#### • Class 11 Physics Demo

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# free answer to math riddles

Question:I need the answer and the steps it took to get the answer to this problem: My digits are three, But just what can they be? Take a third of my first From a half of my third: And a negative, right there, Would, of course be absurd Now add twice my middle To continue the riddle. If you figured it right, Then you can't have got seven; For, believe me, I know That you must have eleven. There's no catch in this, But don't take it amiss When I add one thing more, Just to make it quite clear: You should know that a five Is no part of me here.

Answers:I got it! ok here we go: Let's say the number is abc First, consider the first verse: "Take a third of my first (1/3 a) From a half of my third: (1/2c) And a negative, right there, Would, of course be absurd" This means that : 1/2c - 1/3 a is Not negative So, 1/2c - 1/3a > 0 => 1/2c > 1/3a Alright, there are 6 combinations of a and c that fit this criteria . Assuming a is divisible by 3, a must be 3, 6, or 9. Assuming c is divisible by 2, a must be 2, 4, 6, or 8. Consider the condition 1/2c > 1/3a, The possibilities of (a, c) are as follows: 1. (3,4) => (1<2) 2. (3,6) => (1<3) 3. (6,6) => (2<3) 4. (3,8) => (1<4) 5. (6,8) => (2<4) 6. (9,8) => (3<4) Ok, so now consider this part of the riddle: "Now add twice my middle To continue the riddle." This means that we now have: 1/2c - 1/3a + 2b Then: "If you figured it right, Then you can't have got seven; For, believe me, I know That you must have eleven." So this means that we can set our equation equal to 11: 1/2c - 1/3a + 2b = 11 So, plug each of these sets of values into the equations: 1. (3,4) => 2 - 1 + 2b = 11 2. (3,6) => 3 - 1 + 2b = 11 3. (6,6) => 3 - 2 + 2b = 11 4. (3,8) => 4 - 1 + 2b = 11 5. (6,8) => 4 - 2 + 2b = 11 6. (9,8) => 4 - 3 + 2b = 11 Simplify: 1. (3,4) => 1 + 2b = 11 2. (3,6) => 2 + 2b = 11 3. (6,6) => 1 + 2b = 11 4. (3,8) => 3 + 2b = 11 5. (6,8) => 2 + 2b = 11 6. (9,8) => 1 + 2b = 11 Continue to simplify: 1. (3,4) => 2b = 10 2. (3,6) => 2b = 9 3. (6,6) => 2b = 10 4. (3,8) => 2b = 9 5. (6,8) => 2b = 8 6. (9,8) => 2b = 10 Continue to simplify: 1. (3,4) => b = 5 2. (3,6) => 2b = 9 3. (6,6) => b = 5 4. (3,8) => 2b = 9 5. (6,8) => b = 4 6. (9,8) => b = 5 Now, we know that: "You should know that a five Is no part of me here." So no digits can be 5, which rules out pairs 1, 3, and 6! We are left with: 2. (3,6) => b = 4.5 4. (3,8) => b = 4.5 5. (6,8) => b = 4 Since b can obviously not be 4.5, since it must be a single digit, then # 5 is our choice! 5. (6,8) => b = 4 So, a = 6, c = 8, and b = 4 abc = 648

Question:Please show your work. And if you would need to tell me whether you use multiplication or adding or subtracting or dividing. PROBLEM: Lima beans come in 3-pound and 5-pound bags which cost $1.15 and$1.63 respectively. How many of each should you buy to have at least 17 pounds of lima beans at the lowest cost? Thanks SO much.