My post with Kvantik’s 2012 problems for middle school was a success. So I scanned the 2013 issues and found 7 more problems that I liked. Here are two cute problems I’ve seen before:
Problem 1. In the equation 30 − 33 = 3 move one digit to make it correct.
Problem 2. A patient got two pills for his headache and two pills for his cough. He was supposed to use one of each type of pill today and do the same tomorrow. The pills all looked the same. By mistake, the patient mixed up the pills. How should he use them so that he follows the prescription exactly?
Now logic and information-theoretical problems:
Problem 3. One strange boy always tells the truth on Wednesdays and Fridays and always lies on Tuesday. On other days of the week he can lie or tell the truth. He was asked his name seven days in a row. The first six answers in order are Eugene, Boris, Vasiliy, Vasiliy, Pete, Boris. What was his answer on the seventh day?
Problem 4. Ten people are suspected of a crime. It is known that two of them are guilty and eight are innocent. Suspects are shown to a psychic in groups of three. If there is a guilty person in the group the psychic points him out. If there are two guilty people in the group, the psychic points to one of them. If all of them are innocent, the psychic points at one of the three.
- How would you find at least one guilty person in four séances?
- How would you find both criminals in six séances?
Problem 5. There are four balloons: red, blue, green and black. Some of the balloons might be magical. There is also a detector box that can say how many balloons out of the ones put inside are magical. How can you find all the magical balloons using the detector box not more than three times?
I conclude with two miscellaneous problems.
Problem 6. Three runners started their loops at the same time at the same place on the same track. After some time they ended at their starting point together. The fastest runner passed the slowest runner 23 times. Assuming each runner has a constant speed, how many times was one runner passed by another runner in total?
Problem 7. Given a point inside a circular disk, cut the disk into two parts so that you can put them back together into a new disk such that the given point is the center.
Does the top-half of an ‘=’ resemble 1 well enough to consider it a “digit”?
If so, you could solve Problem 1 by moving it above the ‘-‘ to produce:
30 = 33 – 33 August 2015, 2:10 am
MQ, No. Great idea, but not the intended answer.3 August 2015, 8:43 pm
The first one gave me the inspiration to invent a similar one myself (though it is a bit more of lateral thinking than math)
Move one digit to make this correct:
11 – 33 = 3
Solution:3 August 2015, 10:02 pm
1. Purists would disagree. 🙂 It’s really “move and change font size”.4 August 2015, 1:59 am
Ah, too bad; still, the beginning couple of problems are cute, indeed.
I suppose with a bit of upward mobility we can dispose of the first, and perhaps if the patient’s headache is splitting he’ll be inspired to remedy the second.4 August 2015, 2:54 am
Well did not answer all of them, but these I’ve got:
1. 30 – 3^3 = 27.
3. Pete. Moving the chain FT?T (False, True, ?, True) above the chain EBVVPB?EBVVP and you see he said on a Wednesday Pete, so the ? is the Friday.
5. Put three balloons in the box:
First RBG (red, blue, green)
Second RGZ (red , green, black)
third B, G, Z)
If the first time it reads a 3 or a zero. You know everything the second time.
If the range is 2,2,2 (for two out of tree magical balloons, each): you know it is RBZ
If range is 2,2,1: RG
If range is 2,1,2: BG
If range is 2,1,1: BR
Other possibilities starting with a 2 are not possible.
If range is 1,1,1: G
If range is 1,1,2: BZ
If range is 1,1,0: R
Other possibilities starting with a 1 are not possible.
Question 6 (looks a little like a clock)
Assume the slow runner runs one loop.
The fastest runs 24 then.
The middle one can run: 2,3,4,6,8,or 12 loops.
If 2 or 12 loops: they pass eachother 35 times. (23 given. #1 passes #2 11 times and #2 passes #3 one time, or the other way around)
If 3 or 8 loops: they pass eachother 32 times (23 + 7 + 2)
If 4 or 6 loops: they pass eachother 31 times (23 + 5 + 3)
(or did I misunderstood this question and is only one answer possible?)
pipo4 August 2015, 5:23 am
Another lateral thinking answer to 1:
turn the equation upside down and view it in hexadecimal. you can readily see which digit to move to create the correct equation E0= EE- E
Of course, it also works in any base greater than 14 (not just 16) or any time your teacher will let you substitute a variable for a digit.4 August 2015, 12:04 pm
For Question 1, 30 – 33 = 3, would 33 – 33 = 0 solve the problem?4 August 2015, 2:02 pm
Jim, You are swapping digits, not moving one digit.4 August 2015, 10:09 pm
Problem 6. Remember that the fastest runner passes the middle runner and the middle runner passes the slowest runner.4 August 2015, 10:15 pm
2: Take half of each tablet.5 August 2015, 7:13 am
1. 30 – 3 = 3 power 35 August 2015, 7:18 am
Q3: To solve this problem, we have to try 7 cases in the worst case: assume (a) first day is Monday (b) first day is Tue…..(7) assume first day is Sunday. But wait, the last day answer is asked for. So, it means that we can answer it – which is possible only if the last day is Wednesday or Friday. On the other days, it could be anything and hence we cannot be asked to answer that. (Remember: if the last day is a Tuesday – it has to be a lie – but there could be several lies).
For W to be the last day, first day has to be Thueday.If we look at the answers, we see his answers are the same on Tuesday and Friday – so this combination is ruled out.
