If v is a three digit number less than 250 that has a remainder of 2 when divided by 2, 3 or 5, what is a possible value for v?

is an integer such that

Since 2, 3, and 5 are mutually coprime, you can use the Chinese remainder theorem directly.

Suppose

Then taken mod 2, the last two products vanish and you're left with a remainder of 0 (since 2 divides 2*3*5);

taken mod 3, the first and last products vanish and you're left with a remainder of 2 (since 2*2*5 = 20, and 20 = 2 mod 3);

and taken mod 5, the first two products vanish and you're left with a remainder of 2 (since 2*2*3 = 12, and 12 = 2 mod 5).

So we have , and , so that any integer of the form satisfies the system of congruences, the largest of which is 242.

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

its adding 5.5 each time

step-by-step explanation: