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

Solution:
1. First, the operation inside the parentheses is solved: [tex](6 \nabla 4)[/tex].
2. The rule is applied with [tex]A = 6[/tex] and [tex]B = 4[/tex]: [tex](6 \times 2) - 4 = 12 - 4 = 8[/tex].
3. Now the result is substituted into the original equation, resulting in: [tex]8 \nabla 3[/tex].
4. The rule is applied again with [tex]A = 8[/tex] and [tex]B = 3[/tex]: [tex](8 \times 2) - 3 = 16 - 3 = 13[/tex].

Solution:
1. The problem is solved by working backward using inverse operations.
2. The last step was multiplying by [tex]3[/tex] to get [tex]24[/tex]. The inverse is division: [tex]24 \div 3 = 8[/tex].
3. The previous step was dividing by [tex]2[/tex] to get [tex]8[/tex]. The inverse is multiplying: [tex]8 \times 2 = 16[/tex].
4. The first step was adding [tex]5[/tex] to get [tex]16[/tex]. The inverse is subtracting: [tex]16 - 5 = 11[/tex]. The initial number was [tex]11[/tex].

Solution:
1. If the perimeter of a square is [tex]32[/tex] cm, each side measures [tex]32 \div 4 = 8[/tex] cm.
2. When sewing [tex]3[/tex] squares in a row, the inner sides are hidden and do not count towards the perimeter.
3. The new rectangle has a boundary formed by [tex]8[/tex] small sides ([tex]3[/tex] at the top, [tex]3[/tex] at the bottom, [tex]1[/tex] on the left, [tex]1[/tex] on the right).
4. The perimeter is [tex]8 \times 8 = 64[/tex] cm.

Solution:
1. The false assumption method is used. If all [tex]10[/tex] pieces were Knights ([tex]3[/tex] points each), the total would be [tex]10 \times 3 = 30[/tex] points.
2. However, there are actually [tex]42[/tex] points. There is a shortage of [tex]42 - 30 = 12[/tex] points.
3. Each time a Knight is exchanged for a Queen, [tex]6[/tex] extra points are gained ([tex]9 - 3 = 6[/tex]).
4. Since [tex]12[/tex] extra points are needed, [tex]12 \div 6 = 2[/tex] Knights must be exchanged for Queens. There are [tex]2[/tex] Queens.

Solution:
The options are tested by assuming where the trophy is to see if the “only one truth” rule holds.
1. If the trophy is in the Green safe: The Red safe is telling the truth. The Blue safe is telling the truth. The Green safe is lying. There are TWO truths (this does not hold).
2. If the trophy is in the Blue safe: The Red safe is telling the truth. The Blue safe is lying. The Green safe is telling the truth. There are TWO truths (this does not hold).
3. If the trophy is in the Red safe: The Red safe is lying (it says it’s not there). The Blue safe is lying (it says it’s in the Green safe). The Green safe is telling the truth (it says it’s not there). There is EXACTLY ONE truth. The trophy is in the Red safe.

Solution:
1. A number is divisible by [tex]4[/tex] if its last two digits form a number divisible by [tex]4[/tex].
2. The last two digits are [tex]A6[/tex]. Numbers ending in [tex]6[/tex] are tested for divisibility by [tex]4[/tex]: [tex]06[/tex] (No), [tex]16[/tex] (Yes), [tex]26[/tex] (No), [tex]36[/tex] (Yes), [tex]46[/tex] (No), [tex]56[/tex] (Yes), [tex]66[/tex] (No), [tex]76[/tex] (Yes), [tex]86[/tex] (No), [tex]96[/tex] (Yes).
3. The possible values for [tex]A[/tex] are: [tex]1, 3, 5, 7, 9[/tex].
4. The possible values are added together: [tex]1 + 3 + 5 + 7 + 9 = 25[/tex].

Solution:
1. Let the first number be [tex]X[/tex].
2. The second number is [tex]2 \times X = 2X[/tex].
3. The third number is twice the second: [tex]2 \times (2X) = 4X[/tex].
4. The sum of all of them is [tex]X + 2X + 4X = 7X[/tex].
5. It is known that [tex]7X = 42[/tex]; therefore, the first number is [tex]42 \div 7 = 6[/tex].
6. The value of the second number is requested: [tex]6 \times 2 = 12[/tex].

Solution:
1. Working backwards: There are [tex]12[/tex] marbles left at the end.
2. These [tex]12[/tex] marbles are the [tex]\frac{2}{3}[/tex] of the marbles that were left after giving some to the friend.
3. If [tex]\frac{2}{3} = 12[/tex], then [tex]\frac{1}{3} = 6[/tex]. Therefore, before giving marbles to the friend, there were [tex]12 + 6 = 18[/tex] marbles.
4. Those [tex]18[/tex] marbles represent the half that was left after giving the other half away initially.
5. If half of the original collection was [tex]18[/tex], there were initially [tex]18 \times 2 = 36[/tex] marbles.

Solution:
1. For the first stripe (left), any of the [tex]3[/tex] colors can be chosen. ([tex]3[/tex] options)
2. For the second stripe (center), the color just used in the first cannot be used. There are [tex]2[/tex] colors available. ([tex]2[/tex] options)
3. For the third stripe (right), the center color cannot be used. The left color can be used again, or the remaining one, so there are [tex]2[/tex] available colors. ([tex]2[/tex] options)
4. The possibilities are multiplied: [tex]3 \times 2 \times 2 = 12[/tex] different flags.

Solution:
1. The work done in [tex]1[/tex] minute is calculated. The worker completes [tex]\frac{1}{15}[/tex] of the rocket per minute. The robot completes [tex]\frac{1}{10}[/tex] of the rocket per minute.
2. Their teamwork per minute is added together: [tex]\frac{1}{15} + \frac{1}{10}[/tex].
3. A common denominator is found: [tex]\frac{1}{15} + \frac{1}{10} = \frac{2}{30} + \frac{3}{30} = \frac{5}{30}[/tex].
4. Simplifying the fraction gives [tex]\frac{1}{6}[/tex].
5. Together they complete [tex]\frac{1}{6}[/tex] of the rocket in [tex]1[/tex] minute.
6. Therefore, it will take them [tex]6[/tex] minutes to complete the entire rocket.