Find the smallest possible value of a whole number 'y' if it leaves a remainder of 3 when divided by 5,6 or 9.

We want to find such that

6 and 9 are not coprime, so we split the moduli according to and to get the system

If we take

we can see that

taken mod 2, the last two terms vanish and we're left with ;taken mod 3, the first and last terms vanish, and the remaining term is ;taken mod 5, the first two terms vanish, and we're left with

By the Chinese remainder theorem, we've found that any satisfying

will satisfy each congruence above, and that any solution of the form for any integer will work.

The smallest possible value of these occurs for , so that 3 is the least positive solution.

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12