NeoPets Riddles (Lenny Conundrums) and Answers Here

How many times do you have to get in the top 250 to get a better Lenny Trophy and actually have it count as a score in the high games?

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Lu, that can't be it
Lu-
725 X 5 = 3625 and there can't be a 6 anywhere in the problem.

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Lu wrote:

725
x35
------
25375 ?

That doesn't work for the whole thing though

725
x35
------
3625
21750
-------

In the middle you have numbers that are not 7235

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brisboy76 wrote:

How many times do you have to get in the top 250 to get a better Lenny Trophy and actually have it count as a score in the high games?

It depends where you are in the top 250 which trophy you get. I htink only first 10 are gold.

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Thanks

I can now go and get some sleep...

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ro67 wrote:

MariaWB wrote:
ro67 wrote:
How about:

775
33
------
2325
2325
-------
25575

The answer would be: 25575

How did you get to that answer?

I figured that the last digit of each of the middle numbers HAD to be 5. The meant that the next to last digit in the first middle middle number had to be a 2 or a 7. After that, it was trial and error...

I think I understand.. Thanks =) I like to understand the answers before I submit, especially when I wasn't the one to find it.

Like last week. I was messed up last week because I was ill (I still am, my mom keeps laughing at my voice, so mean of her =p) so I completely forgot the new Lenny came out. I did go inhere to see how it was solved because I wanted to see how it was done, but I was way late for a trophy, and I didn't even try to help solve it.

So I didn't submit for the np's. I just don't think it's fair for those brainy people, like you, who figure the LC's out week after week =)

This time, I learned something more. I always need more math skills, I SUCK at math!

Though, I'll probably forget how it was done in, like two minutes. Or less.. Anyways, I'm off to bed. Night all =)

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lc
2x2=4
2*3=6
2*5=10
2*7=14
3*2=6
3*3=9
3*5=15
3*7=21
5*2=10
5*3=15
5*5=25
5*7=35
7*2=14
7*3=21
7*5=35
7*7=49

By this chart, the last number of the first and last number of the second row can only be 3 and 5, 5 and 5, or 5 and 7.

Assuming that the first number of the answer has to be a 2 (can't be a 1), the first number would have to be at least 700 (500 is too small)

Therefore, the numbers you have for sure are:
(7) (?) (3, 5 or 7)
x (?) (3, 5 or 7)

I tried different combinations until i got the answer.
723, 725, 727, 733, 735, 737, 753, 755, 757, 773, 775, 777
each multiplied by 23, 33, 53, 73, 35, 55, 57, 75, 77
That still leaves 108 different combinations of numbers, but still significantly less than you would have if you didn't do some eliminating.

Hope this method helps, I got it pretty easy after that.

using 775 x 33 works because both lines below would be 2325 and the answer is 25575.

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Hi ro67, you beat me by 22 minutes since I forgot to keep checking. Took me about 7 minutes from the time I remembered to check it. I just wrote a quick script to multiply all 3 digit numbers composed of 2, 3, 5 and 7 against all 2 digit numbers composed of 2, 3, 5 and 7. Made sure the intermediate values and the final value were all made up of 2, 3, 5 and 7 and printed out anything that matched. Only one possibility came up, so that was the answer.

Maybe I'll finally get my avatar, but I wish I'd checked earlier.

I suppose since the answer is already posted, I might as well include the script for anyone that wants it. ASP (VBScript) code.

Code:



'd contains the allowed digits

dim d(4)

'a1, a2, and a3 are the array indexes for the digits of the top number

'b1 and b2 are the array indexes for the digits of the bottom number

dim a1, a2, a3, b1, b2

'top is the top 3-digit number and bot is the bottom 2-digit number

'prod is their product, and s is just the string form of it

'l1 and l2 are the first and second intermediate values, 

'while s1 and s2 are the string forms

dim top, bot, prod, s, l1, l2, s1, s2

t(1) = 2

t(2) = 3

t(3) = 5

t(4) = 7



'Loop through all the possible digits for each of the 5 digit 

'positions in the two numbers to multiply.

for a1 = 1 to 4

for a2 = 1 to 4

for a3 = 1 to 4

for b1 = 1 to 4

for b2 = 1 to 4

top = 100 * d(a1) + 10 * d(a2) + d(a3)

bot = 10 * d(b1) + d(b2)

prod = top * bot

s = cstr(prod)

