The marching band has more than 100 members but fewer than 200 members. When they line up in rows of 4 there is one extra person; when they line up in rows of 5 there are two extra people; and when they line up in rows of 7 there are three extra people. How many members are in the marching band?
I need to find a number that is between 100 and 200, and has remainders of 1, 2, and 3 when divided by 4, 5, and 7 respectively. One way to approach this problem is to use the Chinese remainder theorem, which says that there is a unique solution modulo the product of the divisors, which in this case is 4 * 5 * 7 = 140. That means that the number I'm looking for is of the form 140k + r, where k is some integer and r is the common remainder.
[* to find r, I can use a system of congruences: x = 1 mod 4, x = 2 mod 5, and x = 3 mod 7. To solve this system, I can use the method of back-substitution. Starting with the last equation, I can write x = 7y + 3 for some integer y. Then, plugging this into the second equation, I get 7y + 3 = 2 mod 5, which simplifies to 2y = 4 mod 5, or y = 2 mod 5. This means that y = 5z + 2 for some integer z. Plugging this into the first equation, I get x = 7(5z + 2) + 3 = 1 mod 4, which simplifies to 35z = 0 mod 4, or z = 0 mod 4. This means that z = 4w for some integer w. Finally, plugging this into x = 7y + 3, I get x = 7(5(4w) + 2) + 3, which simplifies to x = 140w + 17. Therefore, r = 17, and the number I'm looking for is 140k + 17. To make this between 100 and 200, I need to choose k = 1, since k = 0 would make it too small and k = 2 would make it too large. So the final answer is 140 + 17 = 157. 