BoS Trials Maths and Business Studies 2024 (1 Viewer)

lqmoney · Oct 9, 2024

Canteen said:
Here's my attempt. I've probably over-complicated it or am just plain wrong:

Base case ( $n=0$ ) – Trivial

Inductive hypothesis – Assume true for some non-negative integer $k$ :
$\text{i.e. } \exists p, q \in \mathbb{Z} : w_{2k} = 11 p \rightarrow u_{2k} = 11q \qquad (1)$

Inductive step – RTP true for $n = k + 1$ :
$\text{i.e. } \exists r,s \in \mathbb{Z}: w_{2k+2} = 11r \rightarrow u_{2k+2} = 11s$

$\begin{aligned} \text{Consider, } w_{2k+2} - u_{2k+2} &= (10^{2k+1}a_{2k+1} + 10^{2k}a_{2k} + w_{2k}) - [(-1)^{2k+1}a_{2k+1} + (-1)^{2k}a_{2k} + u_{2k}], \quad \text{by definition} \\ &= (10^{2k+1} + 1) a_{2k+1} + (10^{2k} - 1) a_{2k} + (w_{2k} - u_{2k}) \\ &= (10+1)(\underbrace{1 + 10 + 10^2 + \dots + 10^{2k}}_{=N})a_{2k+1} + [(11 - 1)^{2k} - 1]a_{2k} + 11(p - q), \quad \text{from (1)} \\ &= 11Na_{2k+1} + \left[11^{2k} - \binom{2k}{1} 11^{2k-1} + \dots - \binom{2k}{2k-1}11 + 1 - 1\right]a_{2k} + 11(p - q) \\ &= 11Na_{2k+1} + 11\underbrace{\left[11^{2k-1} - \binom{2k}{1} 11^{2k-2} + \dots - \binom{2k}{2k-1}\right]}_{=M} a_{2k} + 11(p-q) \\ &= 11Na_{2k+1} + 11Ma_{2k} + 11(p-q) \\ \therefore u_{2k+2} &= 11(\underbrace{r - Na_{2k+1} - Ma_{2k} + q - p}_{=s}), \quad \text{as } w_{2k+2} = 11r \\ &= 11s, \quad \text{as required} \end{aligned}$

i did something similar, problem is I don’t think you can assume w_2k and u_2k are still divisible by 11 when proving for the case of n=k+1, since you are only assuming that there is an implication between the 2. doing it the way you did means the original statement 11| w2n => 11|u2n is only necessarily true if w2, w4 … w2n are all divisible by 11 as well.

Trebla · Oct 9, 2024

I think you've highlighted an interpretation of the question that was different to what was intended. I believe the intention was to suppose that w_2n was divisible by 11 for all n, not for the same value of n as u_2n. I will confirm with the original author if that was the intention. If confirmed as I suspected, the wording will be adjusted in the final release file to remove that potential ambiguity and taken into consideration in the marking (and I will provide a less complicated proof as well

).

nonya2000 · Oct 10, 2024

Canteen said:
Here's my attempt. I've probably over-complicated it or am just plain wrong:

Base case ( $n=0$ ) – Trivial

Inductive hypothesis – Assume true for some non-negative integer $k$ :
$\text{i.e. } \exists p, q \in \mathbb{Z} : w_{2k} = 11 p \rightarrow u_{2k} = 11q \qquad (1)$

Inductive step – RTP true for $n = k + 1$ :
$\text{i.e. } \exists r,s \in \mathbb{Z}: w_{2k+2} = 11r \rightarrow u_{2k+2} = 11s$

$\begin{aligned} \text{Consider, } w_{2k+2} - u_{2k+2} &= (10^{2k+1}a_{2k+1} + 10^{2k}a_{2k} + w_{2k}) - [(-1)^{2k+1}a_{2k+1} + (-1)^{2k}a_{2k} + u_{2k}], \quad \text{by definition} \\ &= (10^{2k+1} + 1) a_{2k+1} + (10^{2k} - 1) a_{2k} + (w_{2k} - u_{2k}) \\ &= (10+1)(\underbrace{1 + 10 + 10^2 + \dots + 10^{2k}}_{=N})a_{2k+1} + [(11 - 1)^{2k} - 1]a_{2k} + 11(p - q), \quad \text{from (1)} \\ &= 11Na_{2k+1} + \left[11^{2k} - \binom{2k}{1} 11^{2k-1} + \dots - \binom{2k}{2k-1}11 + 1 - 1\right]a_{2k} + 11(p - q) \\ &= 11Na_{2k+1} + 11\underbrace{\left[11^{2k-1} - \binom{2k}{1} 11^{2k-2} + \dots - \binom{2k}{2k-1}\right]}_{=M} a_{2k} + 11(p-q) \\ &= 11Na_{2k+1} + 11Ma_{2k} + 11(p-q) \\ \therefore u_{2k+2} &= 11(\underbrace{r - Na_{2k+1} - Ma_{2k} + q - p}_{=s}), \quad \text{as } w_{2k+2} = 11r \\ &= 11s, \quad \text{as required} \end{aligned}$

