0<f<1

Now, I + F + f = i n t e g e r \displaystyle I + F + f = \mathrm{integer} , since the irrational terms cancel out.

For the LHS to be an integer, F + f \displaystyle F+f must also be an integer. We know that 0 < F + f < 2 \displaystyle 0<F+f<2 which implies F + f = 1 \displaystyle F+f = 1 .

Now I + F + f = I + 1 = 2 ( ( 6 0 ) 2 3 + ( 6 2 ) 2 2 + ( 6 4 ) 2 + ( 6 6 ) ) = 198 \displaystyle I+F+f = I+1 = 2({6 \choose 0}2^3 + {6\choose 2}2^2+{6\choose 4}2 + {6 \choose 6})=198 and finally I = 198 − 1 = 197 \displaystyle I=198-1 = \boxed{197}

50 Helpful 3 Interesting 10 Brilliant 2 Confused

Moderator note:

Great! Does this work for odd numbers n n in ( 2 + 1 ) n (\sqrt{2}+1)^n as well? Why or why not? Can you generalize this?

Response to Challenge Master note

Yes! It does work for odd n \displaystyle n . The solution starts off the same, except, instead of finding I + F + f \displaystyle I+F+f (which isn't an integer now) we find I + F − f \displaystyle I+F-f which is an integer. This means F − f \displaystyle F-f is an integer.

We will also have − 1 < F − f < 1 \displaystyle -1<F-f<1 implying that F − f = 0 \displaystyle F-f=0

So by computing I + F − f = I = ( 2 + 1 ) n − ( 2 − 1 ) n \displaystyle I+F-f=I=(\sqrt{2} +1)^n-(\sqrt{2}-1)^n we can find the integral part.

Demonstration:

Find the integral part of ( 2 + 1 ) 7 \displaystyle (\sqrt{2}+1)^7

Let ( 2 + 1 ) 7 = I + F \displaystyle (\sqrt{2}+1)^7=I+F where I I is the integral part and 0 < F < 1 \displaystyle 0 < F<1 is the fractional part.

Consider the number f = 1 ( 2 + 1 ) 7 = ( 2 − 1 ) 7 \displaystyle f = \frac{1}{(\sqrt{2}+1)^7}=(\sqrt{2}-1)^7

Observe that 0 < f < 1 \displaystyle 0<f<1

Now, I + F − f = i n t e g e r \displaystyle I + F - f = \mathrm{integer} , since the irrational terms cancel out.

For the LHS to be an integer, F − f \displaystyle F-f must also be an integer. We know that − 1 < F − f < 1 \displaystyle -1<F-f<1 which implies F − f = 0 \displaystyle F-f = 0 .

Now I + F − f = I = 2 ( ( 7 1 ) 2 3 + ( 7 3 ) 2 2 + ( 7 5 ) 2 + ( 7 7 ) ) = 478 \displaystyle I+F-f = I = 2({7 \choose 1}2^3 + {7\choose 3}2^2+{7\choose 5}2 + {7 \choose 7})=\boxed{478} .

This technique with a minor change can be used to find the integral part of many other numbers involving square roots, for example, ⌊ ( 5 5 + 11 ) n ⌋ \displaystyle \lfloor (5\sqrt{5}+11)^n \rfloor . Here choose f = ( 5 5 − 11 ) n \displaystyle f = (5\sqrt{5} - 11)^n and observe that 0 < f = 4 n ( 5 5 + 11 ) n < 1 \displaystyle 0<f=\frac{4^n}{(5\sqrt{5}+11)^n}<1 and then add/subtract f \displaystyle f to/from I + F \displaystyle I+F depending on the parity of n \displaystyle n to get an integer and the rest of the procedure is identical to the above examples. Basically if you want to find the integral part of ( a b + c ) n = I + F \displaystyle (a\sqrt{b} +c)^n = I+F take f = ( a b − c ) n \displaystyle f=(a\sqrt{b}-c)^n and try to work from there

Danish Mohammed - 6 years, 1 month ago

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why did you use F-f as integer for the second and F+f for the first? I can not choose to use which one when i was working out this.

Hafizh Ahsan Permana - 6 years, 1 month ago

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.For the first, look at the binomial expansion of I + F = ( 2 + 1 ) 6 I+F=(\sqrt{2} +1)^6 and f = ( 2 − 1 ) 6 f=(\sqrt{2}-1)^6 .

