Let pp, qq and rr be 2-digit numbers where p < q < r. If pp + qq + rr = tt0, where tt0 is a 3-digit number…
Let pp, qq and rr be 2-digit numbers where p < q < r. If pp + qq + rr = tt0, where tt0 is a 3-digit number ending with zero, consider the following statements:
1. The number of possible values of p is 5.
2. The number of possible values of q is 6.
Which of the above statements is/are correct?
1 only
2 only
Both 1 and 2
Neither 1 nor 2
Solution
pp = 11p, qq = 11q, rr = 11r, so the sum is $11(p+q+r)$. This must equal a 3-digit number tt0 ending in 0. Enumerating valid distinct digits with $p < q < r$ that make $11(p+q+r)$ a 3-digit number ending in zero, p takes 5 possible values and q takes 6 possible values. Per the official answer key, both statements are correct.