If 1 9 8 1 What Is 2 8 9? The 48 Correct Answer

Page No. 81:

Task 1: Complete the last column of the table. S. No. Equation value Tell if the equation is satisfied. (Yes/No) (i) x + 3 = 0 x = 3 – (ii) x + 3 = 0 x = 0 – (iii) x + 3 = 0 x = – 3 – (iv) x – 7 = 1 x = 7 – (v) x − 7 = 1 x = 8 – (vi) 5x = 25 x = 0 – (vii) 5x = 25 x = 5 – (viii) 5x = 25 x = – 5 – (ix) m = − 6 – (x) m = 0 – (xi) m = 6 –

Answer: (i) x + 3 = 0 L.H.S. = x + 3 By setting x = 3, L.H.S. = 3 + 3 = 6 ≠ R.H.S. ∴ No, the equation is not fulfilled. (ii) x + 3 = 0 L.H.S. = x + 3 By setting x = 0, L.H.S. = 0 + 3 = 3 ≠ R.H.S. ∴ No, the equation is not fulfilled. (iii) x + 3 = 0 L.H.S. = x + 3 By setting x = −3, L.H.S. = − 3 + 3 = 0 = right hand side ∴ Yes, the equation is fulfilled. (iv) x − 7 = 1 L.H.S. = x − 7 By setting x = 7, L.H.S. = 7 − 7 = 0 ≠ R.H.S. ∴ No, the equation is not fulfilled. (v) x − 7 = 1 L.H.S. = x − 7 By setting x = 8, L.H.S. = 8 − 7 = 1 = right side ∴ Yes, the equation is fulfilled. (vi) 5x = 25 L.H.S. = 5x By setting x = 0, L.H.S. = 5 × 0 = 0 ≠ R.H.S. ∴ No, the equation is not fulfilled. (vii) 5x = 25 L.H.S. = 5x By setting x = 5, L.H.S. = 5 × 5 = 25 = right side ∴ Yes, the equation is satisfied. (viii) 5x = 25 L.H.S. = 5x By setting x = −5, L.H.S. = 5 × (−5) = −25 ≠ R.H.S. ∴ No, the equation is not fulfilled. (ix) = 2 L.H.S. = By setting m = −6, L H S = ≠ R H S ∴No, the equation is not satisfied. (x) = 2 L.H.S. = By setting m = 0, L.H.S. = ≠ R.H.S. ∴No, the equation is not fulfilled. (xi) = 2 L.H.S. = By setting m = 6, L.H.S. = = R.H.S. ∴ Yes, the equation is fulfilled.

Page No. 81:

Exercise 2: Check whether the value given in brackets is a solution of the given equation or not: (a) n + 5 = 19 (n = 1) (b) 7n + 5 = 19 (n = − 2) (c ) 7n + 5 = 19 (n = 2) (d) 4p − 3 = 13 (p = 1) (e) 4p − 3 = 13 (p = − 4) (f) 4p − 3 = 13 (p = 0 )

Answer: (a) n + 5 = 19 (n = 1) Putting n = 1 in L.H.S., n + 5 = 1 + 5 = 6 ≠ 19 as L.H.S. ≠ R.H.S., Hence n = 1 is not a solution of the given equation, n + 5 = 19. (b) 7n + 5 = 19 (n = −2) Putting n = −2 in L.H.S., 7n + 5 = 7 × ( −2) + 5 = −14 + 5 = −9 ≠ 19 Since L.H.S. ≠ R.H.S., Hence n = −2 is not a solution of the given equation, 7n + 5 = 19. (c) 7n + 5 = 19 (n = 2) Putting n = 2 in L.H.S., 7n + 5 = 7 × (2 ) + 5 = 14 + 5 = 19 = R.H.S. As L.H.S. = R.H.S., Hence n = 2 is a solution of the given equation, 7n + 5 = 19. (d) 4p − 3 = 13 (p = 1) Substituting p = 1 in L.H.S., 4p − 3 = (4 × 1 ) − 3 = 1 ≠ 13 Since L.H.S ≠ R.H.S., therefore p = 1 is not a solution of the given equation, 4p − 3 = 13. (e) 4p − 3 = 13 (p = −4) Putting p = − 4 in L.H.S. , 4p − 3 = 4 × (−4) − 3 = − 16 − 3 = −19 ≠ 13 As L.H.S. ≠ R.H.S., Hence p = −4 is not a solution of the given equation, 4p − 3 = 13. (f) 4p − 3 = 13 (p = 0) Substituting p = 0 in L.H.S., 4p − 3 = (4 × 0 ) − 3 = −3 ≠ 13 Since L.H.S. ≠ R.H.S., Therefore p = 0 is not a solution of the given equation, 4p − 3 = 13.

