The worlds first riddle!

TTH:

EIPNT = This one is inept
APREPSROPPORINATSEE = inappropriate response

What a Gal!


Anagram:

WOMAN HITLER = Mother-in-law

Thanks for the clue (the anagram is self-explanatory). I may have got one right!



I invite you to find a 5-digit number consisting of five different digits (not starting with zero) which when multiplied by 8 produces another 5-digit number consisting of the other 5-digits; in addition the sum of the digits of your 5-digit number must be greater than the sum of the digits of the product.

Which 5-digit number have you found

(Remember that the answer required is the number as it was before the multiplication.)

5-DIGIT NUMBER
[size=7]10679 * 8 = 85432[/size]

So, thoh
Do I have to keep chasing you
Why are you running anyway :wink:

btw Hi mark and I figured you would know the #

Hi TTH. I've got lots of numbers just laying around the house. Chances are, one of them will be the answer to any puzzle he puts forth.

You're cute mark Really, you are.....
Just don' tell your wife I said that :wink:

"You're cute mark really, you are....."


I'm not saying I'm jealous over Mark's success with the Chicks; it's just that even as a child, my parents had to tie a pork chop round my neck just to get the dog to play with me! But then; adolescence is the awkward age when a child is too old to say something cute and too young to say something sensible.



Mark:

5-DIGIT NUMBER
10679 * 8 = 85432


Does that look right to you?
Any number multiplied by 8 will be even. Therefore, the second 5 digit number must end in an even number, either 0, 2, 4, 6, 8

Possible digits to use are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The sum of these digits is 45. We know that the sum of the first 5 digit number must be higher than the sum of the second 5 digit number (the multiplied one). From this we know two things: The minimum sum of the first 5 digit number is 23, and the maximum sum of the second 5 digit number is 22.

The highest possible number can only be 5 digits, and must be even, so 99998 is the highest. However, since all digits must be unique, it can be lowered to 98764. Even that is not the highest number though. We know that the second 5 digit number has a maximum sum of 22. Therefore our highest number is now 98410. This is both even and has a sum of 22.

If the highest number for the second 5 digit number is 98410, then we know the maximum number for our first 5 digit number is 98410 divided by 8. Which leaves us with 12301.25 obviously that means that 98410 is too high a number, since it does not divide exactly by 8. However, that's not important at the moment. What is important is we know our first 5 digit number must begin with a 1 because we know it can't begin with a 0 and we have just worked out what its highest number in theory could be. The second digit can also only be 0 or 2. So our first 5 digit number is either 10??? or 12???.

We've worked out the highest value of the first 5 digit number. However, in reality it is not a valid number, because it is not whole, nor does its sum add up to 23 or more. So we can in fact work out what our lowest and highest 5 digit number is for the first 5 digit number.
Lowset possible is: 10589

Highest possible is a number lower than 12301, and with a sum of 23 or more. For now we will leave it at 12301. Because from this we can actually work out a bit more of our first 5 digit number. It can either begin 105??, 106??, 107??, 108??, 109?? or 120??

Let's work some more on those possible starts we have above.

105?? adds up to 6 and needs to be minimum of 23, so this leaves 17 to be made up of two numbers. Only possible numbers are 8 and 9. So we have either 10589 or 10598.
10589 x 8 = 84712
10598 x 8 = 84784
Neither of these work because there are repeated digits.

106?? adds up to 7 leaving 16 to be made up. This can only be made from 9 and 7, leaving us with:
10679 x 8 = 85432

And so would you believe it, there is our answer.

So to answer the original question which asks for the original 5 digit number, we have:

10679 which when multiplied by 8 produces 85432.
10679 has a sum of 23 and 85432 has a sum of 22, which satisfies the question's requirement.

Dang, that dude guessed lucky again!




AGRAAIN


FPUAL

Hey Try, You're cute in a blue sorta way. I have told you that in niagara?
painful?

AGRAAIN

[size=7]rain again[/size]

Hi mark
I think your answer is better

TTH:

FPUAL = painful

Cute in an angelic sorta way!


Mark:

AGRAAIN = rain again

I'm saying nothing!



