METHOD
1: Reason using addition number facts.
For
the ones column to sum to a number ending in A,
B must be either 0 or 5.
In the hundreds column, neither A nor B can be
0, so B = 5.
In the thousands column A is either 1 or 2. Since
A + 5 + 5 < 20, in the thousands column A =
1.
Thus, 155 + 515 + 551 = 1221. Use the test of
divisibility for 11: 1221 = 11 × 111 = 11
× 3 × 37. Or use the test of divisibility
for 3: 1221 = 3 × 407 = 3 × 11 ×
37. Either way, the greatest prime factor is
37.
METHOD
2: Use algebra.
The sum
in each place is A + 2B. The entire sum is 100(A
+ 2B) + 10(A + 2B) + (A + 2B) which simplifies to
111(A + 2B). 111 = 3x37, both of which are prime.
In the original sum ACCA, the thousands digit A
is the result of carrying and cannot be larger than
2 (upon further investigation, we can determine
that it must be 1 and that the sum A + 2B must =
11). Even if (A + 2B) is the largest 2digit prime
sum possible beginning with 2 (i.e. 29), this is
still a prime smaller than 37. So the largest prime
factor is still 37.
