answer:Let's call the remainder when 690 is divided by the greatest number "r". According to the problem, when 875 is divided by the greatest number, the remainder is 25. Since the greatest number that divides both 690 and 875 leaving remainders of r and 25 respectively is 170, we can write the following equations: 690 = 170k + r 875 = 170l + 25 where k and l are integers representing the quotient when 690 and 875 are divided by 170, respectively. We need to find the value of r. Since 875 leaves a remainder of 25 when divided by 170, we can subtract 25 from 875 to find a multiple of 170: 875 - 25 = 850 850 is a multiple of 170, so we can divide it by 170 to find the value of l: 850 / 170 = 5 So, l = 5. Now, we need to find the value of k such that when 690 is divided by 170, it leaves the same remainder r as when 875 is divided by 170. Let's find the largest multiple of 170 that is less than 690: 690 / 170 = 4 with a remainder The largest multiple of 170 less than 690 is 4 * 170 = 680. Now, we can find the remainder r when 690 is divided by 170: 690 - 680 = 10 So, the remainder r when 690 is divided by 170 is boxed{10} .

answer:1. **Given Situation and Setup:** We are given a circle and a point M outside the circle. A point C on the circle is such that MC^2 = MA cdot MB, where A and B are the points where a secant, extended through M, intersects the circle. We aim to prove that MC is tangent to the circle. 2. **Expression and Theorem:** Consider another point C_1 that intersects the circle on the extension of MC. According to the Power of a Point Theorem, which states that if two secants intersect at a point outside a circle, the product of the lengths of the segments of one secant is equal to the product of the lengths of the segments of the other secant. Applying that: [ MC cdot MC_1 = MA cdot MB ] 3. **Using the Given Condition:** We know from the problem that: [ MC^2 = MA cdot MB ] 4. **Verification and Simplification:** Since MC cdot MC_1 = MA cdot MB and MC^2 = MA cdot MB, we have: [ MC^2 = MC cdot MC_1 ] Dividing both sides of this equation by MC, we get: [ MC = MC_1 ] Therefore, C and C_1 must coincide. If C and C_1 coincide, then by definition, the point C is both on the circle and on the tangent line intersecting at a single point. 5. **Conclusion:** This implies that MC must be a tangent to the circle at point C. Therefore, we conclude that MC is indeed a tangent to the circle: [boxed{text{MC is tangent to the circle}}.]

answer:Let's denote the total number of bananas as B. If each of the 840 children were present, they would each get 2 bananas. So the total number of bananas can be calculated as: B = 840 children * 2 bananas/child = 1680 bananas However, on that day, 420 children were absent, so the number of children present was: 840 children - 420 children = 420 children Since the same total number of bananas (1680) was distributed among fewer children (420), we can calculate how many bananas each child present got: B / number of children present = 1680 bananas / 420 children = 4 bananas/child Each child was supposed to get 2 bananas, but since they got 4 bananas each, the extra number of bananas each child got is: 4 bananas/child - 2 bananas/child = 2 extra bananas/child So, each child got boxed{2} extra bananas.

answer:1. **Initial Setup and Definitions**: - We are given a graph ( G ) where each vertex represents a bead on a stick, and edges represent elastic bands connecting pairs of beads. - Each bead can be at an integer height from 1 to 2017. - The young ant can traverse any edge, while the old ant can only traverse edges where the height difference between connected beads is at most 1. - A "valid situation" is when there is at least one pair of connected beads with different heights. 2. **Coloring the Graph**: - We color the vertices of ( G ) with colors ( 1, 2, ldots, 2017 ), where the color represents the height of the bead. - Start from a vertex ( v_0 ) and let ( V_1 ) be the maximal set of vertices reachable from ( v_0 ) using edges where the height difference is at most 1 (i.e., the old ant's path). 3. **Partitioning the Vertices**: - Let ( V_2 = V(G) setminus V_1 ), the set of vertices not in ( V_1 ). 4. **Swapping Colors to Expand ( V_1 )**: - Consider any two adjacent colors ( i ) and ( i+1 ). - Let ( W_i ) be the vertices in ( V_2 ) colored with ( i ), and ( W_{i+1} ) be the vertices in ( V_2 ) colored with ( i+1 ). - Swap the colors of ( W_i ) and ( W_{i+1} ), i.e., recolor ( W_i ) with ( i+1 ) and ( W_{i+1} ) with ( i ). 5. **Validity of the New Coloring**: - After swapping, any two connected vertices in ( V_2 ) are still colored differently because we only swapped two colors. - If there were two vertices ( v_1 in V_1 ) and ( v_2 in V_2 ) with the same color before swapping, then ( V_1 ) would not have been maximal, as we could have included ( v_2 ) in ( V_1 ). 6. **Expanding ( V_1 )**: - By performing such swaps, we ensure that eventually, there will be a vertex ( v_2 in V_2 ) that can be added to ( V_1 ) because it will be colored with a color adjacent to one of the vertices in ( V_1 ). - Repeat this process until ( V_1 ) includes all vertices of ( G ). 7. **Conclusion**: - By iteratively swapping colors and expanding ( V_1 ), we can ensure that the old ant can traverse the entire graph ( G ) starting from any bead. - This process guarantees that we can create a valid situation where the old ant can reach every bead. (blacksquare)