# True or false brainteaser

Below are 10 statements, all either true or false (This statement is promised to be true):

1.At least one of statements 9 and 10 is true.
2.This is either the first true statement or the first false stament.
3.There are three consecutive false statements.
4.The difference between the number of the last true statement and the first true statement divides the number which is to be found.
5.The sum of the numbers of the true statements is the number which is to be found.
6.This is not the last true statement.
7.The number of each true statement divides the number which is to be found.
8.The number that is to be found is the percentage of true statements.
9.The number of divisors of the number that is to be found (apart from 1 and itself) is greater than the sum of the numbers of the true statements.
10.There are no three consecutive true statements.

Find the minimum admissible number (which is to be found).

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LeonhardEuler
Gold Member
Here is my solution:

minumum #=42 (funny!)

My reasoning:
If 1 is true then 2=liar's paradox, so 1 is false=>9&10 are false, and 2 is true, so at this point we have 1- false 2- true 9-false 10-false for sure.
If both 7 & 8 are false, 6=liar's paradox, so 7 or 8 is true and 6 is true, so now for sure: 1- false 2- true 6- true (7 or 8)-true 9-false 10-false
If 3 is true, 8 must be false so 7 must be true since either 7 or 8 is true
If 3 is false, 8 must be true and 4 or 5 must be true
So 3<=>~8
Suppose 3 is true: Then we have 7 is true, so the number, x, is divisible at least by 2,3,6 and 7. The least common multiple of these numbers is 42, so it is the smallest number that could work if 3 is true. Checking each statement gives:
1-false
2-true
3-true
4- false
5-false
6-true
7-true
8-false
9-false
10-false
Each of these statements checks out, so 42 is a possibility.
Suppose 3 is false. Then 8 is true, so x is the percentage true. We know at least 4 statements are true: 2, 4 or 5, 6 and 8. This would correspond to 40%. Any additional true statements would give x at least 50>42, so those can't give the minimum number. There are two possiblities for the statements then:
40:1F,2T,3F,4T,5F,6T,7F,8T,9F,10F but 4 is false
40:1F,2T,3F,4F,5T,6T,7F,8T,9F,10F but 5 is false (10 false)
both give contradictions, so x is not 40. So 3 being false does not give the minimum. Therefore 3 being true does. So x=42.

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It seems to me that statement 10 is true now since there are no three consecutive trues.
Here is my solution:

minumum #=42 (funny!)

Checking each statement gives:
1-false
2-true
3-true
4- false
5-false
6-true
7-true
8-false
9-false
10-false
Each of these statements checks out, so 42 is a possibility.
. So x=42.

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This is my solution and logic:

***1f,2t,3t,4t,5f,6t,7t,8f,9f,10f with x=420 ***

Logic:
Find x = minimum admissible number.
If 1 is T then 2 is a paradox so 1F.

1F implies 9f and 10f.

2T if 1F

6T else logic paradox. Implies 7T or 8T

9F implies num of divisors is less than the sum of T
10F implies there are three consec T.

Status:

f,2, , , ,6, , ,f,f (short hand for 1f,2t,3?,4?,5?,6t,7?,8?,9f,10f)

Assume That 8 is T ...eg x is % of T statements. Status:
f,2, , , ,6, ,8,f,f

x=70,60,50,40 possible since minimum of 4T's to satisfy 10f
70=7*5*2
60=2*2*3*5
50=5*5*2
40=2*2*2*5

Now test 4T?
difference between last T and 1st T =8-2=6 must divide into x.
This implies x=60 and only 6 true statements ..therefore we have 7F since 7 will not divide into 60. Status: f,2, ,4, ,6,f,8,f,f

Need 6 true statements so 3t and 5t required. But 3 must be F since there are not three consecutive false statements.

So try 4F. Status:
f,2, ,f, ,6, ,8,f,f
7t required to give 3 consecutive t. So status:
f,2, ,f, ,6,7,8,f,f , now there are either 4,5 or 6 true statements so
x=40,50 or 60 but the least common denominator >=2*2*2*3*7=168 so
7t contradicts 8t and thus with 8t there is no valid state for 4.

Assume now that 8 is F..eg x is not % of T statements.
6T implies 7T since 8f and now 3T also. Status:
f,2,3, , ,6,7,f,f,f

Since 7T then number of each t divides x and diff from last T and 1st T is now 5.
Test 4t : implies x div by 5 status:
f,2,3,4, ,6,7,f,f,f x=lcd*5=2*2*3*5*7=420
Now 5 must be f since sum of t's is much less than 420.

So status:
***f,2,3,4,f,6,7,f,f,f with x=420 Tentative solution***

Now test 4f: then f,2,3,f, , ,7,f,f,f but need 3 consec t therefore status:
f,2,3,f,5,6,7,f,f,f
but this contradicts 5t since sum is 23 .ne. x=420 as implied by 7t.

Therefore, tentative solution is only one.

***1f,2t,3t,4t,5f,6t,7t,8f,9f,10f with x=420 ***

done

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All you have to do
is look at the opening sentence.
Below are 10 statements, all either true or false (This statement is promised to be true):
and the number 10 statement.

opening sentences is the only 100% true statement.
"below are 10 statements" - true
all either true or false..meaning they ALL are true or the ALL are false.
looking at number 10 it says that there can not be 3 trues in a row. therefore they are all false.

All you have to do
is look at the opening sentence.
Below are 10 statements, all either true or false (This statement is promised to be true):
and the number 10 statement.

opening sentences is the only 100% true statement.
"below are 10 statements" - true
all either true or false..meaning they ALL are true or the ALL are false.
looking at number 10 it says that there can not be 3 trues in a row. therefore they are all false.
Ok, but the challenge is to "Find the minimum admissible number", whatever that is.

