Iam a number less than 100. when you divide me by 2, my remainder is 1. when you divide me by 25, my remainder is 2. what numbers could i be?

We want to find such that

Note that both 2 and 25 are coprime, so we can apply the Chinese remainder theorem right away.

Let's start with . Taken mod 2, the 2 vanishes and we're left with 1, as desired. But taken mod 25, we're left with 3. If we multiply 1 by 25, so that , then taken mod 25 the 25 will vanish and we're left with 2, as desired. So is the least positive solution.

The whole range of solutions is given by the CRT to be for all integers . There are only 2 solutions that fall below 100, for and , which are 27 and 77.

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58

the radius is 2 times the diameter.

step-by-step explanation: