Each diferent letter is a diferent number aa+bb=bac What are the numbers?

Answer 1

b must equal 1, bb=11. Why? Two digit #'s added together will not equal a number over 200, so the b in the answer must be a 1. There is a start. Good luck

Answer 2

This can't be solved as a letter substitution problem (e.g. a = 7, b = 8, aa + bb = 77 + 88 = 165) since the carry into the hundreds place would imply b = 1, and then aa + 11 = 1ac would imply that aa >= 100-11 = 89, so a would have to be 9, but 99 + 11 = 110 and bac = 19c do not match.

So it must be understood as an algebraic problem, a^2 + b^2 = bac. The easiest solution is for a = ±b, then the LHS is 2a^2 and the RHS is ca^2, so we get c = 2.

Now suppose p is a prime dividing a, then consider both sides of the equation modulo p. We get b^2 = 0 (mod p) and hence, since p is prime, b = 0 (mod p) and so p divides b too. By symmetry, any prime dividing b must also divide a. Hence b and a must have the same primes in their prime power factorisations.

Write a = p1^(a1) . p2^(a2) . ... pn^(an) and b = p1^(b1) . p2^(b2) ... pn^(bn). Then the equation translates to
p1^(2a1) . ... . pn^(2an) + p1^(2b1) . ... . pn^(2bn) = p1^(a1+b1) ... pn^(an+bn) c.
<=> c = p1^(a1-b1) ... pn^(an-bn) + p1^(b1-a1) ... pn^(bn-an)
For every j from 1 to n, if aj > bj then the second term will be fractional (since p1 to pn are all different primes the other components of the term won't be able to remove a power of pj in the denominator), and if aj < bj the first term will be fractional. So the only possibility for integral solutions is to have aj = bj for every j, in other words a = ± b.

Answer 3

I think you got the equation wrong I think you want

aa + bb + cc = bac

if so

then

99 + 11 + 88 = 198

Answer 4

This is the wrong equation. There is no solution.

Answer 5

the answer is "Brownies"