riddle help please

I don't have time for a closer look at the moment, but from the sum, I deduce

3 in the book would be 1 in real...
Therefore 6 in the book has to be 8 or 9 in real...

Maybe that can set you off...
Gotto go...

This is the Lenny Conundrum from Neopets

The digits can be unscrambled as follows (printed digit -> corrected digit):

0 -> 4
1 -> 7
2 -> 0
3 -> 1 (Way to go Bohne!)
4 -> 5
5 -> 3
6 -> 9
7 -> 2
8 -> 6
9 -> 8

Checking against the original problems:

518 x 70 = 6270 becomes 376 x 24 = 9024
6481 + 3294 = 32847 becomes 9567 + 1085 = 10652

These both check out. Making the same substitutions in the final problem, 7221 x ( 3334 + 3666 ) becomes 2007 x (1115 + 1999) which works out to 6249798.

In the book, that last answer would also have its digits scrambled, so we apply the same substitutions in reverse to get the answer as it would have been printed in the book: 8706169.



BIG thanks to Ann O:

We need to figure out how the digits 0-9 were scrambled, unscramble the third problem, solve it and re-scramble it.

To solve this problem, I started by replacing the digits 1234567890 with letters abcdefghij; by making this switch, I'll be able to fill in the correct digits as I discover them without getting them confused with the incorrect digits. Using the letter-for-digit substitution, the 2 examples are written as:

eah x gj = fbgj
fdha + cbid = cbhdg

I begin with the assumption that none of the leading digits of any number is zero. That applies to the digits c,e, f and g.

Now, let's take a look at the addition problem, fdha + cbid = cbhdg.

We have two 4-digit numbers being added to produce a 5-digit number. This immediately tells us that the 'c' is a '1'. We've already assumed leading digits are not '0', and you can't add two 4-digit numbers and get a result bigger than 9999+9999=19998.
c=1



If the 'c' is a '1', then fdha + 1bid = 1bhdg. The second number begins with a '1' so it's less than 1999, which means the total '1bhdg' is less than 9999 + 1999 = 11998. Now let's look at the digit 'b'. It can't be a '2' or higher; if it could, then the 5-digit sum would be greater than 12000, and we just showed that's not possible. The 'b' can't be a '1' either, because we already used '1' for 'c'. That means 'b' has to be '0'.
b=0



Now we have fdha + 10id=10hdg. The second number begins with '10' so it's greater than 1000 but less than 1099, and the sum begins with '10' so it's less greater than 10000 but less than 10999. That lets us conclude that the first number 'fdha' must be greater than 8901 (10000 - 1099) and less than 9999 (10999 - 1000)...so we know the leading digit 'f' of 'fdha' is either an '8' or a '9'.
f=8 or f=9



Now let's shift attention to the other problem, eah x gj = fbgj. We have a 2-digit number multiplied by a 3-digit number to give a 4-digit number. The 4-digit number begins with 'f', known to be either an '8' or '9'. That's followed by a 'b', known to be a '0', so we know the product is greater than 8000 and less than 9099.

Let's consider those possible values of 'f' one at a time, and let's use what we know about the other digits; specifically, we know different letters don't stand for the same digit. For example, rather than saying 'fbgj' is between 8000 and 9099, if we use everything we know about which digits haven't already been assigned, we can say it's between 8023 and 9087. Similarly, we know 'eah' has to be greater than '234' and 'gj' has to be greater than '23'.

Starting with f=8...the product 'fbgj' is greater than 8023 and less than 8097. If 'eah' has its minimum value '234', then the maximum for 'gj' is '34' (that's 8097/234). On the flip side, if 'gj' has its minimum value '23', then the maximum for 'eah' is 352 (that's 8097/23). Got that? The minimum and maximum values for 'gj' are '23' and '34', and the min/max for 'eah' are '234' and '352'. Well, we've just learned that if 'e' is '2' then 'g' is three; or if 'e' is '3' then 'g' is '2'.

