Wonderful Cryptogram!

Logic Level 3

$\Large\sqrt{\overline{WONDERFUL}}=\overline{OODDF}$

In the above cryptogram if each distinct letter represents a distinct positive digit , Find the value of $W \times L$ .

The answer is 45.

3 solutions

Hadia Qadir
Aug 4, 2015

if O is bigger than 3, then the square of the five digi number becomes a 10 digit number. So, the five digit number should be less than 33000 so O can only be 1 or 2 if O is 1 OODDF squared would be between 11000 squared and 12000 squared which is 121000000 and 144000000 which means that W must be 1, which is contrary to that of each letter representing a different number Therefore, O is 2 OODDF squared would be between 22000 squared and 23000 squared which is 484000000 and 529000000 The second number must also be a 2 as it is a O, which means W = 5 so 22DDF must be bigger than sqrt(520000000) = 22803.xxx so D = either 8 or 9 we try D = 8 that means WONDERFUL would be between 22880 squared and 22990 squared 52349440 and 523952100 This looks pretty plausible, so we look at F and L F cannot be 1 as L would be 1 2 is taken by O, 3 is taken by N 5 would make L 5, 6 would make L 6, So let's try F = 4 22884^2 = 523677456 nope 22887^2 = 523814769

WHOO HOO IT WORKS so answer would be 45

At first I failed to see the logic in "the second number must also be a 2 as it is a O, which means W = 5" as if OODDF were 22113 the squared number would still have started with W=4. Then I realised that you meant the second number in WONDERFUL needs to be a 2 as it is also an O and since the squared number would start with 48 49 50 51 or 52 it could only be 52, hence W is a 5.

And I checked that F could also not be 9 as that would result in a zero in WONDERFUL which is apparently not a positive digit.

I have not checked what impossibilities D=9 would give...

Karen Mulders - 3 years, 4 months ago

Don't be too difficult! An Excel simulation with 165 rows gave an answer although not finished checking.

123 1   1   2   2   3   11223   125955729   1   2   5   9   5   5   7   2   9
124 1   1   2   2   4   11224   125978176   1   2   5   9   7   8   1   7   6
125 1   1   2   2   5   11225   126000625   1   2   6   0   0   0   6   2   5
...
285 2   2   8   8   5   22885   523723225   5   2   3   7   2   3   2   2   5
286 2   2   8   8   6   22886   523768996   5   2   3   7   6   8   9   9   6
287 2   2   8   8   7   22887   523814769   5   2   3   8   1   4   7   6   9

W $\times$ L = 5 $\times$ 9 = 45

Answer: $\boxed{45}$

Lu Chee Ket - 5 years, 4 months ago

yes this one is more easy ! thank you @Lu Chee Ket

Hadia Qadir - 5 years, 4 months ago

No problem.

Lu Chee Ket - 5 years, 4 months ago

Hey! How can you make that ?

M T - 4 years, 12 months ago

Nguyen Tung Anh
Aug 1, 2015

WONDERFUL is 523,814,769 OODDF is 22887

I'm too lazy to check if there is/are other ans as well.

This problem is not strange at all for any secondary students at my country.

P/S: I'm going to be a sophomore.

Here's how I did it:

WONDERFUL has 9 different characters => 9 different numbers

OODDF is somewhere between 10000 and 32??? => O can't be 3 => O must be either 1 or 2

If O is 1 => W is 1 => wrong => O must be 2 => W can be 4 or 5

The rest is cakewalk.

As a computer science major, that "=>" in your comment really threw me off. It looks like greater than or equal to.

Oscar L - 4 years, 5 months ago

Maggie Miller
Jul 22, 2015

$\sqrt{523814769}=22887,$ so $W=5$ and $L=9$ . Thus, the answer is $\boxed{45}$ .

Damn , you found a loophole in the problem. I have edited the problem. Please edit your solution accordingly. Thanks!

Nihar Mahajan - 5 years, 10 months ago

Ok - I thought the other version was purposefully phrased that way so that a clever solution existed, sorry (haha)!

Maggie Miller - 5 years, 10 months ago

Your solution was unexpected and was quite wise! Anyways , you have found out the right answer! ^_^

Nihar Mahajan - 5 years, 10 months ago

@Nihar Mahajan – In which Logic book you found this problem?What is the intended solution?

Adarsh Kumar - 5 years, 10 months ago

Here's how I solved the problem, Python!

from itertools import permutations as p

for i in p('0123456789'):
    k=''.join(i)
    [W,O,N,D,E,R,F,U,L]=list(k)[:9]
    if int(O+O+D+D+F)*int(O+O+D+D+F)==int(W+O+N+D+E+R+F+U+L):
        print int(W+L)

As the out put is 59, answer is $5\times 9 =\boxed{45}$ .

Btw, if somebody wants similar problems, here's as many as I have come across on Brilliant...

A similar problem ,

A similar problem

Aditya Raut - 5 years, 10 months ago

Same idea but didn't use imports. Granted, this code is not a general solution like yours but coincidentally only returned one possible permutation.

for i in range(1,3): # zero < O < 2 because 33000^2 > 1000000000
    for j in range(0,10):
        for k in range(0,10):
            if j!=i and j!=k and k!=i:
                ooddf = i*10000+i*1000+j*100+j*10+k
                wonderful = ooddf*ooddf
                digits = len(set([int(x) for x in str(wonderful)]))
                if digits == 9:
                    print (ooddf,wonderful,digits)

Bryan Hemingway - 5 years, 7 months ago

Wah! How can you make that?

M T - 4 years, 12 months ago

Wonderful Cryptogram!

The answer is 45.

3 solutions

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