Puzzles with numbers and things

[snookersfun]

well, one could have eliminated more than 6 digits (next option would have been 8 digits btw., as we speak of 2 middle digits) by the sum = 1400 info, as the highest possible multiplication would yield 2x699x699=977202 as upper limit.

[moglet]

My approach was vey similar.
Only I didn't miss the 443xxx so I ended up with 2 candidates : 887314 and 443794. Checking divisibility by 13 eliminated the latter.
I was still not sure about my solution because the number could possibly have had more digits. So I checked the other constraints and found out they were all met.

Yes, my simple mind looked no further than the six digits that gave a result and I thought it might get a bit hairy with such large numbers.

[Monique]

My approach was vey similar.
Only I didn't miss the 443xxx so I ended up with 2 candidates : 887314 and 443794. Checking divisibility by 13 eliminated the latter.
I was still not sure about my solution because the number could possibly have had more digits. So I checked the other constraints and found out they were all met.

[snookersfun]

with that we can now officially close R. 367

moglet, bless you for writing all that up. Very well done.
so again congratulations to Mon, moglet and d_g for solving this round!

[moglet]

Aping Primes

This is how I looked at the problem:

The numbers one less and one more each have 5 prime factors all different, the number we are looking for has to have 2 prime as one its factors in order for the sum of the three that make the number end in a zero, all other primes are odd and we can't sum three different primes greater than 2 to reach an even number never mind 1500 or 1400. If we look at the first 10 primes above 2 we can't multiply any 5 and at the same time the other five without coming up with a number less than 5 digits long, so we are looking for a number 5, likely 6, or more digits long and even one.

We know that the number has two middle digits so the number has to have at least 6 digits and be even.

To fit the condition for the middle two digits only three are possible 37, 53 and 73. I assumed here the number was six digits long and to make the first three digits prime in itself we could exclude 53. So now we are looking for a number ??37?? or ??73??. To satisfy the condition: 3 digit number that is prime who's digital sum is also prime only 443 and 887 are available, if we do the second digital sums we get 11 and 23 and then the third digital sums we get 2 and 5, there should be a further digital summing which is where I had a mental block. So, notwithstanding this, we are looking at a number 4437?? or 8873??. (in my first jottings I missed the 443?? number and went straight for the other!)

We know the final digits are not prime and we know that the digital sum of the number is prime and if reversed is also prime, 8873 sums to 26 and the next primes above that that are "reversible" are 31 and 37 so we need to find two non-prime numbers that sum to 5 or 11, only 1 and 4 satisfy, so the number I opted for was 887314 dig sum 31(13).

Set about factoring 887314 and we get 2, 487 and 911, these sum to 1400 (not 1500). The digital sums are 2, 19, 11, the product of these is 418 whose digital sum is 13 (31).

Just checking the other two numbers 887313 and 887315 factoring gives 3, 7, 29, 31, 47 and 5, 11, 13, 17, 73 respectively.

The other number, which I missed first time round was 443794 which is divisible by 13. Here factors 2, 13 need a prime to reach 1500/1400, no prime apart from 5 ends in a 5.

[snookersfun]

Sorry, no answer from me this time. And the next round is above my level too...

Just please post the solutions and let's move on. :snooker:

ok, can close my round then, does anybody feel like explaining how he reached his solutions and put them up?

[abextra]

...
Will leave it open a little longer in case there are any late entrants!

Sorry, no answer from me this time. And the next round is above my level too...

Just please post the solutions and let's move on. :snooker:

[snookersfun]

update: d_g, Mon and moglet all found the number Charlie had in mind

Mon also found the second number Gordon had taken into consideration before doing any actual factoring (apart from the obvious odd or even), moglet as well now and d_g plumped the number now, too.
Very well done everybody!!!

[moglet]

Aaw, shucks, did I mention counselling, definitely need to move on to medication.

[snookersfun]

Haha! Funny question



But I don't understand this bit. Why would Gordon not have realised instantly that one of the numbers was divisible by 13?

Every gorilla I have ever met knows that a number is divisible by 13 if the sum of 4 times its last digit and the number formed by the remaining digits is divisible by 13. E.g. 143 is divisible by 13 because 4x3+14 = 26 is divisible by 13. So what was Gordon thinking?

yep, why indeed, beats me (luckily I didn't think it out, silly people those). Never knew that though, but then I am not a gorilla
But let's just assume he never checked for the divisibility by 13 part, what is the second option he came up with?

Is Charlie sure about this bit? With a bit of guessing, can get what seems to be a fairly obvious answer to satisfy the rest (I think - not fully checked yet) - but this would give sum of prime factors of 1400!

now this is worse, yep not 1500 at all as well, that should read 1400, but one can solve it nicely without knowing this fact anyway, as d_g just proved! So, well done. I now only need to get that second number!

"The middle two digits in the number are its only prime digits and the number formed by the middle two digits is also prime. The number formed by the first three digits in the number is prime and its digital sum is also prime. In fact, the digital sum of the digital sum, and the digital sum of the digital sum of the digital sum of the first three (numbers)- digits are also both prime."

and moglet found another mistake, guess numbers towards the end of the paragraph needs to read digits

will go edit the original question.
brownie points for spotting the mistakes

[davis_greatest]

R366 update

Meanwhile, the list of successful solvers to round 366 reads: snookersfun, moglet and Monique! All found the "amazing answer"! Congratulations to all. Will leave it open a little longer in case there are any late entrants!

