Math Olympiad Problems for Grades 3 and 4: Problems with Solutions

Solution:
The problem is solved by applying inverse operations from the end to the beginning (working backwards).
- The final result was [tex]10[/tex] after dividing by [tex]2[/tex]. Therefore, before dividing, the number was [tex]10 \times 2 = 20[/tex].
- Before obtaining [tex]20[/tex], the machine subtracted [tex]4[/tex]. Therefore, [tex]4[/tex] is added: [tex]20 + 4 = 24[/tex].
- Before obtaining [tex]24[/tex], the machine multiplied by [tex]3[/tex]. Therefore, it is divided by [tex]3[/tex]: [tex]24 \div 3 = 8[/tex].
The initial number was [tex]8[/tex].

Solution:
1. The given values are substituted into the invented rule: [tex]A = 5[/tex] and [tex]B = 4[/tex].
2. First, the multiplication is calculated: [tex](A \times B) = 5 \times 4 = 20[/tex].
3. Then, the sum is calculated: [tex](A + B) = 5 + 4 = 9[/tex].
4. Finally, the second result is subtracted from the first: [tex]20 - 9 = 11[/tex].

Solution:
1. First, the side length of a small square is found. Since the perimeter ([tex]4[/tex] sides) is [tex]24[/tex] cm, one side measures [tex]24 \div 4 = 6[/tex] cm.
2. By joining [tex]3[/tex] squares in a row, a rectangle is formed. The width of this rectangle will be one side of a square ([tex]6[/tex] cm). The length will be formed by [tex]3[/tex] sides ([tex]6 \times 3 = 18[/tex] cm).
3. The perimeter of the new rectangle is calculated by adding its [tex]4[/tex] sides: [tex]18 + 18 + 6 + 6 = 48[/tex] cm.

Solution:
1. The divisibility rule for [tex]9[/tex] states that the sum of all the digits in a number must be a multiple of [tex]9[/tex].
2. The digits of the code are added together: [tex]5 + A + 2 + A = 7 + 2A[/tex].
3. A digit from [tex]0[/tex] to [tex]9[/tex] for [tex]A[/tex] is sought that makes [tex](7 + 2A)[/tex] equal to a multiple of [tex]9[/tex] (such as [tex]9[/tex], [tex]18[/tex], or [tex]27[/tex]).
4. If [tex]A = 1[/tex] is tested, the sum is [tex]7 + 2(1) = 9[/tex]. Nine is a multiple of nine.
5. Testing the following values of [tex]A = 2, 3, 4, 5, 6, 7, 8, 9[/tex] yields sums of [tex]11, 13, 15, 17, 19, 21, 23[/tex], and [tex]25[/tex], none of which are multiples of [tex]9[/tex]. The only valid digit is [tex]1[/tex]. The code is [tex]5121[/tex].

Solution:
1. Let the first number be [tex]X[/tex].
2. The second number will be [tex]X + 3[/tex].
3. The third number will be [tex](X + 3) + 3[/tex], which is the same as [tex]X + 6[/tex].
4. The sum of the three is: [tex]X + (X + 3) + (X + 6) = 36[/tex].
5. Grouping the terms gives: [tex]3X + 9 = 36[/tex].
6. Subtracting [tex]9[/tex] from the total: [tex]3X = 27[/tex].
7. Dividing by [tex]3[/tex]: [tex]X = 9[/tex]. The first number is [tex]9[/tex]. (Verification: [tex]9 + 12 + 15 = 36[/tex]).