answer:1. **Initial Setup**: - There are 10 plus signs and 15 minus signs initially written on the board. - Each operation allowed is: a. Erasing two identical signs and writing a plus. b. Erasing two different signs and writing a minus. 2. **Operation Analysis**: - If two identical signs (either two plus signs or two minus signs) are erased, a plus sign is written, which does not change the evenness of the number of minus signs. - If two different signs (a plus and a minus) are erased, a minus sign is written. In this operation, one minus sign is removed, but another one is added, keeping the evenness of the number of minus signs unchanged. 3. **Parity Consideration**: - The total number of minus signs initially is 15, which is an odd number. - According to the given operations, each step will either: a. Not change the number of minus signs. b. Reduce the number of minus signs by 2, which also does not affect the parity (evenness or oddness) of the count of minus signs. - Therefore, after any number of these operations, the parity (odd or even character) of the number of minus signs will remain unchanged. 4. **Final Step**: - Since 24 operations are performed, we need to consider how the operations affect the number of symbols. - With each operation: - The total number of symbols on the board decreases by 1. - Initially, there are 25 symbols (10 plus signs + 15 minus signs). - After 24 operations, there will be: [ 25 - 24 = 1 ; text{symbol left} ] 5. **Conclusion**: - Given that the number of minus signs must remain odd throughout the operations and there is only one symbol left, this symbol must be a minus sign. [boxed{text{Minus}}]

answer:1. **Define the problem and setup:** - We are given a circle ( k ) with points ( A, B, C, ) and ( D ) lying on it in that order. - ( t ) is the tangent to ( k ) at point ( C ). - ( s ) is the reflection of line ( AB ) across line ( AC ). - ( G ) is the intersection of lines ( AC ) and ( BD ). - ( H ) is the intersection of lines ( s ) and ( CD ). - We need to show that ( GH ) is parallel to ( t ). 2. **Reflecting ( AB ) across ( AC ):** - Let ( s ) be the reflection of ( AB ) across ( AC ). This means that ( s ) is symmetric to ( AB ) with respect to ( AC ). 3. **Intersection points:** - Let ( G ) be the intersection of ( AC ) and ( BD ). - Let ( H ) be the intersection of ( s ) and ( CD ). 4. **Cyclic quadrilateral:** - Consider the reflection property and the cyclic nature of the quadrilateral ( AGDH ). Since ( s ) is the reflection of ( AB ) across ( AC ), the points ( A, G, D, H ) form a cyclic quadrilateral. 5. **Angles in cyclic quadrilateral:** - In a cyclic quadrilateral, opposite angles sum up to ( 180^circ ). Therefore, we have: [ angle AGH + angle ADH = 180^circ ] - Since ( H ) lies on the reflection line ( s ), we have: [ angle HGC = angle HDA ] 6. **Parallel lines:** - To show that ( GH ) is parallel to ( t ), we need to show that the angle ( angle HGC ) is equal to the angle between ( t ) and ( AC ), which is ( angle HDA ). 7. **Tangent and angle properties:** - The tangent at ( C ) makes an angle with the chord ( AC ) equal to the angle subtended by the chord ( AC ) at the circumference of the circle on the opposite side. This is known as the Alternate Segment Theorem. - Therefore, ( angle HGC = angle HDA ) implies that ( GH ) is parallel to ( t ). 8. **Conclusion:** - Since ( angle HGC = angle HDA ), we have shown that ( GH ) is parallel to ( t ). (blacksquare)

answer:The question states that Rebecca will create 3 groups of eggs, each group containing 6 eggs. To find out the total number of eggs, we need to multiply the number of groups by the number of eggs in each group. 3 groups * 6 eggs/group = 18 eggs Therefore, Rebecca has boxed{18} eggs.

answer:- **Factorization and Constraints**: As previously known, 1001 = 7 times 11 times 13. For 10^{j} - 10^{i} = 10^{i} (10^{j-i} - 1) to be a multiple of 1001, the factorization implies gcd(10^i, 1001) = 1 and thus 1001 mid 10^{j-i} - 1. - **Effective modulus constraint**: From earlier analysis, 1001 mid 10^{6k} - 1 for any integer k. Therefore, j-i must meet the condition j-i equiv 0 pmod{6}. - **Counting Valid Pairs**: Count for each valid j-i is (150 - (j-i)). Initially for j-i = 6, there are 144 pairs (i, j) from i = 0 to i = 143. For j-i = 12, there are 138 pairs from i = 0 to i = 137, and so on. - **Sum of arithmetic sequence**: This is an arithmetic sequence where initial term a=144, common difference d=-6, and last term l = a - 23d = 144 - 23 times 6 = 2. Using the formula for the sum of arithmetic sequence text{sum} = frac{n}{2} (a + l), where n=24, we get text{sum} = 12 times (144 + 2) = 12 times 146 = 1752. Final conclusion with boxed answer: boxed{1752} valid combinations.