Solution

Part 1

Suppose that the answer is a number in the form of

n = 15x + 21y + 35z

When divided by 3, n will leave the same remainder as 35z (because the other two terms are divisible by 3), but it is given that the remainder that n leaves is 2, so we have the following equation for z:

35z = 2 (mod 3)

which is a mathematical notation for "35z - 2 is divisible by 3" (see some information on Modular Notation if this is new to you). But 35z = 2z (mod 3), so we have

2z = 2 (mod 3)

or

z = 1 (mod 3), so z = 1, 4, 7, ...

We are only interested in the smallest z, so z = 1.

We now repeat the same procedure for x and y.

21y = 3 (mod 5) =>

y = 3 (mod 5), so y = 3.

15x = 2 (mod 7)

x = 2 (mod 7), so x = 2.

Thus, n = 15(2) + 21(3) + 35(1) = 128. However, notice that any number n + 105k or  n - 105k would also work (because adding 105 = 3*5*7 would not change the remainders upon division by 3, 5, and 7). Therefore, the smallest such number is 128 - 105 = 23.

Part 2.

We use the same method to solve this problem. For convenience, the problem will be restated in a more compact form:

 Divided By Leaves a Remainder Of 2 1 3 2 4 3 5 4 6 5

Note that the third statement implies the first, and that the fifth statement implies the second, so the problem is equivalent to:

 Divided By Leaves a Remainder Of 4 3 5 4 6 5

Now suppose that the number is in the form

n = 20x + 24y + 30z

So we can say that

20x = 5 (mod 6), 24y = 4 (mod 5), and30z = 3 (mod 4)

which is equivalent to

2x = 5 (mod 6), y = 1 (mod 5), and2z = 3 (mod 4)

Note that the first equation is the same as 2x = 6k + 5 for an integer value of k. The left-hand side is always even, but the right-hand side is always odd. So this equation does not have a solution, which means that the problem has no solution.

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