- Joined
- Jan 30, 2018
This is the correct logic to solve the problem. Josh got both questions correct.After watching todays MATI I feel like I should give Josh and especially Tiny some hints for the LSAT question. Its like a general problem solving question asked all around the place and I have a bit of experience from just doing a fuckton of these timed, enough for 5 lifetimes.
View attachment 7042113
You start working backwards from the facts at hand instead of listing all the possible combinations of the first two digits cause thats a waste of time.
If we were to take the number as mnxyz, then n=2m and x<z
If z in this case is 1 then theres only one number x can take considering theres only one number smaller than 1 and each number can occur only once, so x would be 0
The remaining numbers would be 2, 3, 4. To satisfy n=2m only one pair is possible, m is 2 and n is 4.
The given number would be 24031
So the correct option would be A.
Im gonna preempt and say, yes I know you dont care, I am sorry and I will fuck off.
The third rule explicitly states that the value of the second digit must be exactly twice that of the first digit. This forbids zero from being in either the first or second position because:
- If zero is in the first position, there is no other potential digit that is exactly twice the value of zero, since twice the value of zero is zero. The only digit that could fulfill this rule is another zero, which violates the no repeating digit rule.
- Let n = 1, 2, 3, or 4
- 0 * n = 0
- If zero is in the second position, there is no other potential digit that can be in the first position that can have its value doubled to make zero
- 1 * 2 = 2
- 2 * 2 = 4
- 3 * 2 = 6
- This also eliminates 3 from being able in the first position, since 6 is not a valid digit
- 4 * 2 = 8
- This also eliminates 4 from being able to be in the first position, since 8 is not a valid digit
If I'm logic-ing correctly, there are five possible codes:
- 1-2-0-3-4
- 1-2-0-4-3
- 1-2-3-0-4
- 2-4-0-3-1
- 2-4-1-0-3