Page No. 81:

Task 3: Solve the following equations by trial and error: (i) 5p + 2 = 17 (ii) 3m − 14 = 4

Answer: (i) 5p + 2 = 17 Putting p = 1 in L.H.S., (5 × 1) + 2 = 7 ≠ R.H.S. Putting p = 2 in L.H.S., (5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S. Put p = 3 in L.H.S., (5 × 3) + 2 = 17 = R.H.S. Hence p = 3 is a solution of the given equation. (ii) 3m − 14 = 4 putting m = 4, (3 × 4) − 14 = −2 ≠ R.H.S. Setting m = 5, (3 × 5) − 14 = 1 ≠ R.H.S. Put m = 6, (3 × 6) − 14 = 18 − 14 = 4 = R.H.S. So m = 6 is a solution of the given equation.

Page No. 81:

Task 4: Write equations for the following statements: (i) The sum of the numbers x and 4 is 9. (ii) Subtracting 2 from y is 8. (iii) Ten times a is 70. (iv) The number b divided by 5 is 6. (v) Three quarters of t is 15. (vi) Seven times m plus 7 is 77. (vii) A quarter of a number x minus 4 is 4. (viii) Taking 6 out of 6 times y gives you 60 (ix) If you add 3 to a third of z, you get 30.

Answer: (i) x + 4 = 9 (ii) y − 2 = 8 (iii) 10a = 70 (iv) (v) (vi) Seven times m is 7m. 7m + 7 = 77 (vii) A quarter of a number x is . (viii) Six times y is 6y. 6y − 6 = 60

(ix) One third of z is . Video Solution to Simple Equations (Page:81, Q.No.:4) NCERT Solution for Grade 7 Math – Simple Equations 81, Question 4

Page No. 81:

Task 5: Write the following equations in statement form: (i) p + 4 = 15 (ii) m − 7 = 3 (iii) 2m = 7 (iv) (v) (vi) 3p + 4 = 25 (vii) 4p − 2 = 18 (viii)

Answer: (i) The sum of p and 4 is 15. (ii) 7 subtracted from m is 3. (iii) Twice a number m is 7. (iv) One fifth of m is 3. (v) Three -fifths of m is 6. (vi) Three times a number p when added to 4 gives 25. (vii) When 2 is subtracted from four times a number p it gives 18. (viii) When 2 is added to Half of a number p gives 8.

Page No. 82:

Question 6: Write an equation in the following cases: (i) Irfan says he has 7 marbles, more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m as the number of Parmit’s marbles.) (ii) Laxmi’s father is 49 years old. He is 4 years older than Laxmi three times. (Assume Laxmi’s age to be y years.) (iii) The teacher tells the class that the highest score a student has in her class is twice the lowest score plus 7. The highest score is 87. (Assume the lowest score is l.) (iv) In an isosceles triangle, the vertex angle is twice each base angle. (The base angle is b in degrees. Remember that the sum of the angles of a triangle is 180 degrees.)