[davis_greatest]

... The sum of the prime factors of the number is 1500."

Is Charlie sure about this bit? With a bit of guessing, can get what seems to be a fairly obvious answer to satisfy the rest (I think - not fully checked yet) - but this would give sum of prime factors of 1400!

[davis_greatest]

Haha! Funny question

Charlie frowns and says, "One of your numbers is divisible by 13, silly."

Gordon smiles sheepishly ...

But I don't understand this bit. Why would Gordon not have realised instantly that one of the numbers was divisible by 13?

Every gorilla I have ever met knows that a number is divisible by 13 if the sum of 4 times its last digit and the number formed by the remaining digits is divisible by 13. E.g. 143 is divisible by 13 because 4x3+14 = 26 is divisible by 13. So what was Gordon thinking?

[snookersfun]

I found this one quite interesting, so I'll pass it on

R.367 aping around with primes

As we all know, Charlie and Gordon have this ongoing competition to stump each other. This time Charlie tells Gordon he's thinking of a number.

"The numbers one less and one more than the number are both the product of five prime numbers. The three numbers together have thirteen prime factors, all different. The sum of the prime factors of the number is 1400."

"OK," says Gordon after a moment. "That's probably enough information to find the number using excel or programming, you don't want me to do that, do you?"

"No, no, no," replies Charlie. "I don't want you to do anything as inelegant as that. All the puzzles on this thread, at least when posed by d_g, can basically be done in one's head, so here's some more information."

"The digital sum of each of the number's prime factors is prime, as is digital sum of the product of these sums. In fact, if you reverse the digits in the product's digital sum you get a different prime number that is the digital sum of the number I'm thinking of."

Gordon takes out a pad of scrap paper (as the in the head approach doesn't quite work out) and starts jotting down some notes.

"The middle two digits in the number are its only prime digits and the number formed by the middle two digits is also prime. The number formed by the first three digits in the number is prime and its digital sum is also prime. In fact, the digital sum of the digital sum, and the digital sum of the digital sum of the digital sum of the first three digits are also both prime."

"What number am I thinking of?" asks Charlie.

Gordon jots down a few more notes and then does a couple of calculations. He says, "OK, I know it's one of two numbers, but so far I didn't need much calculating power and I don't feel like really factoring these two to figure out which one is the right one."

Charlie frowns and says, "One of your numbers is divisible by 13, silly."

Gordon smiles sheepishly and tells Charlie the number.

What were the two numbers and which one was the one Charlie was thinking of?

Hint: don't use brute force calculations, or even calculator, no need to factor big numbers nor find huge primes

[davis_greatest]

Right, some hints then to round 366 .... in hidden text, so just select the text to look at the hints if you want them!

Hint 1:
(
Ignore Gordon’s crates at first. First work out how many crates Oliver must have. )


Hint 2: Find the red herring
(
“Oliver had bought the maximum possible number of crates consistent with the above information.” is a red herring. Why? )


Hint 3:
(
It is a red herring because there can only ever be one possibility for the number of crates Oliver had – and you can work that out even without needing to know the total number of balls there are! What is the only possibility? )


Hint 4: Simple expression for number of balls Charlie has

(
Suppose Charlie’s grid has x rows and y columns. Then it has xy balls (OK – not much of a hint there!) So number of balls in a crate = xy )


Hint 5: How many balls does Gordon steal?

(
First he takes a row of y balls. The he takes a further ?-? balls from one of columns left.

So the number of rows in the top layer of Oliver’s triangle is y+?-? and the number of rows in the bottom layer is ?+? = n( say) )


Hint 6: What is the form of the number of red balls of Oliver?

(
Take a triangular pack with n rows and then take another slightly smaller pack with n-1 rows. Move the balls in each triangle a little so that each triangle becomes right-angled – now you can fit the two triangles together to form a _QU_R_ with ? rows.

Or show algebraically that n(n+1)/2 + n(n-1)/2 = … )


Hint 7: finding the number of crates for Oliver (part 1)

(
So Oliver has n^2 = (x+y)^2 balls which must divide the number of that Charlie had in his one crate.

If (x+y)^2 is divisible by xy, what does this tell you about x and y? )



Hint 8: finding the number of crates for Oliver (part 2)

(
Divide x and y by any common factors, to get x’ and y’
Then we know (x’+y’)^2 is divisible by x’y’
So x’^2 + y’^2 is divisible by x’y’, and hence also by x’
Since the first term x’^2 is divisible by x’, y’^2 must be divisible by x’. So y’ is divisible by x’, as they have no common factors.

Similarly, by symmetry, x’ is divisible by y’, so x’=y’ (=1).

Hence x=y and Charlie’s grid is a _QU_R_! )


Hint 9:

(
You should now be able to work out the number of crates Gordon has, and the number in each crate is a square.)

Hint 10 : Working out how many crates Gordon has (part 1)

(
If the number of crates Gordon has is as small as possible, the number of balls in each crate must be as ----- as possible! )


Hint 11 : Working out how many crates Gordon has (part 2)

(
What are the factors of 147 million? What is the highest square that divides 147 million? If that is too big, what is the next biggest square that divides 147 million?)