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RE: Riddles, brain puzzles and mathematical problems - MyRedNeptune - 12-13-2012

Yay, I was close! ^^

RE: Riddles, brain puzzles and mathematical problems - BAndrew - 12-22-2012

I will post my riddle on a new seperate thread, but I will keep this thread for general riddles.

Stay tuned. The first part will be online soon.

RE: Riddles, brain puzzles and mathematical problems - BAndrew - 03-23-2014

Almost forgot about this thread. Here goes:

Find the number w with the help of the following sentences:

1.At least one of the sentences 9 and 10 are true.
2.This is either the first true or the first false sentence.
3.There are 3 successive false sentences.
4.The difference between the numbers of the last true sentence and the first true sentence devides w exactly. (eg. 2 devides 10 exactly)
5.w is the sum of the numbers of all the true sentences.
6.This is not the last true sentence.
7.The number of every true sentence devides w exactly.
8.w is the percentage (%) of all true sentences.
9.The quantity of the positive numbers that devide w exactly (excluding/without 1 and w) is greater than the sum of the numbers of all true sentences.
10.There aren't 3 successive true sentences.

RE: Riddles, brain puzzles and mathematical problems - MyRedNeptune - 03-23-2014

(03-23-2014, 05:53 PM)BAndrew Wrote: Almost forgot about this thread. Here goes:

Find the number w with the help of the following sentences:

1.At least one of the sentences 9 and 10 are true.
2.This is either the first true or the first false sentence.
3.There are 3 successive false sentences.
4.The difference between the numbers of the last true sentence and the first false sentence devides w exactly. (eg. 2 devides 10 exactly)
5.w is the sum of the numbers of all the true sentences.
6.This is not the last true sentence.
7.The number of every true sentence devides w exactly.
8.w is the percentage (%) of all true sentences.
9.The quantity of the positive numbers that devide w exactly (excluding/without 1 and w) is greater than the sum of the numbers of all true sentences.
10.There aren't 3 successive true sentences.

w = 168

RE: Riddles, brain puzzles and mathematical problems - BAndrew - 03-23-2014

@MyRedNeptune
Just the answer won't do it. You have to describe how you found it.

RE: Riddles, brain puzzles and mathematical problems - Wapez - 03-23-2014

We have a 3-digit number, that we call abc.

If we shuffle the numbers around, and get the combination acb, the number grows by 9.
If we instead shuffle them around and get the combination bac, the number grows by 90.

How much larger/smaller would the combination cba be, in relation to the original number, abc?

Mathematical proof is needed!

RE: Riddles, brain puzzles and mathematical problems - BAndrew - 03-24-2014

(03-23-2014, 11:30 PM)Wapez Wrote: We have a 3-digit number, that we call abc.

If we shuffle the numbers around, and get the combination acb, the number grows by 9.
If we instead shuffle them around and get the combination bac, the number grows by 90.

How much larger/smaller would the combination cba be, in relation to the original number, abc?

Mathematical proof is needed!

Spoiler below!

acb = abc + 9:

100a + 10c + b = 100a + 10 b + c + 9 <=>
10c + b = 10b + c + 9 <=>
9c = 9b + 9 <=>
c = b + 1 <=>
b = c - 1 (1)


bac = abc + 90:
100b + 10 a + c = 100a + 10 b + c + 90 <=>
90b = 90a + 90 <=>
b = a + 1 (2)

(1),(2) => c-1 = a+1 <=>
c-a = 2(3)



cba = 100c + 10b + a = 100a +10b + c - 99a + 99c =
100a + 10b + c + 99(c-a) = with (3)
= 100a + 10b + c + 198

cba = abc + 198

(I hope I haven't made any stupid mistake here)




RE: Riddles, brain puzzles and mathematical problems - MyRedNeptune - 03-24-2014

(03-23-2014, 10:36 PM)BAndrew Wrote: @MyRedNeptune
Just the answer won't do it. You have to describe how you found it.

Sorry, was in a hurry so didn't have time to type it up. Tongue

I noticed that you have edited the text of sentence 4. I will adjust my answer accordingly.