Now we go through the same logic if we take 'f' equal to '9'.
If 'f' is '9', then the product 'fbgj' is greater than 9023 and less than 9087. If 'eah' has its minimum value '234', then the maximum for 'gj' is '38' (that's 9097/234). On the flip side, if 'gj' has its minimum value '23', then the maximum for 'eah' is 395 (that's 9097/23). So this time, the minimum and maximum values for 'gj' are '23' and '38', and the min/max for 'eah' are '234' and '395'. Once again, 'e' and 'g' must have the values '2' and '3'.
Either e=2 and g=3, or e=3 and g=2.



Now let's shift attention to the other problem, eah x gj = fbgj. Notice that the digit 'j' is the final digit of the factor 'gj' and the product 'fbgj'. Take a quick spin through the multiplication table; how many products end in the same digit as one of the factors...with the added constraint that the two factors are different digits? We can rule out '0' and '1'. We can also rule out 5x5, 6x6, and we know that '2' and '3' are already assigned to 'e' and 'g', so we'll rule out those digits too. What are we left with? I find four possibilities: 6x4=24, 6x8=48, 7x5=35, 9x5=45. The pair (h,j) is one of {(6,4), (6,8), (7,5), (9,5)}


Now let's apply all this knowledge and do a little trial-and-error. We'll consider the possible values for 'f' and the possible values of 'e' and 'g' all together; in each case, we'll also look at the possible values for 'h' and 'j' and see how many produce a valid equation.

1. Take f=8, e=2, g=3
2ah x 3j = 803j
Will this problem work with j=4 or j=5? (we omit j=8 because we've already used 8 for 'f')
8034/34 does not divide evenly.
8035/35 doesn't either.

2. Take f=8, e=3, g=2
3ah x 2j = 802j
Will this problem work with j=4 or j=5? (again we omit j=8)
8024/24 does not divide evenly.
8025/25 = 321...but '2' and '1' are already assigned.

3. Take f=9, e=2, g=3
2ah x 3j = 903j
Will this problem work with j=4 or j=5 or j=8?
9034/34 does not divide evenly.
9035/35 doesn't either.
9038/38 doesn't either.

4. Take f=9, e=3, g=2
3ah x 2j = 902j
Will this problem work with j=4 or j=5 or j=8?
9024/24 = 376
9025/25 = 361...but '1' is already assigned.
9028/28 does not divide evenly.

Well, there we are. We looked at all potential values for f, c and g, and found only one set of values that produces a working equation:
f=9, e=3, g=2 and j=4



We've also deduced the values of 'a' and 'h' from the only equation that works: 376 x 24 = 9024.
a=7 and h=6



Plug in all our now-known digits into the other equation, fdha + cbid = cbhdg, and we have:
9d67 + 10id = 106d2

We quickly discover 'd' is '5' because the sum '7 + d' has a last digit of 2. Now we have:
9567 + 10i5 = 10652, so we know 'i' must be 8.
d=5 and i=8



Our final substitution table is as follows:

Published Digits
1 2 3 4 5 6 7 8 9 0

Substituted Letters
a b c d e f g h i j

Corrected Digits
7 0 1 5 3 9 2 6 8 4


We've now figured out all the substitutions, so we go back to the original problem - what would the printed answer be for '7221 x (3334 + 3666)' assuming the same substitutions?

We need to rewrite this problem with the appropriate substitutions, solve it, and reverse-substitute the digits in the answer.

'7221 x (3334 + 3666)' becomes '2007 x (1115 + 1999)' -- that's 2007 x 3114, or 6249798.

Switching back to the publisher's mixed-up digits, the answer would have been printed as: 8706169 and that's the answer.

Geez, so much work!

OK, so I suppose, I don't have to take a closer look at the problem any more now!

OMG try,
I can never understand you
I still think you are weird. :wink:

Hello?
There is a new on and the answer, just search BRIDGE in the neopian adical, and the only on is the awsner

Is there any particular reason you didn't post this Lenny to the Neopets "Lenny" topic...? Did you have some objection to the answer that was posted there...?

Thank you.

Eliz.

Maybe they didn't know. I know I didn't.