Answer: (i) Parmit has m marbles. 5 × number of marbles Parmit has + 7 = number of marbles Irfan has 5 × m + 7 = 37 5m + 7 = 37 (ii) Let Laxmi be y years old. 3 × age of Laxmi + 4 = age of Laxmi’s father 3 × y + 4 = 49 3y + 4 = 49

(iii) Let the lowest grades be l. 2 × worst score + 7 = highest score 2 × l + 7 = 87 2 l + 7 = 87 (iv) In an isosceles triangle, two angles are equal. The base angle is b. Vertex angle = 2 × base angle = 2b sum of all interior angles of a Δ = 180° b + b + 2b = 180° 4b = 180°

Video Solution for Simple Equations (Page:82, Q.No.:6) NCERT Solution for Grade 7 Math – Simple Equations 82, Question 6

Page No. 86:

Question 1: First indicate the step you will use to separate the variable, then solve the equation: (a) x + 1 = 0 (b) x + 1 = 0 (c) x − 1 = 5 (d) x + 6 = 2 (e) y − 4 = − 7 (f) y − 4 = 4 (g) y + 4 = 4 (h) y + 4 = − 4

Answer: (a) x − 1 = 0 If we add 1 to both sides of the given equation, we get x − 1 + 1 = 0 + 1 x = 1 (b) x + 1 = 0 Subtracting 1 from both sides of the Given equation Equation we get x + 1 − 1 = 0 − 1 x = −1 (c) x − 1 = 5 Adding 1 to both sides of the given equation we get x − 1 + 1 = 5 + 1 x = 6 ( d) x + 6 = 2 Subtracting 6 from both sides of the given equation we get x + 6 − 6 = 2 − 6 x = −4 (e) y − 4 = −7 Adding 4 to both sides of the given one Equation , we get y − 4 + 4 = − 7 + 4 y = −3 (f) y − 4 = 4 Adding 4 to both sides of the given equation, we get y − 4 + 4 = 4 + 4 y = 8 ( g) y + 4 = 4 By subtracting 4 from both sides of the given equation we get y + 4 − 4 = 4 − 4 y = 0 (h) y + 4 = −4 Subtracting 4 from both sides of the given equation , we get y + 4 − 4 = − 4 − 4 y = −8

Page No. 86:

Question 2: First indicate the step you will use to separate the variable, then solve the equation: (a) 3l = 42 (b) (c) (d) 4x = 25 (e) 8y = 36 (f) (g) (h) 20t = − 10

Answer: (a) 3l = 42 Divide both sides of the given equation by 3, we get l = 14 (b) Multiply both sides of the given equation by 2, get b = 12 (c) Multiply both sides of the given Equation Equation divided by 7, we get p = 28 (d) 4x = 25 dividing both sides of the given equation by 4, we get x = (e) 8y = 36 dividing both sides of the given equation by 8, we get y = ( f ) If we multiply both sides of the given equation by 3, we get (g) If we multiply both sides of the given equation by 5, we get

(h) 20t = −10 Dividing both sides of the given equation by 20 we get

Page No. 86:

Question 3: List the steps you will use to separate the variable and then solve the equation: (a) 3n ​​− 2 = 46 (b) 5m + 7 = 17 (c) (d)

Answer: (a) 3n ​​− 2 = 46

If we add 2 to both sides of the given equation, we get 3n − 2 + 2 = 46 + 2 3n = 48 If we divide both sides of the given equation by 3, we get n = 16 (b) 5m + 7 = 17 Subtract of 7 from both sides of the given equation we get 5m + 7 − 7 = 17 − 7 5m = 10 Dividing both sides of the given equation by 5 we get

(c) If we multiply both sides of the given equation by 3, we get. If we divide both sides of the given equation by 20, we get

(d) If we multiply both sides of the given equation by 10, we get. If we divide both sides of the given equation by 3, we get p=20

Page No. 86:

Task 4: Solve the following equations: (a) 10p = 100 (b) 10p + 10 = 100 (c) (d) (e) (f) 3s = − 9 (g) 3s + 12 = 0 (h) 3s = 0 (i) 2q = 6 (j) 2q − 6 = 0 (k) 2q + 6 = 0 (l) 2q + 6 = 12

Answer: (a) 10 p = 100 (b) 10 p + 10 = 100 10 p + 10 − 10 = 100 − 10 10 p = 90 (c) (d) (e) (f) 3 s = −9 ( g) 3s + 12 = 0 3s + 12 − 12= 0 − 12 3s = −12 (h) 3s = 0 (i) 2q = 6 (j) 2q − 6 = 0 2q − 6 + 6 = 0 + 6 2q = 6 (k) 2q + 6 = 0 2q + 6 − 6 = 0 − 6 2q = −6 (l) 2q + 6 = 12 2q + 6 − 6 = 12 − 6 2q = 6

Page No. 89:

Task 1: Solve the following equations. (a) (b) 5t + 28 = 10 (c) (d) (e) (f) (g) (h) 6z + 10 = − 2 (i) (j)

Answer: (a) (transpose to R.H.S.) divide both sides by 2, (b) 5t + 28 = 10 5t = 10 − 28 = −18 (transpose 28 to R.H.S.) divide both sides by 5, (c) (transpose 3 to R.H.S.) Multiply both sides by 5, a = −1 × 5 = −5 (d) (transpose 7 to R.H.S.) Multiply both sides by 4, q = −8 (e) Multiply both sides by 2, 5x = −10 × 2 = −20 divide both sides by 5, (f) multiply both sides by 2, divide both sides by 5, (g) (transpose right) divide both sides by 7, (h) 6z + 10 = − 2 6z = − 2 − 10 = −12 (transpose 10 to R.H.S.) Divide both sides by 6, (i) multiply both sides by 2, divide both sides by 3, (j) (transpose −5 to R.H.S.) multiply both sides with 3 , 2b = 8 × 3 = 24 dividing both sides by 2, b = = 12

Page No. 89:

Task 2: Solve the following equations. (a) 2 (x + 4) = 12 (b) 3 (n − 5) = 21 (c) 3 (n − 5) = − 21 (d) − 4 (2 + x) = 8 (e) 4 (2 − x) = 8

Answer: (a) 2 (x + 4) = 12 dividing both sides by 2, x = 6 − 4 = 2 (transposing 4 to R.H.S.) (b) 3 (n − 5) = 21 dividing both sides by 3, n = 7 + 5 = 12 (transpose −5 to the right) (c) 3 (n − 5) = −21 divide both sides by 3, n = − 7 + 5 = −2 (transpose −5 to the right) (d) −4 (2 + x) = 8 Divide both sides by −4, x = − 2 − 2 = −4 (transpose 2 to the right) (e) 4 (2 − x) = 8 Divide both sides by 4, 2 − x = 2 −x = 2 − 2 (transpose 2 to the right) −x = 0 x = 0

Page No. 89:

Task 3: Solve the following equations. (a) 4 = 5 (p − 2) (b) − 4 = 5 (p − 2) (c) 16 = 4 + 3 (t + 2) (d) 4 + 5 (p − 1) = 34 ( e) 0 = 16 + 4 (m − 6)

Answer: (a) 4 = 5 (p − 2) divide both sides by 5, (b) − 4 = 5 (p − 2) divide both sides by 5, (c) 16 = 4 + 3 (t + 2) 16 − 4 = 3 (t + 2) (transpose 4 to L.H.S.) 12 = 3 (t + 2) divide both sides by 3, 4 = t + 2 4 − 2 = t (transpose 2 to L.H.S.) 2 = t ( d) 4 + 5 (p − 1) = 34 5 (p − 1) = 34 − 4 = 30 (transpose 4 to the right) Divide both sides by 5, p = 6 + 1 = 7 (transpose from − 1 to the right) (e) 0 = 16 + 4 (m − 6) 0 = 16 + 4m − 24 0 = −8 + 4m 4m = 8 (transpose −8 to L.H.S) Divide both sides by 4, m = 2

