1. At the end, Sue is on rung number = 18. Since this is also the top rung, there must be 18 rungs in total.

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1 Grade 7 Solutions. At the end, Sue is on rung number = 8. Since this is also the top rung, there must be 8 rungs in total.. The answer is.. We apply the divisibility rule for. The sum of the digits of 45678, which is 6, is divisible by, so is itself divisible by. Thus, it leaves a remainder of 0 when divided by. 4. Notice that + 54 = ( + 7). The answer is thus 4 = = 5 6. Let the palindrome be abccba. There are 9 choices for a (-9), and 0 choices each for b and c (0-9). This gives a total of = digit palindromes. 7. Split the 6 rabbits into two groups of three rabbits. In three minutes, each of the two groups of rabbits willl have eaten carrots, so after minutes, they will have eaten 6 carrots in total = ( + 45) + ( + 4) + + ( + 5) + = (46) + = = They would have made 9500 = 0000 dollars. 0. There are 0 jelly beans in total, with 4 of them blue. The probability of the first not being blue is 6 5, the probability of the second not being blue is, and the probability of the 0 9 third not being blue is 4. Thus the probability of getting three non-blue jelly beans is = g() =, f(g()) = f() =, g(f(g()) = g() = 6, f(g(f(g()))) = 7, g(f(g(f(g())))) = 4, so f(g(f(g(f(g()))) = = 006 = 50. Since 50 is prime, our answer is just + 50 = Each cycle of the pattern consists of forward steps and then back step, which is a net gain of forward steps. After 99 cycles, or 99 backward steps, the baby chick has made a net of 99 = 98 steps, making it steps away from the traffic light. Therefore, the net step forward will bring it to its destination, so it will have made a total of 99 backward steps. 4. After every 0 minutes, the number of parasprites doubles: each of the original parasprites remains, and each of them spawns a new parasprite. Since si sets of 0 minutes pass in two hours, there will be a total of 6 = 64 parasprites after hours. 5. Since y is constant for all values of and y, we have that y is always equal to 6 4 = 44. Therefore, we need y such that 4 y = 44, so y = Note that 49 7 = (7 ) 7 = As a result, = Since this is equal to 49 = 7, we have that = 45.

2 7. We know that all squares of real numbers are nonnegative. Thus, if two squares add to 0, both of them must be zero. We must therefore have +y = 5 and y = 7. Tripling the second equation and adding it to twice the first equation gives us 7 = = = The resulting figure has one circular face, of surface area πr, one lateral surface of a cylinder, with surface area πrh, and one-half of a sphere, which has surface area (4πr ) = πr. Thus, the total surface area is πr + πrh = π(5 ) + π(5)(0) = 75π. 9. A number has eactly divisors if and only if it is the square of a prime number. The third smallest number of this form is the square of the third smallest prime, 5, so the answer is 5 = It can be easily verified that the statement holds true when n =, 4, or 5. For n > 5, it can easily be seen that the longest diagonal is longer than the shortest. Thus, our answers are, 4, and 5.. If + 5y = 00, since and 00 are both even, 5y must be even too, so y must be even. Since 6, 5y 00 88, so y 7. From here, we can list all the solutions: (, y) = (5, 6), (0, 8), (5, 0), (0, ), (5, 4), (0, 6). There are 6 solutions in this list. (. In the binomial epansion of + ) 0 ( ) n, all the terms are of the form c m, with m + n = 0. In order to attain the term, we must have m n =. Solving these equations gives (m, n) = ( 0, 009 ). But since m and n must be integers, the term must have coefficient 0.. The number 9 n ends with if n is even, and ends with 9 if n is odd. Since is even, the last digit of is. 4. First of all, we must consider that 0 is a leap year so we must consider the etra day (hence 0 will have 66 days instead of 65). We must also consider that October 6, 0, the day of the contest, is a Sunday. One year from now, which is 66 days later, October 6, 0 will be in 5 weeks and days, so October 6 will be a Tuesday. Therefore, October 7 will be a Wednesday. 5. The epression on the left hand side will be equal to only when the eponent is equal to 0 or if + is equal to - or. The eponent will be equal to 0 when = or, and the base is equal to when = or when = 0. We verify that = 0 and = satisfy the equation, while = yields an epression of the form 0 0, which is undefined, so the only satisfying the equation are 0 and -, so our answer is. 6. Subtract the second equation from the first to get 8 = 5 = 8 5 = 0. This factors as ( + 7)( 5) = 0 = = 5 or = 7. However, only one of these is a solution to the first equation, namely = After drawing the figure, we find that half the chord (which has a length of 6), the radius of the circle, and the line from O to AB form a right triangle with legs and 6. Applying