'First check the product to see if it has any of the 6 digits we

'don't want.

if instr(s, "0") = 0 and instr(s, "1") = 0 and instr(s, "4") = 0 and _

   instr(s, "6") = 0 and instr(s, "8") = 0 and instr(s, "9") = 0 then

l1 = top * d(b2)

l2 = top * d(b1)

s1 = cstr(l1)

s2 = cstr(l2)

'Now check the intermediate values

if instr(s1, "0") = 0 and instr(s1, "1") = 0 and instr(s1, "4") = 0 and _

   instr(s1, "6") = 0 and instr(s1, "8") = 0 and instr(s1, "9") = 0 then

if instr(s2, "0") = 0 and instr(s2, "1") = 0 and instr(s2, "4") = 0 and _

   instr(s2, "6") = 0 and instr(s2, "8") = 0 and instr(s2, "9") = 0 then

'Print the values when we find a good one.

Response.Write top & ", " & bot & ", " & l1 & ", " & l2 & ", " & prod & "<br>"

end if

end if

end if

next

next

next

next

next

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So, anyone solve it yet? or is it still confusing?? *goes off to look at previous posts* well, i'm here to work on it if it isn't solved..

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bookworm514 wrote:

So, anyone solve it yet? or is it still confusing?? *goes off to look at previous posts* well, i'm here to work on it if it isn't solved..

ok from what i figure out, the answer is 25575.

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ok... i normally would NEVER ever do this... but it's so SIMPLE and people aren't GETTING IT that it drives me crazy! I just have to let it out. It is NOT guessing, OR trial and error.

First of all... we must know what the last numbers in the "ones" place are, because this number cannot be modified by a carry. That is the KEY that unlocks this problem. It must be end in a 2, 3, 5, or 7 and also be DIVISIBLE by a 2, 3, 5, or 7.

The only thing we can get out of 2, 3, 5, or 7 that ends in one of those is 15! so for the "ones" place, we have 3 x 5 (well, 25, 35 too. actually. I missed that one. BUT it should still be easy to figure out systematically)

But that has a 1 in it, so we have to change that 1 to a 2, 3, 5 or 7 by carrying that 1 to something that adds up to one of those, but is still divisible by 2, 3, 5, or 7. What can you multiply out of those numbers to get an answer that has a difference of 1 from the required numbers? 7! or in the "tens" place, 70. 3 x 70 = 210. put the 15 into that, and you get 225. So far so good!

So now we have
75
x 3
----
225

Now, we need something that can carry a 2 within the required numbers. We should already know how to do this by now, and were getting towards the "front" of this part of the problem, so the answer to this part must BEGIN with a 2, 3, 5, or 7 and be modified by a carry so the number in the hundreds place also falls into one of those numbers. We should see very quickly that the answer is again, 21, or in the hundreds place 2100.

So now we have
775
x 3
-------
2325

Now, we do the same again for the "tens" place of the multiplier. If you're sharp though, you will see the answer already, all you have to do is bump the place over one.

775
x 33
---------
2325
2325
----------
25575

And done!
Basically, I guess there can be some trial and error in there, but not nearly as much as people think. If you do it this way, systematically, you can SEE where you went wrong, instead of throwing numbers around and wondering why it didn't work, and not do it again. Something that is not possible is not a possibility.
It boils down to three "possible" starts, 3x5, 5x5 or 7x5, (six when reversed) but you can see which two are wrong very quickly, if you're sharp, you'll see it as soon as you hit the tens place.

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Theres mroe then one solution.?

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yuka wrote:

Theres mroe then one solution.?

No. But it may appear that way early on.

Like, it must start with one of these.

3x5, 5x3, 5x5, 7x5, or 5x7.
When you first start the problem, some of those seem to have solutions, at first glance, but they really don't. Like 755x5 = 3775, but we can figure out that is wrong because for the next part, 755x20 = 15100, wrong.
755x30 = 22650, wrong
755x50 = 37750, wrong (37750 + 3775 = 41525)
755x70 = 52850, still wrong. So it can't even begin with 5x5, because what you have to do to fit it eventually becomes impossible within the rules.

Also, it must start with 3x5, 5x3, 5x5, 7x5, or 5x7 for a reason, it can save you work of doing things that you should know will be wrong. like 7x3 = 21, so anything with 7x3 is immediately eliminated. Theres no need to do 77 x 3 or 777 x 3 or anything with a 7 on the end of it over a 3 because they are automatically wrong.

