Lesson Math circle level problems on evaluating expressions
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<H2>Math circle level problems on evaluating expressions</H2> <H3>Problem 1</H3>If f(2a-b) = f(a)*f(b) for all "a" and "b", and the function is never equal to zero, find the value of f(5). <B>Solution</B> Let solve the problem in 4 (four) steps. <pre> (a) Let a = 1, b = 1. Then, according to the given formula, f(2*1-1) = f(1)*f(1), or f(1) = f(1)*f(1). Since f(1) =/= 0 (as it is given), it implies f(1) = 1 (we can cancel the common factor f(1) in both sides). (b) Let a = 1, b = 2. Then, according to the given formula, f(2*1-1) = f(1)*f(2), or f(1) = f(1)*f(2). Since f(1) = 1 (we just know it from (a)), it implies f(2) = 1. (c) Let a = 2, b = 1. Then, according to the given formula, f(2*2-1) = f(2)*f(1), or f(3) = 1*1 = 1 (since we just know from (b) and (a) that f(2) = 1, f(1) = 1 ). (d) Let a = 3, b = 1. Then, according to the given formula, f(2*3-1) = f(3)*f(1), or f(5) = 1*1 = 1 (since we just know from (c) and (a) that f(3) = 1, f(1) = 1 ). Thus we just found out that f(5) = 1, and the solution is complete at this point. </pre> <H3>Problem 2</H3>A function f is defined for integers m and n as given: f(mn) = f(m)f(n)-f(m+n)+ 1001, where either m or n is equal to 1, and f(1)=2. a) Prove that f(x) = f(x-1) + 1001. b) Find the value of f(9999). <B>Solution</B> <pre> In the given formula, take x = m, n = 1. You will get f(x*1) = f(x)*f(1) - f(x+1) + 1001, or f(x) = 2*f(x) - f(x+1) + 1001, which is the same as 0 = f(x) - f(x+1) + 1001, f(x+1) = f(x) + 1001. (1) Thus part (a) is just solved and completed. From the formula (1), we conclude that the sequense (1) is an arithmetic progression with the first term f(1) = 2 and the common difference of 1001. So f(9999) = f(1) + (9999-1)*1001 = 2 + (9999-1)*1001 = 10,008,000. Thus part (b) is completed, too. </pre> <H3>Problem 3</H3>Given that {{{f(x)=5x^2-3x+7}}} and {{{f(g(x))=(5x^4/9)+(17x^2/3)+21}}}, find all possible values for the sum of the coefficients in the quadratic function g(x). <B>Solution</B> The key idea to solve this problem is to use the fact that the sum of coefficients of any polynomial p(x) = {{{a[n]*x^n + a[n-1]*x^(n-1) + ellipsis + a[1]*x + a[0]}}} is the value of this polynomial p(1) at x = 1, which is quite OBVIOUS. <pre> Based on this idea, the sum of the coefficients in the quadratic function g(x) is g(1). We don't know this value g(1), but from the problem we can calculate the composition f(g(1)) by substituting x= 1 into the given formula for f(g(x). We have then f(g(1)) = {{{(5*1^4)/9}}} + {{{(17*1^2)/3}}} + {{{21}}} = {{{5/9 + 17/3 + 21}}} = {{{5/9 + 51/9 + 21}}} = {{{56/9 + 21}}} = 27 2/9. Now we can state that the value g(1), which we are seeking for, is one of two possible roots of the equation f(x) = 27 2/9, or one of the two possible roots of the equation 5x^2 - 3x + 7 = 27 2/9. In standard quadratic form, this equation is 5x^2 - 3x - 20 {{{2/9}}} = 0, or, multiplying all the terms by 9, for convenience, 45x^2 - 27x - 182 = 0. To find the roots, use the quadratic formula. The roots are {{{210/90}}} = {{{7/3}}} and {{{-156/90}}} = {{{-26/15}}}. So we conclude that the sum of the coefficients in the quadratic function g(x) is EITHER {{{7/3}}} = 2{{{1/3}}} OR {{{-26/15}}} = -1{{{11/15}}}. <U>ANSWER</U> </pre> <H3>Problem 4</H3>Five different positive integers added two at a time give the following sums: 16, 20, 22, 23, 25, 28, 29, 30, 34 and 37. Find the product of the five integers. <B>Solution</B> <pre> 1. Let the numbers be a < b < c < d < e (in this order/ordering). Four pair-wise sums of "a" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}. Four pair-wise sums of "b" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}. Four pair-wise sums of "c" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}. Four pair-wise sums of "d" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}. Four pair-wise sums of "e" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}. Therefore, if we sum up all the given numbers, we will get 4*(a+b+c+d+e): 4*(a + b + c + d + e) = 16+20+22+23+25+28+29+30+34+37 = 264. It implies that a + b + c + d + e = {{{264/4}}} = 66. (1) 2. Further, 16 is the sum of the two smallest numbers: 16 = a + b. (2) 37 is the sum of the two largest numbers: 37 = d + e. (3) It implies (by subtracting (2) and (3) to (1)) than c = 66 - 16 - 37 = 13. So, we just found one of the five numbers, namely, the middle number c = 13. 