First, let's look at sentence 2. If it is true, then it must be the first true sentence, therefore sentence 1 must be false. If it is false, then it cannot be the first false sentence. Therefore, sentence 1 must be false. It follows, then, that sentence 1 is false at all times.

If sentence 1 is false, then both sentences 9 and 10 are false.

Sentence 6 can only be true. If it were false, then it would be not the last true sentence, and therefore true, which is a paradox.


What we have now is this initial setup:

1. f
2.
3.
4.
5.
6. t
7.
8.
9. f
10. f


In other words:

-there are 3 successive true sentences
-the number of positive (and, I'm assuming, natural) divisors of w, excluding 1 and w, is greater than the sum of the numbers of all true sentences
-there is at least one true sentence after sentence 6


With this in mind, we can move on to the different cases.

Case I.
Spoiler below!

Let's assume that 2 is false.

a) If 2 is false then 3 cannot be false and is therefore true.
b) Because 3 is true, 8 must be false or it is impossible to satisfy the condition.
c) Because 6 is true and 8, 9 and 10 are false, then 7 is true.

Now we have this:

1. f
2. f
3. t
4.
5.
6. t
7. t
8. f
9. f
10. f

d) Because 10 is false, then 5 must be true. If 5 is true, then w is either 21 or 25, neither of which satisfy the condition of sentence 7.

Therefore, 2 cannot be false.


Case II.
Spoiler below!

Since we have discovered that 2 cannot be false, we now consider it to be true. Let's assume that 3 is false.

a) If 3 is false then 8 must be true. This means that w is a multiple of 10.

1. f
2. t
3. f
4.
5.
6. t
7.
8. t
9. f
10. f

b) If 3 is false then at least one out of 4 and 5 must be true.
c) If 4 is true, then w must be a multiple of 6 and 10 - the only possible numbers being 30 or 60. However, if 4 is true, then w > 30. If w = 60, then 5 and 7 must also be true. 5 and 7 cannot be true when w = 60.
d) If 5 and not 4 is true, then w is either 21 or 28, which are not multiples of 10.

Therefore, 3 cannot be false.


Case III.
Spoiler below!

2 and 3 are true.

a) Because 3 is true, 8 must be false.
b) Because 8 is false and 6 is true, 7 must be true.

1. f
2. t
3. t
4.
5.
6. t
7. t
8. f
9. f
10. f

c) because 10 is false, at least one out of 4 and 5 must be true.
d) If 5 is true, then w is either 23 or 27, which does not satisfy the condition of sentence 7. Therefore, 5 is false and 4 is true.

5*7*4*3 = 420

The number 420 has 22 divisors (2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210) and the sum of the numbers of all true sentences is 22. This is consistent with sentence 9.

Therefore, w = 420


Also,
(03-24-2014, 12:51 AM)BAndrew Wrote: (I hope I haven't made any stupid mistake here)

(03-24-2014, 12:51 AM)BAndrew Wrote: 9c = 9b + 9 <=>
c = b + 9

Tongue

RE: Riddles, brain puzzles and mathematical problems - BAndrew - 03-24-2014

@MyRedNeptune
Congratulations! You got it! Smile



Do you want to post a problem? Let me know, because I have something in mind Wink



PS. Oh my God! I am an idiot. (facepalm)
Nevermind I corrected it, thanks.

RE: Riddles, brain puzzles and mathematical problems - Wapez - 03-24-2014

(03-24-2014, 12:51 AM)Titanomegistoterastiotatos Wrote:
Spoiler below!

acb = abc + 9:

100a + 10c + b = 100a + 10 b + c + 9 <=>
10c + b = 10b + c + 9 <=>
9c = 9b + 9 <=>
c = b + 1 <=>
b = c - 1 (1)


bac = abc + 90:
100b + 10 a + c = 100a + 10 b + c + 90 <=>
90b = 90a + 90 <=>
b = a + 1 (2)

(1),(2) => c-1 = a+1 <=>
c-a = 2(3)



cba = 100c + 10b + a = 100a +10b + c - 99a + 99c =
100a + 10b + c + 99(c-a) = with (3)
= 100a + 10b + c + 198

cba = abc + 198

(I hope I haven't made any stupid mistake here)


Well done, correct! Tongue

Please spoiler it though, so other people have the same chance.