Maths challenge
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dewtell
Posts: 29

Posted: 12:55PM Feb 5, 2015 

Ok, lets try for an easier one this time. The number 21 isn't a prime, but it could be changed into one by either changing the first digit (to 11) or the second digit (to 23). What is the smallest positive integer that is neither a prime itself, nor can it be changed into a prime by changing one digit?

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dewtell
Posts: 29

Posted: 04:49PM Feb 5, 2015 

And JQ, to answer the question you intended to ask, the discriminant of that quadratic is 16 cos^2 θ  24 sin θ, and the denominator is 6 sin θ. To have distinct real roots, we need the discriminant > 0 and the denominator nonzero. Rewriting the discriminant, and factoring out 8, we have 2 sin^2 θ  3 sin θ +2 > 0, or 2 sin^2 θ + 3 sin θ  2 < 0. Factoring the LHS, we get: (2 sin θ 1)(sin θ + 2) < 0. The second factor will always be positive, so this reduces to 2 sin θ  1 < 0, or sin θ < 1/2. Combine with the requirement that sin θ =/= 0, and we should get solutions where (7π/6 + 2πk < θ < π/6 + 2πk) and (θ not an integral multiple of π).

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JQPublic
Posts: 1895

Posted: 08:45AM Feb 6, 2015 

According to the prime number list (hope that doesn't count as cheating ...), there is at least one prime for every 10number interval below 200.
200 seems to be the answer. Since the last digit is zero, obviously we can't get any primes by changing the first two digits. 201 (3), 202 (2), 203 (7), 204 (2), 205 (5), 206 (2), 207 (3), 208 (2), 209 (11) are all composite.
P.S. The 'trap' intended by the book is sin θ =/= 0, since it's easy to forget that.
This message was edited on 08:46AM Feb 6, 2015
'An idea, like a ghost, must be spoken to a little before it will explain itself.'  Charles Dickens 
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dewtell
Posts: 29

Posted: 11:09AM Feb 6, 2015 

200 is correct. As you noted, the key is to see that you are looking for the first primefree decade, which you can get to by noting that the last digit of the number must be 0 (because if it wasn't, then you could get a smaller number from whatever the answer was by changing the last digit to 0), and then noting that if the last digit is 0, the only digit that is a candidate for changing is that last digit.
I was able to solve it in my head without a prime list by noting that the key numbers in any decade are those ending in 1, 3, 7, and 9, because the others will be nonprime for any decade but the first. 3 will divide at most 2 of those (either 1 and 7 or 3 and 9), because any number divisible by 3 must have a digit sum divisible by 3, and 7 will divide at most 1 of those, so we need to have at least one of the numbers with a smallest prime factor of 11 or greater. So the first possible decade is the one including 121. 123 and 129 are divisible by 3, but 127 isn't divisible by 7, so it won't work. The next possible decade is the 140s, where we add 22 to 121 to get 143. 141 and 147 are divisible by 3, but 149 isn't divisible by 7, so it won't work. The 160s has 169 divisible by 13, but otherwise is completely unpromising, since both 3 and 11 hit 165 instead of the four numbers we care about. 13 is a nonfactor for the 180s (182), 11 hits 187, and 3 gets 183 and 189, but 181 isn't a multiple of 7. 13 won't be any help in the 190s (195), so we skip to the 200s, where 11 gets 209, 3 gets 201 and 207, and 203 *is* divisible by 7. Bingo!
Your turn to pose.

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JQPublic
Posts: 1895

Posted: 10:42AM Feb 7, 2015 

Hmm. I gave this some thought and eventually decided to copy from a book again.
CDE is a triangle where CE = DE. B is a point on DE and F is a point on CE. BF is produced to A such that AB = AC. Angle BAC and angle DEC are equal. Prove that AE and CD are parallel.
'An idea, like a ghost, must be spoken to a little before it will explain itself.'  Charles Dickens 
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dewtell
Posts: 29

Posted: 05:19PM Feb 8, 2015 

I tried to see if I could create an example of your triangle problem using right angles (to make the problem a little easier), and wound up finding a counterexample. On a standard cartesian grid, let C(4,0), D(0,4), E(0,0), and B(0,2) be points for this problem. Then there are two possible locations for A satisfying the constraints of the problem: A(3,3), implying F(6,0), and A(1,1), implying F(2/3,0). With the first location, length(AB) = length(AC) = sqrt(10), and angle BAC is a right angle congruent to DEC, but AE and CD are perpendicular, not parallel. Only the second location for A results in them being parallel.
It might be the case that if F is constrained to lie between E and C, then the lines must be parallel, but it's not true in general. For a given set of locations for C, D, E, and B, there will be two possible locations for A that have the angles match: both located on the perpendicular bisector of BC, one on each side of BC. At most one of those will yield parallel lines for AE and CD.

