All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Sum, Times, I Wonder..(3) (Posted on 2011-08-27) Difficulty: 3 of 5
Each of the small letters in bold represents a different nonzero base n digit from 1 to n-1 to satisfy this cryptarithmetic equation.

a.b + c.d = (a.b)*(c.d)

Determine all possible positive integer values of n ≤ 36 such that:

(I) The above equation has at least two valid solutions.

(II) a+b+c+d is a perfect square.

Notes:

(A) Solve parts (I) and (II) separately.

(B) Adjacent numerals are multi-digit base n numbers, and not the product. For example, if n = 36, a=1, b=2, c=3 and d=4, then a.b represents the base-36 number 1.2 and, c.d represents the base 36 number 3.4.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution Comment 1 of 1
Let a.b = x, and c.d = x * y, where y>1 (a.b and c.d can't be equal). We are left with the equality:

x + xy = x*(xy) which reduces to 1 + y = xy. If y=1, then x=2, which is the max value for x (ie, y=2 gives x=1.5, y=3 produces x=1.3333, etc). Therefore, for both parts of the problem, a=1.

Looking further into the original equation, a.b + c.d = a.b * c.d becomes:

1 + c + (b + d) / n = c + d / n + b * c / n + b / n * d / n
1 + (b + d) / n = d / n + bc / n + bd / n^2
n + (b + d) = d + bc + bd / n
n + b = bc + bd/n
n^2 + bn = bcn + bd
n^2 = b(cn - n + d)
(n^2)/(cn - n + d) = b

Now we have a=1, and b in terms of the other 2 variables, and n. As c < n, (cn - n + d) will not divide n^2 evenly for any prime n.

This narrows it down to 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 28, 30, 32, 33, 34, 35, 36  for possible values of n (n > 4 in order for a, b, c, and d to be distinct non-zero numbers). Also, 8 = 2^3, 9 = 3^2, 16 = 2^4, 25 = 5^2, and 32 = 2^5. As they are all perfect powers of a prime, n^2 will only be divisible by perfect powers of the prime, as will n, leaving cn - n + d without a solution (d would have to equal n).

So our potential list of candidates for n is: 6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36.

<meta http-equiv="content-type" content="text/html; charset=utf-8">One final thing to note is that only cn - n + d which are not a multiple of n can work. So, if n = 18 for example, that's 2*3*3. cn - n + d must not have (2*3*3) in its factorization. In this case, that means cn - n + d must be either 3*3*3 = 27, or 3*3*3*3 = 81.

Armed with this knowledge, we can quickly list the cn - n + d values needed for a solution to be possible.

n = 6: {9}
n = 10: {25}
n = 12: {16, 18}
n = 14: {49}
n = 15: {25}
n = 18: {27, 81}
n = 20: {25, 50}
n = 21: {49}
n = 22: {121}
n = 24: {32, 36, 64}
n = 26: {169}
n = 28: {49, 98}
n = 30: {36, 45, 50, 75, 100, 225}
n = 33: {121}
n = 34: {289}
n = 35: {49}
n = 36: {48, 54, 81, 162}

Of those values. there is only 1 which fails to work. For n = 14, (cn - n + d) = 49, c=4, d=7, but 196 / 49 = b = 4. As c and b are the same, this is not a solution. So, for part I, we look at the values of n for which multiple factors of n^2 will work.

Part I:
n = {12, 18, 20, 24, 28, 30, and 36}

Part II:
Going through the values above and solving for b, c, and d we get just 2 values such that a+b+c+d is a perfect square.

n = 12 {1, 9, 2, 4}
n = 26 {1, 4, 7, 13}

Edited on October 27, 2011, 9:10 am
  Posted by Justin on 2011-10-22 14:15:50

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information