Puzzles with numbers and things

[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]

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?

[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]

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!

[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

[moglet]

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

[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!!!

[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]

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?

[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]

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!

[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.

[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.

[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.

[abextra]

Thank you all for explanations, very well done.