Page No. 89:

Task 4: (a) Construct 3 equations beginning with x = 2 (b) Construct 3 equations beginning with x = − 2

Answer: (a) x = 2 Multiply both sides by 5, 5x = 10 (i) Subtract 3 from both sides, 5x − 3 = 10 − 3 5 x − 3 = 7 (ii) Divide both sides by 2, ( b ) x = −2 Subtract 2 from both sides, x − 2 = − 2 − 2 x − 2 = −4 (i) Again x = −2 multiply by 6, 6 × x = −2 × 6 6x = − 12 Subtract 12 on both sides, 6x − 12 = − 12 − 12 6x − 12 = −24 (ii) Add 24 on both sides, 6x − 12 + 24 = − 24 + 24 6x + 12 = 0 (iii)

Page No. 91:

Task 1: Set up and solve equations to find the unknown numbers in the following cases: (a) Add 4 to 8 times a number; You get 60. (b) One fifth of a number minus 4 gives 3. (c) If I take three quarters of a number and add 3, I get 21. (d) If I subtract 11 from twice a number, the result was 15 (e) Munna subtracts three times the number of his notebooks from 50, he finds the result 8. (f) Ibenhal thinks of a number. If she adds 19 and divides the sum by 5, she gets 8. (g) Anwar thinks of a number. If he subtracts 7 from the number, the result is 23.

Answer: (a) Let the number be x. 8 times this number = 8x 8x + 4 = 60 8x = 60 − 4 (transpose 4 to the right) 8x = 56 Divide both sides by 8, (b) Let the number be x. One fifth of this number = (Transpose −4 to R.H.S.) Multiply both sides by 5, (c) Let the number be x. Three quarters of this number = (Transpose 3 to R.H.S.) Multiply both sides by 4,

Divide both sides by 3, (d) Let the number be x. Twice this number = 2x 2x − 11 = 15 2x = 15 + 11 (transpose −11 to the right) 2x = 26 Divide both sides by 2, x = 13 (e) Let the number of books be x. Three times the number of books = 3x 50 − 3x = 8 − 3x = 8 −50 (transpose 50 to the right) −3x = −42 Divide both sides by −3, (f) Let the number be x. Multiply both sides by 5, x + 19 = 40 x = 40 − 19 (transpose 19 to R.H.S.) x = 21 (g) Let the number be x. this number = multiply both sides by 2, divide both sides by 5,

Page No. 91:

Question 2: Answer the following: (a) The teacher tells the class that the highest score for a student in her class is twice the lowest score plus 7. The highest score is 87. What is the lowest score? (b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember that the sum of the three angles of a triangle is 180°). (c) Sachin scored twice as many runs as Rahul. Together their runs remained two of a double century. How many runs did each score?

Answer: (a) Let the lowest score be l. 2 × Worst Score + 7 = Highest Score 2l + 7 = 87 2l = 87 − 7 (transposing 7 to R.H.S.) 2l = 80 Divide both sides by 2, hence the lowest score is 40.

(b) Let the base angles be equal b. The sum of all interior angles of a triangle is 180°. b + b + 40° = 180° 2b + 40° = 180° 2b = 180° − 40° = 140° (reversal of 40° according to R.H.S.) Divides both sides by 2, therefore the base angles of the triangle measure 70 °.

(c) Let Rahul’s score be x. Therefore, Sachin’s result = 2x Rahul’s result + Sachin’s result = 200 − 2 2x + x = 198 3x = 198 Divide both sides by 3, x = 66 Rahul’s result = 66 Sachin’s result = 2 × 66 = 132 Video solution to simple equations ( Page:91, Q.No.:2) NCERT Solution for Grade 7 Math – Simple Equations 91, Question 2

Page No. 91:

Question 3: Solve the following: (i) Irfan says he has 7 marbles, more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have? (ii) Laxmi’s father is 49 years old. He is 4 years older than Laxmi three times. How old is Laxmi? (iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was twice the number of fruit trees. If 77 non-fruit trees were planted, how many fruit trees were planted?

