This is a whole new level of Trial and Error

Level pending

In a parallel universe, you've forgotten the pin code of your credit card. Though fortunately you remember the characteristics of your pin code. Which are-

-It is a 4 digit numerical -The first digit is 1/2 of the 4th digit. -The second and the third digits are the same. -The first 2 digits combined are 1/2 of the last 2 digits.

Example for further clearance-

1224 We see that the 2nd and the 3rd conditions are fulfilled as the 2nd and the 3rd digits are equal and the first 2 digits combined are 1/2 of the last 2 digits combined (12=1/2(24)), though as the first condition is not fulfilled this answer is incorrect.

0000 Sadly this answer, despite of fulfilling all the conditions is wrong, as 0000 isn't considered a 4 digit number.

The answer is 4998.

1 solution

Rick B
Jan 9, 2015

Let the number be $\overline{abcd}$

We have

$\begin{cases} d = 2a \\ b = c \\ 2(10a+b) = 10c+d \end{cases}$

Substituting $d$ by $2a$ and $c$ by $b$ in the third equation:

$20a+2b = 10b + 2a \implies 18a = 8b \implies \dfrac{18}{8}a = b$ or $\dfrac{2a}{2^3} \times 9 = b$

Considering that $b$ must be an integer and that $1 \leq a \leq 9$ , we have $2$ possible values for $a$ : $4$ and $8$

But $a = 8$ would make $d = 16$ , so $a = \dfrac{d}{2} = 4 \implies b = c = \dfrac{2 \times 4}{2^3} \times 9 = 9$

$\implies \overline{abcd} = \boxed{4998}$

Eh bro, it was supposed to be trial and error with logic. You know, simply the first digit can either be 1, 2, 3, or 4 and the last ones 2, 4, 6 or 8. Now you can use trial and error for the correct combination. That is it ._.

Devang Srivastava - 6 years, 5 months ago

I really dislike trial and error. It's unreliable and would only work for this problem because there aren't many possibilities. I'm showing the reliable method. And my solution is long just because I explained it with detail, but the system of equations is unnecessary and the rest is, at least for me, faster than trial and error.

Rick B - 6 years, 5 months ago

That might be one of the reasons which could explain that why you're level 1 in Computer Science :3

Devang Srivastava - 6 years, 5 months ago

@Devang Srivastava – No, it's because I use Brilliant to prepare myself for math olympiads (and also because I like it), so there is no reason for me to practice Computer Science. Plus, I'm not interested in it as of now. But I don't see how it could be based on trial and error, like you implied... For example, in inequality problems, you can keep trying everything, but you need to plan ahead to get an answer.

Rick B - 6 years, 5 months ago

@Rick B – You see, on brilliant, problem tagged in the Computer Sciences category are mostly problems related to logic and case studies. You can check it out, I myself was confused when I saw that.

Devang Srivastava - 6 years, 5 months ago

I used trial and error by a little like henrique berger( some of his algebra) well trial and error by just trying the first digit as 1,2,3 or4

Samuel Samuel - 6 years, 4 months ago

This is a whole new level of Trial and Error

The answer is 4998.

1 solution

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