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JQPublic
Posts: 1895

Posted: 07:55AM Feb 9, 2015 

dewtell wrote: I tried to see if I could create an example of your triangle problem using right angles (to make the problem a little easier), and wound up finding a counterexample. On a standard cartesian grid, let C(4,0), D(0,4), E(0,0), and B(0,2) be points for this problem. Then there are two possible locations for A satisfying the constraints of the problem: A(3,3), implying F(6,0), and A(1,1), implying F(2/3,0). With the first location, length(AB) = length(AC) = sqrt(10), and angle BAC is a right angle congruent to DEC, but AE and CD are perpendicular, not parallel. Only the second location for A results in them being parallel.
It might be the case that if F is constrained to lie between E and C, then the lines must be parallel, but it's not true in general. For a given set of locations for C, D, E, and B, there will be two possible locations for A that have the angles match: both located on the perpendicular bisector of BC, one on each side of BC. At most one of those will yield parallel lines for AE and CD.
Sorry, my bad again. the original problem was a diagram and I translated it into words. Clearly, my translation was quite muddy (no oxymoron intended there). When I said BF was produced to A, I thought it would be understood that BF is extended in the direction from B to F, but I guess this isn't the case. When I said F is a point on CE, I meant the line segment CE, not the line CE, but again I guess I wasn't clear there. Hope this clears things up.
'An idea, like a ghost, must be spoken to a little before it will explain itself.'  Charles Dickens 
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ultimatecub
Posts: 1679

Posted: 09:10AM May 31, 2015 

I'll start this topic back up.
You have 4 circles arranged so they each touch the ones next to it. (Hope you get what I mean.) There is an area in the middle of the 4 circles that kind of looks like a square with the sides bending inwards.
Each circle has the radius of 20cm.
WITHOUT using pi or anything, how do you work out the area of the centre section and one of the circles collectively?
A bird in the hand is worth all the birds in all the bushes of the world, because if the other birds are in the bush, how will you sell them? 
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dewtell
Posts: 29

Posted: 03:39PM May 31, 2015 

If you draw a square connecting the centers of the four circles, you have a square 40 cm on a side that includes the whole of the center area plus 1/4 of each circle. This is obviously equal to the area of the center area + one whole circle, and is 1600 cm^2.

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dewtell
Posts: 29

Posted: 04:11PM May 31, 2015 

Here's an old one: Three men, A, B, and C, prepare to fight a threeway duel. A is a perfect marksman who always kills his target, B is the second best, hitting and killing his target 2/3 of the time, and C is the worst, hitting and killing only 1/3 of the time. To compensate, they agree that C will take the first shot, followed by B (if he survives), and then A, continuing in CBACBA order with single shots until there is only one survivor. To maximize his chance of survival, who should C target with his first shot, and what is his probability of surviving the duel?

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ultimatecub
Posts: 1679

Posted: 04:27AM Jun 1, 2015 

Did you look that up
Sorry I know this already, in the ground because B will go for A not him and if he wins C gets to shoot B, if he misses A will kill B not him as he is better.
Will post another one later.
A bird in the hand is worth all the birds in all the bushes of the world, because if the other birds are in the bush, how will you sell them? 
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Hydra1234
Posts: 2116

Posted: 06:07AM Jun 1, 2015 

Sorry is this is a bit rude, but isn't thaat logic? I once was going to post it, but I saw the exact same teaser in a logic section. Here is a easy one that I found in a old textbook my son had.
100 fish are caught, tagged and released. Later, 70 fish are caught and 8 of them are tagged. Approximately how many fish are there in the lake?
When life gives you lemons, chuck a rabid hydra to life and enjoy the lemonade. 
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ultimatecub
Posts: 1679

Posted: 07:31AM Jun 1, 2015 

8/70 x 100 = about 1100
(this might be wrong, I remember seeing this in a biology textbook but am at school now and they don't have the same one)
A, B and C work together to sow a field. The field has 9 parts. They were meant to do 3 each but C falls down the stairs before the work day so A and B cover him. C will pay them for that.
A does 6 parts.
B does 3 parts.
C has 9 coins. How many should he give to each of A and B?
Also, this is NOT ridiculously easy.
@Hydra on dewtell's puzzle: logic is similar to maths
A bird in the hand is worth all the birds in all the bushes of the world, because if the other birds are in the bush, how will you sell them? 
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mnkgyl
Posts: 700

Posted: 08:02AM Jun 1, 2015 

C will pay only A the amount for his 3 parts. I don't know the cost for doing each part.
Que Sera Sera 
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ultimatecub
Posts: 1679

Posted: 10:14AM Jun 1, 2015 

So sorry it should have been 5 parts and 4 parts not 6 and 3
A gets 6 and B gets 3. They were supposed to do 3 each but C would not pay them for doing their OWN work only for his work.
Subtract each from 3 giving 2 and 1 so it should be 2:1 ratio
SORRY I WROTE THE PROBLEM WRONG AND A SHOULD HAVE DONE FIVE PARTS AND B FOUR PARTS. YOU GET THE POINT BECAUSE I WROTE THE QUESTION WRONG.
A bird in the hand is worth all the birds in all the bushes of the world, because if the other birds are in the bush, how will you sell them? 
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celtichero
Posts: 804

Posted: 10:16AM Jun 1, 2015 

Hydra1234 wrote: Sorry is this is a bit rude, but isn't thaat logic? I once was going to post it, but I saw the exact same teaser in a logic section. Here is a easy one that I found in a old textbook my son had.
100 fish are caught, tagged and released. Later, 70 fish are caught and 8 of them are tagged. Approximately how many fish are there in the lake?
This one is definitely 875.
I'm just an ordinary average guy with nothing to lose. 
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dewtell
Posts: 29

Posted: 11:15PM Jun 1, 2015 

@ultimatecub & Hydra  there was a second part to the question (what is his probability of surviving)?

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