# Thread: [SOLVED] Difficult binomial theorem question(involving greatest integer function)?

1. ## [SOLVED] Difficult binomial theorem question(involving greatest integer function)?

If (3√3 + 5)^9 = R and [R] denotes the greatest integer less than or equal to R, then:

A) [R] is divisible by 10
B) [R^2] is divisible by 512
C) [R] is divisible by 15
D) [R] is an even number

More than one options may be correct.

2. ## Binomial

Hello again fardeen_gen

As with your previous posting, I'm not sure how you're supposed to tackle this question. I have used a calculator, and found [R] to be 1191041440. So A and D are true. The others are not.

3. Originally Posted by fardeen_gen
If (3√3 + 5)^9 = R and [R] denotes the greatest integer less than or equal to R, then:

A) [R] is divisible by 10
B) [R^2] is divisible by 512
C) [R] is divisible by 15
D) [R] is an even number

More than one options may be correct.
Here is another approach:

$\displaystyle (3 \sqrt{3} + 5)^9 = \sum_{i=0}^9 \binom{9}{i} 3^{3(9-i)/2} 5^i$
$\displaystyle (3 \sqrt{3} - 5)^9 = \sum_{i=0}^9 (-1)^i \binom{9}{i} 3^{3(9-i)/2} 5^i$

Subtracting, the even-numbered terms cancel, yielding

$\displaystyle (3 \sqrt{3} + 5)^9 - (3 \sqrt{3} - 5)^9 = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}$

So
$\displaystyle (3 \sqrt{3} + 5)^9 = (3 \sqrt{3} - 5)^9 + 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}$

$\displaystyle [(3 \sqrt{3} + 5)^9] = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}$
because
$\displaystyle 0 < 3 \sqrt{3} - 5 < 1$ and the summation is an integer.

4. ## Binomial

Originally Posted by awkward
Here is another approach:

$\displaystyle (3 \sqrt{3} + 5)^9 = \sum_{i=0}^9 \binom{9}{i} 3^{3(9-i)/2} 5^i$
$\displaystyle (3 \sqrt{3} - 5)^9 = \sum_{i=0}^9 (-1)^i \binom{9}{i} 3^{3(9-i)/2} 5^i$

Subtracting, the even-numbered terms cancel, yielding

$\displaystyle (3 \sqrt{3} + 5)^9 - (3 \sqrt{3} - 5)^9 = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}$

So
$\displaystyle (3 \sqrt{3} + 5)^9 = (3 \sqrt{3} - 5)^9 + 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}$

$\displaystyle [(3 \sqrt{3} + 5)^9] = 2 \sum_{i=0}^4 \binom{9}{2i+1} 3^{3(4-i)} 5^{2i+1}$
because
$\displaystyle 0 < 3 \sqrt{3} - 5 < 1$ and the summation is an integer.
That is pretty neat! It's clear, then, that R is a multiple of 10. And it's not a multiple of 3, since all the terms are except the last. Therefore R is not a multiple of 15. Any idea how this might determine the truth of statement B? (Other than by evaluating [R], of course.)

PS I have just looked again at statement B, and noticed that it is $\displaystyle R^2$, not just R. Since $\displaystyle 512 = 2^9$, this will be true if R is a multiple of 32. And it is. So B is true after all!

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