a and b are consecutive, positive integers such that a2−b2>22. What is the minimum possible value of a+b?
Solution 1: Since a2>b2+22, thus a>b. Since a and b are consecutive positive integers, we have a=b+1. Substituting in the above expression, we have (b+1)2=b2+2b+1>b2+22⇒b>22−12⇒b≥11. Thus a+b=2b+1≥2×11+1=23.
Solution 2: Since a2>b2+22, thus a>b which implies that a=b+1,a−b=1. Hence a+b=a2−b2a−b>221 ⇒a+b≥23 (since they are integers). We verify that a=11+1=12,b=11 satisfies the inequality, so the minimum possible value of a+b is indeed 23.
23
To find the minimum possible value of a+b, we need to consider the condition a^2 - b^2 > 22.
We know that a and b are consecutive positive integers, which means that b is the integer that comes before a. So, we can express a and b mathematically as follows:
a = b + 1
Substituting this into the inequality, we get:
(b + 1)^2 - b^2 > 22
Expanding and simplifying the equation:
b^2 + 2b + 1 - b^2 > 22
2b + 1 > 22
2b > 21
b > 10.5
Since b must be an integer, the smallest value that satisfies b > 10.5 is 11. Therefore:
b = 11
a = b + 1 = 11 + 1 = 12
The minimum possible value of a+b is 11 + 12 = 23.