The worlds first riddle!

Although even if your reading is correct...I do not see how your answer is correct.

5-digit number: 3*(100000 + x) = 10x + 1
300000 + 3x = 10x + 1
299999 = 7x
x = 299999/7
x = 42857
Highlight to reveal.

What am I missing? 14257 is 5 digits...x 3 = 42771 = 5 digits...

I am going to work the puzzle using SM reading...with just 4 numbers. I'll submit a PM.

Ahh ok, so 5 different digits? thank you...

14287 x 3 = 42861 /3 = 14287...5 digits each...Is this correct?

Since you guys are intent on posting here...

...my answer is 42857.

If you put a 1 in front of that number, you get 142857. Multiply that by 3...and you get 428571.

The 42857 in the middle is the key if this truly is the reading of the puzzle. Put a one in front of it...and it is 3 times less than the number if you put a 1 at the end of it.

Use the cursor

This is as close as I am going to go...Because I am beat and want to lay down...If I wanted to sit here...I think I could get it...

12857 x 3= 38571

I think you were correct Bennett...I can only see it being done with the number 142,857 x 3 = 428,571...That number is a cyclic number...

I am too beat to figure out 5...g'night all...

I deleted my first response because markr beat me to it. There is a brute force method to solve this which I've just figured out, if you're interested. So bare with me, it is somewhat long.

It's a five digit number so give each digit in the perspective place value a variable: _ _ _ _ _ = a s d f g.

And here is the simple equation we have to work with.

1 a s d f g
x______3
a s d f g 1

This is the logic:

g x 3 ends in 1 so g must be 7 since 7x3 = 21, and carry the 2
f x 3 ends in 7 minus the remainder 2 so f must be 5 since 5x3 = 15, and carry the remainder 1 from g
d x 3 ends in 5 minus the remainder 1 so d must be 8 since 8x3=24 and carry the remainder 2
s x 3 ends in 8 minus the remainder 2 so s must be 2 since 2x3=6, no carry)
a x 3 ends in 2 so a must be 4 since 4x3=12 and carry the 1 and since x 3 ends in 4-1, This is true.

Or to put it more elgantly. (Carried numbers in BOLD)

_____2
1 a s d f 7
x______3
a s d f 7 1

____1
1 a s d 5 7
x______3
a s d 5 7 1

___2
1 a s 8 5 7
x______3
a s 8 5 7 1


1 a 2 8 5 7
x______3
a 2 8 5 7 1

1
1 4 2 8 5 7
x______3
4 2 8 5 7 1

Run the cursor on my post above. That is what I got also.

I knew the last number had to be a 1…and the only number that ended in a 1 when multiplied by 3 was 7. That left 2 remainder…and I knew the next number (from the end) had to end in a 7…so 3 x 5 = 15 + the 2 got that. The next number had to be a 5 and had to account for the 1 remainder…so 3 x 8 = 24 + 1 = 25. The next number had to be an 8…so 3 x 2 + 2 + 8. The next number had to be a 2…(there was no remainder) so 3 x 4 = 12. The next number had to be a 4 (there was a 1 remainder)…so 3 x 1 + 1= 4.



So the numbers first to last became: 4 2 8 5 7. The check was: 3 x 142857 = 428571

Hi Frank, what a delight to see you here again. I still remember our exchange of about a decade ago when you were ruminating on a puzzle while getting in and out of the shower; one of the funniest exchanges ever.


I must congratulate you all on the ways you found to come up with the answer, I am sure it will inspire many others to think in creative ways to reveal their true potential.


Using an easy equation:
3(100000 + x) = 10x+1
(Why? Well, adding 100000 puts a 1 at the front of a five-digit number, and multiplying by 10 and adding 1 puts a 1 at the end of a number)
Solving this gives:
10x+1 = 3(100000 + x)
10x+1 = 300000 + 3x
10x = 299999 + 3x
7x = 299999
x= 299999/7 = 42857
The answer is 42857 as many have proved.



This question also requires some clever thinking:

I was on Tee Vee the other day; although I doubt that was her real name; more of a stage name – she asked me to recite 99 different numbers from 1 to 100 in a random order and she said she would be able to name the only number in that range that I will have left out.

Of course, she will have to perform the task in her head, legs in the air and without taking any notes.

What is the best way for her to accomplish such a feat?