D H
Staff Emeritus
all either true or false..meaning they ALL are true or the ALL are false.
looking at number 10 it says that there can not be 3 trues in a row. therefore they are all false.
If all are false then there are three consecutive false statements, making #3 true.

"All either true or false" means they are all decidable: No liar's paradox stuff going on here.

D H
Staff Emeritus
I see three solutions -- and that is assuming that the number is positive.

- Setting statement 1 true leads to a liar's paradox for statement 2. Therefore statement 1 is false, and therefore both 9 and 10 are false.

- With statement 1 false, statement 2 can be set to either true or false and be consistent with itself (and with statement 1 being false). 1=F,2=T is obviously consistent with statement 2. So is 1=F,2=F.

- A similar thing happens with statement 6. Setting it true means at least one of 7-10 must be true. Setting statement 6 false says absolutely nothing.

- For statement 10 to be false means that at a minimum one of the runs 2,3,4; 3,4,5; 4,5,6; 5,6,7; or 6,7,8 are all true.

- Statements 7 and 8 cannot both be true because 7 divides neither 40 nor 80. This eliminates the 6,7,8 run (or any longer run that ends in 6,7,8).

- Statement 8 is not consistent with statement 4 and a run of at least three. Statement 8 is false.

- Statement 7 is not consistent with statements 4 and 5.

- The potentially true statements are thus 2,3,4; 2,3,4,7; 2,3,4,6,7; and 3,4,5.

Solution #1:
2,3,4 are true, all the rest are false, and the number is 2.

- 10 F because 2,3,4 are true
- 9 F because 2 has zero divisors other than 1 and itself.
- 1 F because of the above.
- 8 F because the number is 2, not 30.
- 7 F because 3 and 4 do not divide 2.
- 6 F because setting statement 6 false doesn't say anything.
- 5 F because 2+3+4=9, not 2
- 2 T because it can be.
- 3 T because 5-10 are false.
- 4 T because it must be to make statement 10 false.

The only constraints on the number are from statements 4 and 9. The number must be a multiple of 2, but not so huge a multiple of 2 as to have more than 9 divisors. Since the problem is to find the minimal number, the answer for this solution is 2.

Solution #2:
2,3,4,6, and 7 are true, all the rest are false, and the number is 420.

- 10 F because 2,3,4 are true
- 9 F because 420 has 22 divisors other than 1 and itself, and 2+3+4+6+7=22.
- 1 F because of the above.
- 8 F because the number is 420, not 50.
- 5 F because 2+3+4+6+7=22, not 420.
- 2 T because it can be.
- 3 T because 8-10 are false.
- 4 T because it must be to make statement 10 false.
- 6 T because 7 is the last true statement.
- 7 T because 2,3,4,6, and 7 divide 420.

The desired number must be divisible by 2,3,4,6, and 7 per statement 7 and by 5 per statement 4. 12*7*5 is the minimal such number, and is also the maximal such number to avoid elevating statement 9 to true.

Solution #3:
3,4,5 are true, all the rest are false, and the number is 12.

- 10 F because 3,4,5 are true
- 9 F because 12 has 4 divisors other than 1 and itself, and 4<3+4+5=12.
- 1 F because of the above.
- 2 F because it can be.
- 8 F because the number is 12, not 30.
- 7 F because 5 does not divide 12.
- 6 F because setting statement 6 false doesn't say anything.
- 3 T because 6-10 are false
- 4 T because it must be to make statement 10 false.
- 5 T because it must be to make statement 10 false.

Note that 5 being true requires that the number be 3+4+5=12, which it is.

The solution:
The minimal admissible number is 2, from solution #1.

Well, you have arrived at 2 solutions that do not include 6T. It still seems to me that 6 must be T since if F it leads to a contradiction. not(this is not the last true statement)-> to me that (this is the last true statement) , but the statement itsself is F . Maybe you can convience me otherwise:)

D H
Staff Emeritus
Yikes! I just realized this thread is *old*. The OP did not respond in 2007 when the thread was current and most likely will not respond to the new set of posts this week.

The OP messed up if this was supposed to be a trick question (post #5). All answers cannot be false due to question #3, and all answers cannot be true due to question #10. I'll assume this isn't a trick question.

It still seems to me that 6 must be T since if F it leads to a contradiction. not(this is not the last true statement)-> to me that (this is the last true statement) , but the statement itsself is F . Maybe you can convience me otherwise:)
I read the question differently. However, I see your point. If the last true statement is 5 or less, statement 6 is not the last true statement, and thus statement 6 must be true: contradiction. A contradiction also results if the last true statement is 6. Therefore, the last true statement is 7 or more, and as statement 6 is not the last true statement, statement 6 must be true (as well as at least one of 7 to 10).

The answer is 420, which is what you said in post #4.

A much harder puzzle: Why didn't dontdisturbmycircles say this was the right answer two years ago?

DaveC426913
Gold Member
Here is my solution:

minumum #=42 (funny!)

Creeeepyyyy!

I am currently watching The Unusuals, wherein one of the main characters is obsessed with the number 42 and sees it on street signs, birthdates, bus routes and everywhere else he turns.

Creeeepyyyy!

Statement 2 is false for both cases per the opening statement, so it is false.

Statement 2 is false for both cases per the opening statement, so it is false.
One could argue that the opening statement is included in the logic set. So there would now be 11 statements. the opening is true, 1 is F and 2 would then be F... But my friend..you must now solve the puzzle under thiis assumption or are you asing us to do the work:)

D H
Staff Emeritus