I + F = ( 6 0 ) ( 2 ) 6 + ( 6 1 ) ( 2 ) 5 + ( 6 2 ) ( 2 ) 4 + ( 6 3 ) ( 2 ) 3 + ( 6 4 ) ( 2 ) 2 + ( 6 5 ) ( 2 ) 1 + ( 6 6 ) I+F={6 \choose 0}(\sqrt{2})^6+{6\choose1}(\sqrt{2})^5+{6\choose2}(\sqrt{2})^4+{6\choose3}(\sqrt{2})^3+{6\choose4}(\sqrt{2})^2+{6\choose5}(\sqrt{2})^1+{6\choose6} f = ( 6 0 ) ( 2 ) 6 − ( 6 1 ) ( 2 ) 5 + ( 6 2 ) ( 2 ) 4 − ( 6 3 ) ( 2 ) 3 + ( 6 4 ) ( 2 ) 2 − ( 6 5 ) ( 2 ) 1 + ( 6 6 ) f={6 \choose 0}(\sqrt{2})^6-{6\choose1}(\sqrt{2})^5+{6\choose2}(\sqrt{2})^4-{6\choose3}(\sqrt{2})^3+{6\choose4}(\sqrt{2})^2-{6\choose5}(\sqrt{2})^1+{6\choose6}

2 \sqrt{2} raised to any even power is an integer so you can see every odd term in I + F I+F and f f is an integer and every even term in them is irrational Observe that taking I + F − f I+F-f in this case is a bad idea, because it becomes irrational; all the integer terms cancel out and you can't tell what the value of F − f F-f is. Taking I + F + f I+F+f does give us an integer (since those irrational terms cancel out) and we can easily tell the value of F + f F+f . This works for even n

For odd n you should take I + F − f I+F-f . If you look at the power 7 demonstration in my comment, You can see that the binomial expansions of both I + F I+F and f f starts off as ( 7 0 ) ( 2 ) 7 … {7\choose0}(\sqrt{2})^7\dots with f f having alternating signs and I + F I+F having plus signs only. You can see that every odd term is irrational so to cancel them out you need I + F − f I+F-f .

Hope this helps.

Danish Mohammed - 6 years, 1 month ago

Well this might be novice question. But could u please explain or give any link for followings: 1. how I+F+f : is integer 2. How the last equation forms? ie. 2((0 6)2^3...

Thanks

omar faru - 6 years, 1 month ago

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Everything I've done relies on the Binomial Theorem. Look it up on wikipedia if you don't know what it is.

These are the expansions of I + F I+F and f f

I + F = ( 6 0 ) ( 2 ) 6 + ( 6 1 ) ( 2 ) 5 + ( 6 2 ) ( 2 ) 4 + ( 6 3 ) ( 2 ) 3 + ( 6 4 ) ( 2 ) 2 + ( 6 5 ) ( 2 ) 1 + ( 6 6 ) \displaystyle I+F={6 \choose 0}(\sqrt{2})^6+{6\choose1}(\sqrt{2})^5+{6\choose2}(\sqrt{2})^4+{6\choose3}(\sqrt{2})^3+{6\choose4}(\sqrt{2})^2+{6\choose5}(\sqrt{2})^1+{6\choose6} f = ( 6 0 ) ( 2 ) 6 − ( 6 1 ) ( 2 ) 5 + ( 6 2 ) ( 2 ) 4 − ( 6 3 ) ( 2 ) 3 + ( 6 4 ) ( 2 ) 2 − ( 6 5 ) ( 2 ) 1 + ( 6 6 ) \displaystyle f={6 \choose 0}(\sqrt{2})^6-{6\choose1}(\sqrt{2})^5+{6\choose2}(\sqrt{2})^4-{6\choose3}(\sqrt{2})^3+{6\choose4}(\sqrt{2})^2-{6\choose5}(\sqrt{2})^1+{6\choose6}

Add them, and you'll see

I + F + f = 2 ( ( 6 0 ) ( 2 ) 6 + ( 6 2 ) ( 2 ) 4 + ( 6 4 ) ( 2 ) 2 + ( 6 6 ) ) \displaystyle I+F+f=2({6 \choose 0}(\sqrt{2})^6+{6\choose2}(\sqrt{2})^4+{6\choose4}(\sqrt{2})^2+{6\choose6})

= 2 ( ( 6 0 ) 2 3 + ( 6 2 ) 2 2 + ( 6 4 ) 2 + ( 6 6 ) ) = i n t e g e r = 198 \displaystyle =2({6 \choose 0}2^3 + {6\choose 2}2^2+{6\choose 4}2 + {6 \choose 6})=\mathrm{integer}=198

I took I + F + f \displaystyle I+F+f to make sure it's an integer. For odd powers you would have to take I + F − f \displaystyle I+F-f to make sure it's an integer. See my comments above for a demonstration with an odd power (7).