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xhaan wrote:

ok... i normally would NEVER ever do this... but it's so SIMPLE and people aren't GETTING IT that it drives me crazy! I just have to let it out. It is NOT guessing, OR trial and error.

Aw, but when you write a computer program to do the work for you, trial and error is faster. I wrote a simple one and got the answer in 7 minutes from first looking at the problem. Seriously, I agree with what you've done - narrow the possibilities way down before trying them out and I probably would have gone that route if I didn't spend my days writing programs anyway.

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edstock wrote:

xhaan wrote:
ok... i normally would NEVER ever do this... but it's so SIMPLE and people aren't GETTING IT that it drives me crazy! I just have to let it out. It is NOT guessing, OR trial and error.

Aw, but when you write a computer program to do the work for you, trial and error is faster. I wrote a simple one and got the answer in 7 minutes from first looking at the problem. Seriously, I agree with what you've done - narrow the possibilities way down before trying them out and I probably would have gone that route if I didn't spend my days writing programs anyway.

Wow, that's pretty cool actually. I got it in a couple minutes, part of that was the luck of finding 5x3 before trying some of the other start combinations. I knew it had to end with 5, and that there were only a few ways to do that, and each of those few ways are almost self deducing.

And actually, I discovered now that it was even easier than I thought in the earlier posts. What is true for the "ones" digit of the multiplier is also true for the "tens" digit of the multiplier... which means if it's wrong in the ones place, then it's also wrong in the tens place. Like it's discovered that 5x7 is wrong, so 5x70 is also wrong, so you never need to use a 7 in the multiplier again after 5x7. Lets say you tried 5x7, find it wrong, then you try 5x5, you find that wrong also, and trying the 7 again wont change that, so 5x75 is still wrong because 5x7 and 5x5 are both wrong.

I also think there's a way to solve the problem in one go, without any trial at all... I'm still looking at that, I think it may have to do with prime numbers (they're all prime numbers, and all single digits) and the fact that 1 is a special number in math, it's a "unit", what all other numbers are made with, and the correct "beginning" to the problem, 5x3, equals 15, which has the 1, and a 5, which is a prime number. I also find it interesting that if you added another digit to the problem, it would mess it up, like 7775 x 33 no longer works, you get 256575. Also, it's interesting because when you multiply all the prime numbers known, then add 1, you get another prime number. 2*3*5*7 = 210 + 1 = 211, prime number, also 210 is part of the correct answer, as in my first post, 70x3 = 210...

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Because we seem to need to point out the obvious here, don't wait up for a Lenny this week--it's Thanksgiving, after all Confused

Go eat!

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Jen Aside wrote:

Because we seem to need to point out the obvious here, don't wait up for a Lenny this week--it's Thanksgiving, after all Go eat!

Seriously? arghs! when WILL it be there?i was waiting up all night...bleh

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how to make money
the best way to make neopets money is to make more accounts and play games. is the only cheat that actually works.

here is a quick url to neopets. create a new account. but you have to play it if this cheat is to work

Edit [Moderator]: Link removed

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Re: how to make money

stretchystuff wrote:

the best way to make neopets money is to make more accounts and play games. is the only cheat that actually works.

here is a quick url to neopets. create a new account. but you have to play it if this cheat is to work

Edit [Moderator]: Link removed

Except that is illegal. The only reason you are allowed to have extra accounts is for more gallery and/or shop space, and the money to expand those you have to get from your main account.

And linking to your own account for some kind of benifit is just too low! Someone should kick your *ss out of here.

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Re: how to make money

MariaWB wrote:

stretchystuff wrote:
the best way to make neopets money is to make more accounts and play games. is the only cheat that actually works.

here is a quick url to neopets. create a new account. but you have to play it if this cheat is to work

Edit [Moderator]: Link removed

Except that is illegal. The only reason you are allowed to have extra accounts is for more gallery and/or shop space, and the money to expand those you have to get from your main account.

And linking to your own account for some kind of benifit is just too low! Someone should kick your *ss out of here.

Yap2! and the only one who DOES benefit is YOU! tsktsk..you wanna make nps then finish the daily dare. You get a split lenny morph potion..after some time it might be worth a lot..it is going for around 20k now..

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