Well one easy way to accomplish this feat, assuming she is comfortable with basic arithmetic, is to sum all the numbers from one to a hundred: 1 + 2 + 3 +...+100 = (N+1)(N/2) = (101)(50) = 5050.
Now as the first number is randomly picked and recited from the range of one to a hundred, she must subtract it from 5050 and then as the second number is picked recited, that second number must be subtracted from the difference of the first operation. And so on and so on until the 99th number After 99 generated numbers, all the operation of subtraction of the 99 numbers from 5050 will tell the number which will have been left out.

If she's a programmer, she could start with 100 and exclusive or (XOR) each number (binary representation) with the previous result. The final result will be the missing number.

OOps.

I sent a PM to tryagain saying she should add the numbers as they were being recited...and subtract them from 5000. The resulting number would be the one missing. But if the numbers add to 5050...I gotta go with your answer.

Or she could start with 100! (100 factorial) and divide the previous result by each number. Again, the final result will be the missing number.

You guy’s sure nailed that one; I think you covered every solution to the problem – amazing!

Ps. no offense Frank, but I don’t do PM’s – not since the whipped cream scandal of ‘05.


She sure did a number on me:
Solution Since the sum of all the consecutive integers from 1 to 100 is equal to S = 1 + 2 + • • • + 100 = 100 × 101/2 = 5050,
One can find the missing number m by subtracting the sum of the numbers recited, J = 1 + 2 + • • • + (m − 1) + (m + 1) + • • • + 100, from this sum: m = S − J.

Therefore she can sum up the numbers recited by me and then compute this difference.

For example, if the missing number is 10, J = 5040 and m is obtained as the difference 5050−5040.

More formally, the above formulas can be derived from the following property of the modulo division:

(S − J) mod 100 = (S mod 100 − J mod 100) mod 100, where S mod 100 = 50 and j = J mod 100 are the remainders of the division of S and J by 100, which are the numbers formed by the last two decimal digits of S and J, respectively. Then for the missing number m’s values between 1 and 99, we have the formula:

m = m mod 100 = (S − J) mod 100 = (50 − j) mod 100 50 − j if 0 ≤ j ≤ 49,
150 − j if 51 ≤ j ≤ 99, and if m = 100, we can still use the second of the above formulas, because then J = 5050 − 100 = 4950, j = 50, and m = 150 − j = 100.


At this point I was gonna ask about topography, prominence, also known as autonomous height, relative height, shoulder drop (in North America), or prime factor (in Europe), which characterizes the height of a mountain's or hill's summit by the vertical distance between it and the lowest contour line encircling it and no higher summit. It is a measure of the independence of a summit….

But I got in trouble at the local park for lining all the squirrels up in order of height - apparently they didn't like me critter sizing.
So…..

Consider strings composed of the three symbols M, I, and U that can be obtained by starting with the string MI and then applying the following transformation rules a finite number of times:

Rule 1. Add U at the end of any string ending in I, for example, changing MI to MIU.

Rule 2. Double any string after the M (that is, change Mx to Mxx), for example, changing MIU to MIUIU.

Rule 3. Replace any III with a U, for example, changing MUIIIU to MUUU.

Rule 4. Remove any UU, for example, changing MUUU to MU.

Is it possible to obtain the string MU using these rules?

Is this a trick question? I thought I was going crazy after going in circles. I can't find any way to get MU from MI using the four rules given. The idea of course is to get I count to a multiple of 3. And the rule that adds to the letters is Rule 2, but the thing is Rule 2 doubles I, and this can not create a multiple of three from a non multiple of three to start with. Rule 3 can't help you because I with multiples of 2, the minimum of 4 I's will only convert three of the I's to U. For example:

1) We start with M I and apply Rule 2 and Double the string I after the M to get M I I.
2) We have M I I and then apply Rule 2 to the strings I I after M to get M I I I I.
3) We are left with M I I I I and then apply Rule 1 to add U at the end of any string ending in I, so we are left with M I I I I U.
4) Apply rule 3 to M I I I I U to either get M U I U or M I U U
5) And we are back to where we started if pick M I U U and apply Rule 4.

Mind you this is not the only way to start out with. You can start with applying Rule 1 to MI, but it doesn't matter, you aren't going to get near the solution.

It is safe to conclude that MU can not be obtained from MIU theorems given. It's simply not part of the system.

Yep, just as Bennet said. Ignore U and focus on I. You start with 1 and must get to 0 (a multiple of 3). The only operations that affect I are double and subtract 3. Each can only preserve multiple of 3 status. Neither can convert a non-multiple of 3 to a multiple of 3. Therefore, it is impossible.