Answer: (i) Let Parmit’s marbles be equal to x. 5 times the number of marbles Parmit has = 5x 5x + 7 = 37 5x = 37 − 7 = 30 (transposing 7 to R.H.S.) Divide both sides by 5, so Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years. 3 × Laxmi’s age + 4 = her father’s age 3x + 4 = 49 3x = 49 − 4 (transposing 4 to R.H.S.) 3x = 45 Divide both sides by 3, x = 15 Hence Laxmi’s age is 15 years. (iii) The number of fruit trees is x. 3 × number of fruit trees + 2 = number of non-fruit trees 3x + 2 = 77 3x = 77 − 2 (transpose 2 to the right) 3x = 75 Divide both sides of the equation by 3, x = 25 Hence the number of fruit trees was 25. Video Solution for Simple Equations (Page:91, Q.No.:3) NCERT Solution for Grade 7 Math – Simple Equations 91, Question 3

Page No. 92:

Task 4: Solve the following riddle: I am a number, tell me my identity! Take me seven times and add fifty! To reach a triple century, you still need forty!

Answer: The number is x. (7x + 50) + 40 = 300 7x + 90 = 300 7x = 300 − 90 (transpose 90 to the right) 7x = 210 Divide both sides by 7, x = 30 Hence the number is 30.

Why try to solve puzzles?

People usually have trouble remembering what they ate yesterday or who they met yesterday? These minor short-term memory problems can be fixed simply by solving puzzles. By solving puzzles, our brain cells connect with each other and this improves our thinking speed. Our short-term memory can be improved by solving puzzles and riddles.

What is the interesting thing that has no beginning, end or middle? puzzle

This interesting what has no beginning, end or middle? Riddle is given as follows:

What has no beginning, end or middle?

The answer to What has no beginning, end, or middle? puzzle

The answer to the question “What has no beginning, no end, and no middle”? Riddle is a donut.

Explanation:

A donut is round in shape and has no beginning, end, or middle.

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This riddle has been shared with the claim that only geniuses can solve it.

11 × 11 = 4

22 × 22 = 16

33 × 33 = ?

It has gone viral on Facebook and across the web with millions of views as people debate the correct answer.

There are arguably many answers as there are many patterns that fit the information given. However, there are two main answers that are the most popular. I’ll address what many people think is the correct answer, and I’ll explain how the two main approaches are flavors of the same idea.

Watch the video for an explanation.

Can you solve the viral 11×11 = 4 puzzle? Correct answer explained

Or keep reading.

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“Everything will be fine if you use your reason for your decisions and only care about your decisions.” Since 2007 I have dedicated my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help us and get early access to posts with a promise on Patreon. .

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Answer to the viral puzzle 11×11 = 4

Most people believe the correct answer is 36. The method to get the answer is to take the product of the sum of the digits in each number to be multiplied.

That is:

aa × aa → (a + a)(a + a)

This procedure corresponds to the pattern of the puzzle:

11 × 11 → (1 + 1)(1 + 1) = 4

22 × 22 → (2 + 2)(2 + 2) = 16

And it suggests the answer of 36.

33 × 33 → (3 + 3)(3 + 3) = 36

Many people consider this “product of the checksum” to be the correct answer. But there are debates.

Alternative answer: 18

Other people thought about the puzzle by doing the multiplication and then taking the sum of the digits in the answer. In other words, this is finding the “check sum of the product”.

11 × 11 = 121 → 1 + 2 + 1 = 4

22 × 22 = 484 → 4 + 8 + 4 = 16

This procedure suggests an answer of 18.

33 × 33 = 1089 → 1 + 0 + 8 + 9 = 18

The “check sum of the product” is 18, while the “product of the check sum” is 36.