Hope this helps

Danish Mohammed - 6 years, 1 month ago

You can just use Pascal's Triangle

Arulx Z - 6 years, 1 month ago

can you tell me how you know 0<F + f<2? Thanks!!!!

Willia Chang - 5 years, 1 month ago

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0<F<1 0<f<1

adding them up

0<F+f<2

The Physicist Cuber Mauro - 3 years, 4 months ago

I did similarly

Maunil Chopra - 2 years, 2 months ago
Otto Bretscher
Apr 23, 2015

Consider the "Fibonacci style" sequence recursively defined by x n + 2 = 2 x n + 1 + x n x_{n+2}=2x_{n+1}+x_n with x 0 = x 1 = 2 x_0=x_1=2 . Arguing as in the case of the Fibonacci sequence, we find that x n = ( 1 + 2 ) n + ( 1 − 2 ) n x_n=(1+\sqrt{2})^n+(1-\sqrt{2})^n . Now − 1 < 1 − 2 < 0 -1<1-\sqrt{2}<0 , so that the integral part of ( 1 + 2 ) n (1+\sqrt{2})^n is x n − 1 x_n-1 for positive even n n and x n x_n for odd n n .

Now x 0 = x 1 = 2 , x 2 = 6 , x 3 = 14 , x 4 = 34 , x 5 = 82 , x 6 = 198 x_0=x_1=2, x_2 = 6, x_3=14, x_4=34, x_5= 82, x_6=198 so the integral part of ( 1 + 2 ) 6 (1+\sqrt{2})^6 is 197 \boxed{197}

This method allows us to find the integral part of ( a + b ) n (a+\sqrt{b})^n as long as ∣ a − b ∣ < 1 |a-\sqrt{b}|<1 .

Basic version, sans Fibonacci \textbf{Basic version, sans Fibonacci} (abreviated version of Danish's work)

Define x n = ( 1 + 2 ) n + ( 1 − 2 ) n = 2 ∑ k = 0 n / 2 ( n 2 k ) 2 k x_n=(1+\sqrt{2})^n+(1-\sqrt{2})^n= 2\sum_{k=0}^{n/2}\binom{n}{2k}2^k , an integer.

Now − 1 < 1 − 2 < 0 , -1<1-\sqrt{2}<0, so that the integral part of ( 1 + 2 ) n (1+\sqrt{2})^n is x n − 1 x_n-1 for positive even n n and x n x_n for odd n n .

In particular, x 6 = 2 ∑ k = 0 3 ( 6 2 k ) 2 k = 198 x_6=2\sum_{k=0}^{3}\binom{6}{2k}2^k=198 so that the integral part of ( 1 + 2 ) 6 (1+\sqrt{2})^6 is 197 \boxed{197}

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Moderator note:

Yes! Newton's sum is hidden in here.

Pardon my doubt if it's silly - How did you get to know that a Fibonacci style series is at work here? Was it by making a pattern of increasing n?

Giri V K - 5 years ago
Louis W
Apr 23, 2015

Using binomial expansion for anything greater than 3 I always find too complicated to just start with. But in this particular instance you can use 2, and then 3.

( 2 + 1 ) 6 = ( ( 2 + 1 ) 2 ) 3 = ( 2 + 2 2 + 1 ) 3 (\sqrt{2}+1)^{6} = ((\sqrt{2}+1)^{2})^{3}=(2+2\sqrt{2}+1)^{3} = ( 3 + 2 2 ) 3 = 27 + 54 2 + 72 + 16 2 = 99 + 70 2 =(3+2\sqrt{2})^{3}=27+54\sqrt{2}+72+16\sqrt{2}=99+70\sqrt{2} ≈ 99 + 70 ∗ 1.414 ≈ 197.98 \approx99+70*1.414\approx197.98

The integral part is what's to the left of the decimal (the terminology was actually new to me), so the answer is 197 □ \Box

I haven't seen a calculator in a while that can't handle a square root, so it would probably be simple for most to just type this into a calculator, but an approximation of the square root of 2 is probably common knowledge in case you don't have a calculator you can use.

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Moderator note:

This solution has been marked incomplete. You need to know the fact that 2 ≈ 1.414 \sqrt 2 \approx 1.414 but you did not specify the error term.

Ervyn Manuyag
Jun 22, 2018

Use a calculator

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