3 the Pythagorean Theorem, we find that the hypotenuse is 5, which is also the radius. The ( area is therefore π radius = π 5) = 45π. 8. If n is a number, it satisfies the property if and only if each of the numbers divides n +. Thus, we just need to find the least common multiple of 4, 5, and 6 and subtract. The lowest common multiple is 60, so our answer is If the first roll is a, we have si choices for the second roll. If it is a, that gives us three choices for the second roll (, 4, 6). If it is a, then there are only two choices (, 6), and for 4, 5, and 6, there is only one choice (itself). That gives us a probability of = By the factor theorem, factors as ( r)( s)( t). Thus, ( r)( s)( t) = = = 5.. ((!)!)! = (6!)! = 70!; the number of zeroes at the end of 70 is = = Let = BD and y = DC. Clearly, +y = 5. Also, by the angle bisector theorem, y = 4 6 =, so = y = (5 ), so 5 = 0, so = BD = = + ( + )( ) + (5 + 4)(5 4) + + (5 + 50)(5 50). (Note: a b = (a + b)(a b).) This is equal to = 0 = Using the Pythagorean Theorem on the diagonal of the rectangle gives us that it has length = 6. Since the diagonal is the hypotenuse of a right triangle inscribed in a circle, it is also a diameter. Thus the radius is and its area is π = 69π. 5. Note that the line segments A B, B C, C A divides triangle ABC into four identical smaller triangles, and similarly A B, B C, C A divides triangle A B C into four even smaller identical triangles. Therefore, the area of triangle ABC is four times that of triangle A B C, which in turn has area four times that of A B C. Hence, the ratio of the area of triangle A B C to the area of triangle ABC is Since Archimedes position in the line is fied, it is enough to find the number of ways to arrange Bernoulli, Cauchy, Descartes, and Euler in a line. Because Descartes and Euler must stand net to each other, we temporarily view them as a single unit. The number of ways to arrange Bernoulli, Cauchy, and the Descartes-Euler unit is = 6, and the number of ways to arrange Descartes and Euler in the unit is, so our answer is 6 =. 7. Let y be the length of the side parallel to the river, and let be the lengths of the other two sides. We must have + y = 00. We seek to maimize the area, y, that is enclosed by the fence and the river. Rearranging, we get y = 00, so the area is y = (00 ) = 00, which is a quadratic in. The verte, or maimum (because the coefficient of is negative), occurs at = 00 = 50. When = 50, y = = 00. The area is ( ( )) then y = = 5000.

4 8. The epression factors to ( + 4)( ), which can only be a prime number if one of the two epressions is equal to or - and the other one is a prime number or a negative number whose absolute value is prime. The only values of that work are 5 and. 9. Call the roots of + 4 = 0 a and b. From Vieta s formulas, we know that ab = and a + b = 4. Write the desired quadratic as q + r, for some q and r. Again by Vieta s forumlas, we have that r = (ab) and q = a +b. r is simply ( ) = 4. To find q, we note that (a+b) = a +ab+b. Rearranging, we have that a +b = (a+b) ab = ( 4) ( ) = 0. The quadratic equation we seek is therefore = There are 50 cards from which Steven can draw. The cards he can draw to keep from going above are each of the four aces, twos, threes, fours, and fives, the three remaining sies, the three remaining sevens, and the four eights. There are 0 such cards in total, so the probability is 0 50 = Consider the attached diagram. If we choose, y randomly in this square, the probability that their sum is less than is the probability that (, y) lies in the shaded region. The area of the shaded region is 8, while the area of the whole region is, so our answer is Let a, t, r be the percentage of the apples that Applejack, Twilight Sparkle, and Rainbow Dash can pick in one day, respectively. We are given that (a + t) = 4(a + r) = 6(t + r) =, so a + t =, a + r = 4, and t + r = 6. Adding these up and dividing by gives a + r + t = 8, so it will take a + r + t = 8 days for them to pick all the apples. 4. Let us consider n mod. When n 0 mod, n + n mod. When n mod, n + n mod. Finally, when n mod, n + n mod. Thus, n + n + eactly when n is congruent to 0 or mod. There are 670 numbers that are 0 mod, namely,,..., 670 = 00. There are 67 numbers that are mod, namely 0 +, +,..., = 0. Thus there are = 4 numbers that satisfy the restrictions. 4

5 = ( + ) + ( + ) + ( 4 + ) + + ( 00 + ) = This is the same as the sum of positive integers up to 0 without n(n + ) and. The sum of positive integers from to n can be epressed as. The sum is therefore 0(0) ( + ) = Consider the top left hand corner square of an n n square. Notice that when n =, there are 6 choices for the top left hand corner square, when n =, there are 5 choices, and, in general, for an k k square there are (7 k) choices for the top left hand corner square. Thus, since k ranges from to 6, the total number of squares is = Consider the balls Bob picks. The only way one of your balls will not be among Bob s balls is if you pick the two he did not pick. This only happens with probability ( 5 = ), so the 0 probability one of your balls will be among Bob s chosen ones is 0 = If ( =, then ) = 9, or + = 9, so + =. 48. The ratio of the areas of two similar polygons is the square of the ratio of similitude between ( ) B B the two polygons, so we need to find. If we let M be the midpoint of B B, we A A see that A B M is a triangle, so B B = B M =, so the desired ratio is A A B A ( ) = First, notice that 5 = Letting = 0 0 gives us that the subtrahend is ( ). Net, we realize that 0 00 = , or a string of 00 9 s, which has a sum of 900. Since we are subtracting the number ( ), which is composed of a total of 5 s and many 0 s, and since no borrowing is done, the answer is just = The area of a polygon with 0,000 sides can be approimated by the area of a circle. The polygon has a diagonal of length 0, which corresponds to a diameter of 0 in the approimating circle. A circle with a diameter of 0 has a radius of 0 for an area of π 0 = 00π, which is approimately 4. 5