how does one write in latex

lqmoney · Oct 10, 2024

lqmoney said:
i did something similar, problem is I don’t think you can assume w_2k and u_2k are still divisible by 11 when proving for the case of n=k+1, since you are only assuming that there is an implication between the 2. doing it the way you did means the original statement 11| w2n => 11|u2n is only necessarily true if w2, w4 … w2n are all divisible by 11 as well.

A proof for this interpretation

carrotsss · Oct 10, 2024

nonya2000 said:
how does one write in latex

[TEX]normal latex code [/TEX]

Trebla · Oct 10, 2024

Confirmed with author that the intended solution was to take as given that w_2n is divisible by 11 for all n so going to adjust wording accordingly. Hence, the outline of the solution would be:

From the assumption n=k
w_2k = 11P and u_2k = 11Q

Given w_2k+2 = 11R then
w_2k+2 - w_2k = 11(R-P)
=> a_2k - a_2k+1 = 11(R-P)

Consider
u_2k+2
= u_2k + 10^2ka_2k + 10^2k+1a_2k+1
= 11Q + 10^2k(a_2k+10a_2k+1) by assumption
= 11Q + 10^2k(11(R-P)+11a_2k+1)
= 11S

lqmoney · Friday at 5:57 PM

do you have to account for this in ii)? because if w_2n is divisble by 11 for all n then a0=a1, a2=a3 etc which wouldnt allow the implication in i) to be applied to palindromic numbers without some manipulation atleast

yanujw · Saturday at 3:21 PM

lqmoney said:
do you have to account for this in ii)? because if w_2n is divisble by 11 for all n then a0=a1, a2=a3 etc which wouldnt allow the implication in i) to be applied to palindromic numbers without some manipulation atleast

I think you've confused a few implications. Firstly, don't proceed by assuming w_2n is divisible by 11, but apply the palindromic nature of the coefficients to w_2n first. If u_2n is palindromic, a_k = a_(2n-1-k) for all k, then w_2n = 0, and then you can make the argument u_2n is divisible by 11.

lqmoney · Saturday at 8:50 PM

yanujw said:
I think you've confused a few implications. Firstly, don't proceed by assuming w_2n is divisible by 11, but apply the palindromic nature of the coefficients to w_2n first. If u_2n is palindromic, a_k = a_(2n-1-k) for all k, then w_2n = 0, and then you can make the argument u_2n is divisible by 11.

But if the proof in i) relies on w_2k being divisble by 11 for all integers up to n, then showing w_2n is divisible by 11 is not satisfactory to show that u_2n is also divisible by 11, since to apply the proof from i) you must also show w_2, w_4 … are also dividble by 11. For 11|w2, the only solution is w2 = 0 since w2 is at most 9 and at-least -9, so a1 = a2. then for 11|w4 similarly a3 = a4 and so on.

Trebla · Saturday at 8:59 PM

lqmoney said:
But if the proof in i) relies on w_2k being divisble by 11 for all integers up to n, then showing w_2n is divisible by 11 is not satisfactory to show that u_2n is also divisible by 11, since to apply the proof from i) you must also show w_2, w_4 … are also dividble by 11. For 11|w2, the only solution is w2 = 0 since w2 is at most 9 and at-least -9, so a1 = a2. then for 11|w4 similarly a3 = a4 and so on.

As soon as you show that w_2n = 0 in general for any palindrome, regardless of the value of n (within the positive integers that is), then it must be true for all positive integers of n.

lqmoney · Saturday at 9:30 PM

Trebla said:
Confirmed with author that the intended solution was to take as given that w_2n is divisible by 11 for all n

i think im misunderstanding something. with this change to the q, in part i. it is proven that u_2n is divisble by 11 if w_2k is divisble by 11 for all integer k from 1 to n, right? so even if w_2n is 0, it is not necessarily true that u_2n is divisbile by 11? (it is true but only working from the proof from i. we cannot assume it is true from this step).

Trebla · Saturday at 11:19 PM

In part (i), you are asked to show that if X is true then Y follows.

In part (ii), you need to show that X is actually true for this specific scenario, so that you can then deduce that Y follows.

In this case, you would show that w_2n = 0 which is an integer that is divisible by 11 and holds for all values of n. This means your "if" condition in part (i) has been satisfied so the deduction can be made.

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