3. OK. Let's continue our analysis. So, the two smallest numbers a and b give the sum of 16, and the third number is 13. It means that the next sum, 20 (see the condition) is a+c. It can not be nothing else. (Because b + c must be just greater than a + c). If a + c = 20 and c = 13, then a = 20 - 13 = 7. It implies b = 16-7 = 9. So, the first three numbers are a = 7, b = 9 and c = 13. 4. OK. Now we can make the similar analysis from the other end. So, the two largest numbers d and e give the sum of 37, and the third number in the descending order is 13. It means that the next from the largest sum, 34 (see the condition) is e+c. It can not be nothing else. (Because d + c is just less than e + c). If e + c = 34 and c = 13, then e = 34 - 13 = 21. It implies d = 37-21 = 16. 5. Thus the numbers are 7, 9, 13, 16 and 21. Then their product is 275184. <U>Answer</U>. The product of the numbers is 275184. </pre> <H3>Problem 5</H3>There are 5 sacks, and they are weighed 2 at a time. Their weights are 11, 11.2, 11.3, 11.4, 11.5, 11.6, 11.7, 11.8, 12 and 12.1. This is the weight of all the possible outcomes. How heavy are each of the sacks? <B>Solution</B> <pre> Let us call these sacks by the names "a", "b", "c", "d" and "e". And let "a" be the lightest, "b" is the next by the weight, "c" is the next after "b", "d" is next after "c", and "e" is the next after "d" (and the last). Since all 10 given outcomes are different numbers, all "a, "b", "c", "d" and "e" have different weights. (Indeed, had some of the sacks have equal weights, some numbers in the outcome data would be equal, which is not the case.) If you add all outcomes, you will get the sum 11 + 11.2 + 11.3 + 11.4 + 11.5 + 11.6 + 11.7+ 11.8 + 12 + 12.1 = 115.6. It is the sum of all pairs (a+b), (a+c), (a+d), . . . , (d+e) taken 2 at a time. By grouping and regrouping, you can easily understand that it is nothing else as 4*(a+b+c+d+e), i.e four times the sum of (a+b+c+f+e). It implies that the sum a + b + c + d + e = {{{115.6/4}}} = 28.9. (1) Now, it is clear that the very first outcome 11 (the smallest outcome) is the sum of "a and "b", the two lightest sacks. It can not be nothing else. So, we have a + b = 11. (2) Similarly, it is clear that the very next outcome 11.2 is the sum of "a" and "c". It simply can not be nothing else. So, we have a + c = 11.2. (3) Next, it is clear that the very last outcome 12.1 (the greatest outcome) is the sum of "d" and "e", the two heaviest sacks. It can not be nothing else. So, we have d + e = 12.1. (4) Similarly, outcome 12 is the sum of "c" and "e". It simply can not be nothing else. So, we have c + e = 12. (5) Take the sum of (2) and (3). You will get a + b + d + e = 11 + 12.1 = 23.1. (6) Now distract (6) from (1). You will get c = 28.9 - 23.1 = 5.8. So, we just found "c". Now we can easily find all remaining unknowns. From (3), a = 11.2 - c = 11.2 - 5.8 = 5.4. From (2), b = 11 - a = 11 - 5.4 = 5.6. From (5), e = 12 - c = 12 - 5.8 = 6.2. From (4), d = 12.1 - e = 12.1 - 6.2 = 5.9. <U>Answer</U>. The sacks are 5.4, 5.6, 5.8, 5.9 and 6.2, from lightest to heaviest. </pre> <H3>Problem 6</H3>If {{{x^2 + y^2 + z^2}}} = {{{2*(5X - 8Y + 6Z) - 125}}}, where x, y and z are real numbers, find the value of x+y+z. <B>Solution</B> <pre> {{{x^2 + y^2 + z^2}}} = 2*(5x - 8y + 6z) - 125 ====> {{{x^2 + y^2 + z^2}}} = 10x - 16y + 12z - 125 {{{x^2-10x + y^2+16y + z^2-12z}}} = -125 {{{(x^2-10x)}}} + {{{(y^2+16y)}}} + {{{(z^2-12Z)}}} = -125 Complete the square in each parentheses as follows {{{(Xx^2-10x+25)}}} + {{{(y^2+16y+64)}}} + {{{(z^2-12z+36)}}} = -125 + 25 + 64 + 36 Write each parenthetical expression as a perfect square and combine the terms on the right: {{{(x - 5)^2}}} + {{{(y + 8)^2}}} + {{{(z - 6)^2}}} = 0. Left side is the sum of the three squares of real numbers, and it is equal to zero, according to the equation. It is possible if and only if parenthetical member is equal to zero. So, x = 5, y = -8, z = 6. 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