For F to be the last day, first day has to be Saturday. Now we see he answered Pete on Wednesday and it does not conflict with the other conditions. Hence on the last day (that is Friday), he should have said Pete again.5 August 2015, 7:29 am
2. grind the pills into dust, mix them in water/some other drink, and drink half today and half tomorrow!5 August 2015, 1:24 pm
Kvantik 2013 | Beyond Solutions:
[…] Selection of problems from 2013 […]5 August 2015, 8:14 pm
loved 7, very simple yet tricky.
cut a small disc such that both the center of the original disc and the random point are on the circumfrence6 August 2015, 7:21 am
It is pete.7 August 2015, 6:00 am
6: Suppose the numbers of laps completed by the three runners are a<b<c. We know c-a-1 = 23. We want (c-a-1) + (c-b-1) + (b-a-1). b cancels, leaving 2c-2a-3 = 2(c-a-1) – 1 = 45.8 August 2015, 8:44 pm
8 part 1. Line up ABC, then DEF, then GHI, (omitting suspect J altogether). Call the answers X, Y and Z. At least one of them is guilty. Line up XYZ to find them.8 August 2015, 8:47 pm
8 part 2. Continuing from my answer for 8 part 1, suppose without loss of generality that C was found to be guilty. Line up ABJ, and call the result X’. One of X’, Y and Z is the remaining guilty person. Line them up to find out who.8 August 2015, 8:50 pm
7, alternatives to david’s answer.8 August 2015, 9:02 pm
– Cut out any shape that is symmetrical in the line that is equidistant from the given point and the centre, and turn it over.
– Cut out any shape that is 2-fold rotationally symmetrical about the point half way between the given point and the centre, and rotate it.
– Cut out an equilateral triangle with the given point and the centre as two of its corners, and rotate it. Obviously this generalises to other polygons and other orders of rotational symmetry, and david’s answer is the limiting case.
– If you only have scissors, and no pen or ruler or compass, and if you insist on simply-connected parts, then it is still possible. Fold the disk so that the given point aligns with the centre, and cut along the edge of the disk. This gives a leaf shape and a crescent. Rotate the leaf shape.
For 2, since you only have four pills in total, couldn’t you cut each pill in half and have one half of each pill today and the remaining halves tomorrow? I may be wrong..9 August 2015, 12:21 am
Yes, cutting in half works too.10 August 2015, 2:03 pm
Nobody has posted anything about the problem involving the psychic, so I’ll take a stab at it.
Part a) is quite simple I think. Take the ten people and divide them into three groups of 3 with one person remaining. Show each of the three groups to the psychic. The three people selected from the three groups then constitute the fourth seance that is shown to the psychic. The person selected in the fourth seance should be guilty.
I’m less certain about part b), but I’ll post my solution anyway. From our solution for part a), we have already found one guilty suspect. This person can be removed from the pool of suspects, leaving nine people. Finding the second guilty person should conform to one of the following cases
1. Both guilty persons were included in the fourth seance (i.e. two of the three groups that we split the ten people into contained guilty persons).
2. The same group that produced the first guilty person contained the second guilty person, so only one guilty person was included in the fourth seance (i.e. one of the three groups contained both of the guilty persons).
3. The remaining person is the second guilty person.
We can use these cases to figure out the second guilty person. Take the two persons remaining from the fourth seance and form a new group that includes the remaining person. This new group will constitute the fifth seance. The person selected from the fifth seance will then form a new group with the remaining two people from the group that initially contained the first guilty person (this group will constitute the sixth seance). The person selected in the sixth seance should be the second guilty person.13 August 2015, 8:34 am
In other words, for part a we create three groups (call them groups A, B and C). One person is left over (call this person R). Show each of groups A, B and C to the psychic (this will use up three seances). Then take the persons selected from A, B and C to the psychic as part of the fourth seance. The person selected by the psychic is guilty (for simplicity’s sake, we’ll say the person selected from group A was guilty).
For part B, you take the two people selected from B and C and form a new group including person R. This group is shown to the psychic as the fifth seance. For the sixth seance, take the two remaining people from part A (the people who weren’t chosen) and the person selected from the fifth seance and show these people to the psychic. The person the psychic selects in the sixth seance is the second guilty person.
I’m not sure if this is correct, but I think it makes sense (hopefully).13 August 2015, 8:41 am
For problem 6, the answer depends on the meaning of “passed”. The answer given by apt1002 includes the simultaneous arrival at the finish line in the passing count. If we instead only count times during the running when one runner goes on past another runner, then the number of “passings” is (c-a-2) + (c-b-2) + (b-a-2), giving 44 total passings. This also assumes that the distances run differ by 2 or more laps. If the middle speed runner runs one less lap than the fastest, then the fastest doesn’t pass him. (c-b-2 would be -1). This situation (25, 24, 1 laps) gives 45 passings. Likewise (25, 2, 1) gives 45 passings. If the middle speed runner runs just as fast as the fastest, (25, 25, 1 laps) then there are 46 passings. Likewise (25, 1, 1) gives 46 passings.14 August 2015, 2:15 pm
Oops: the answer given by apt1002 does NOT include the simultaneous arrivals. I think the bit about the middle speed runner that is just as fast as another is okay, though. So it could be 45 or 46.15 August 2015, 1:55 am
for 1. it may b.15 August 2015, 8:00 am
30 – 3 = 3^3