Each of the small letters in bold represents a different nonzero base n digit from 1 to n1 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 multidigit 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 base36 number 1.2 and, c.d represents the base 36 number 3.4.
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 nonzero 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 httpequiv="contenttype" content="text/html; charset=utf8">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 20111022 14:15:50 