In a math test the candidates were given four different three-figure numbers with no two differing by more than 20. They then had to label them (a), (b), (c) and (d) such that:

(a) Was a perfect cube;
(b) Was a product of four different primes;
(c) Was the sum of some consecutive integers from 1 upwards, and;
(d) Was a perfect square.

Try guessed and allocated the numbers so that (a), (b), (c) and (d) were in increasing order. He got none right.

What were the correct answers

Mark, thoh or someone else....Try is playing with #'s again :wink:

4 NUMBERS
[size=7]343, 330, 325, 324[/size]

Mark, I knew you would know
I am just assuming you got them right, I didn't check the #'s

We must accept finite disappointment, but we must never lose infinite hope; because the truth does not depend on a consensus of opinion.



Mark:

4 NUMBERS
343, 330 , 325, 324


Thanks buddy, that's my assignment wrapped up!



Letter A - All possible 3 digit cubes are:
125, 216, 343, 512, 729

We can eliminate 125 from this because of the letter C having a minimum of 210. And also 729 because it is also a possible answer for the letter D, leaving us with:
216, 343, 512

Letter B - All possible primes to use are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

Lowest possible number is 2x3x5x7 = 210, therefore, for the letter A we can eliminate 125 as an answer.

Letter C - From the letters A and D and B, we now have a maximum and minimum number possible, which are 532 and 196. So we now list all the possible answers for C that are in between those numbers, which is all the numbers between 20 and 32 (by this I mean if you did 1+2+3....+20, all the way up to 32). The answers each gives is:
210, 231, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528

If we then eliminate any that are more than 20 away from any of our answers to A and D it leaves us with:
210, 231, 325, 351, 496, 528

Letter D - All possible 3 digit squares from 196 to 749 are:
196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784

We can then eliminate any of these numbers that are more than 20 away from the 4 possible numbers we have for the letter A:
This leaves us with the following:
196, 225, 324, 361, 529

Now we can look at all the possible A, C and D combinations:

1. a) 216 c) 210 d) 196
2. a) 216 c) 210 d) 225
3. a) 216 c) 231 d) 225
4. a) 343 c) 325 d) 324
5. a) 343 c) 351 d) 361
6. a) 512 c) 528 d) 529

Straight away we can rule out numbers 5 and 6 because in those, A is the lowest number and D is the highest. Therefore, no matter where you inserted a value for b), either A would still be the lowest or D would still be the highest.

Lets take a look at the 4 possibilities in more detail:
1. B can only go in between C and A, therefore it must equal between 211 and 215. We know the lowest possible value for B is 210, and the next lowest possible is 2x3x5x11 = 330. Therefore it can't be this solution.

2. B can only go after D for this one to work, meaning it must equal between 226 and 230. Again this is not possible as we know from possibility number 1 above.

3. B can only go before A for this to work, so it can be between 211 and 215. We already know this isn't possible.

4. B can actually in two places here. It could go after A, in which case it must equal 344, or it can go after C, in which case it must be in between 326 and 342. Well we have our answer as we know that the next lowest value of B was 330, which fits in perfectly.

So this leaves us with Mark's answer.



TTH, I am sure your encouragement is appreciated by all.





CBALTTUERDIEESD


CLNINOETD

Double post.

[size=7]batteries included
not inclined[/size]

Uh, is this a riddle or am I staring at the words for nothing

Mark:A triangular-based pyramid is constructed of three tiers of blocks, lettered sequentially A to C, D to I and J to R.

Mark and Thoh were to replace the letters with digits, so that A, B and C were all different, A+B+C were even, and AC was to be a prime.

All sequential digit pairs AB, BC, ..., QR were to be primes, except DE. Similarly, all sequential digit triads ABC, BCD, ..., PQR were to be primes, except CDE. On the 'A' pyramid face AD, DJ, ADJ, AE, EL and AEL were all to be primes, as were ADE, DJK and EKL; the same was to be true for the corresponding digit pairs and triads on the B and C faces.

Thoh and Mark each found a solution, differing only in one of the eighteen block faces, Thoh's having the lower digit.

What were:

(i) The letter of the position with the different digits in the two solutions

(ii) Thoh's and Mark's digits for this position

I didn't realize that I did that. I notice I am double clicking more due to being on my sister's pc