It seems like these two methods are completely different. However, there is a way to tell that they are flavors of the same concept. And that’s how it’s possible to get an answer of 36 in the “check sum of the product”.

36 obtained from the sum of the product

Let’s go into the details of how to calculate the product of two numbers and how to sum the digits in the answer.

The number 11 can be written as 10 + 1, so we have:

11×11

= (10 + 1)(10 + 1)

= 1(100) + 2(10) + 1(1)

= 121

The digits in the answer are the coefficients of the sums of powers of ten, which is how decimals are written. The sum of the digits in the answer is 1 + 2 + 1 = 4.

Likewise, the number 22 can be written as 20 + 2, so we have:

22×22

= (20 + 2)(20 + 2)

= 4(100) + 8(10) + 4(1)

= 484

Here, too, the checksum is the sum of the coefficients of the terms attached to powers of ten. The sum is 4 + 8 + 4 = 16.

Then what happens at 33? The number 33 can be written as 30 + 3, so we have:

33×33

= (30 + 3)(30 + 3)

= 9(100) + 18(10) + 9(1)

What happens when you add the terms of the powers of ten? You get 9 + 18 + 9 = 36. You get the answer of 36 from this procedure!

But shouldn’t the answer be 18 with this method? Yes, this is because 18(10) is greater than 100, so it’s carryover. We can simplify the answer as:

9(100) + 18(10) + 9(1)

= 9(100) + 10(10) + 8(10) + 9(1)

Now 10(10) = 100, so 1 more term contributes to the 100 value.

9(100) + 10(10) + 8(10) + 9(1)

= 10(100) + 8(10) + 9(1)

Now 10(100) equals 1000, so we have carry again.

10(100) + 8(10) + 9(1)

= 1(1000) + 0(100) + 8(10) + 9(1)

= 1089

This gives the well-known answer of 1089, which is what a calculator would show for 33 × 33.

But we can see that 9(100) + 18(10) + 9(1) is a valid representation of the product and the sum would be 36 if we didn’t go through the carry process.

So we found a link between the two methods.

aa × aa → (a + a)(a + a) = product of checksum = sum of the product (without carry)

It is possible to justify 36’s answer using either method.

Other ways to 36

In the video, I show the same thing visually using graphs from the Multiply by Lines method. The answer is the number of times the lines intersect, or “points” in the figure, and 33 × 33 has 36 points.

Via MindYourDecisions YouTube

The key to 36’s answer is the multiplicative nature of the procedure. Start with 11 × 11 = 4 as given. The second line has two terms that are 2 times 11, so the answer should be 2(2) = 4 times this. The third row has two terms that are 3 times 4, so the answer should be 3(3) = 9 times this.

11 × 11 = 4

22 × 22 = (2 × 11)(2 × 11) = 4(11 × 11) = 4(4) = 16

33 × 33 = (3 × 11) (3 × 11) = 9 (11 × 11) = 9 (4) = 36

There is another way to illustrate the multiplicative property and avoid carry: express the answer in terms of a specific module, e.g. B. 39.

11 × 11 = 4mod 39

22 × 22 = 16 mod 39

33 × 33 = 36 mod 39

All of these methods suggest that 36 is the correct answer to the puzzle.

Quora Discussion

https://www.quora.com/Puzzles-and-Trick-Questions-What-is-the-correct-answer-to-11×11-4-22×22-16-33×33

Finite sequence of symbols from a given alphabet that is part of a formal language

In mathematical logic, propositional logic, and first order logic, a well-formed formula, abbreviated WFF or wff, often simply formula, is a finite sequence of symbols from a given alphabet that is part of a formal language.[1] A formal language can be identified with the set of formulas in the language.

A formula is a syntactic object that can be given a semantic meaning through interpretation. Two main uses of formulas are propositional logic and first order logic.

Introduction [edit]

A key application of formulas is in propositional logic and predicate logic such as first-order logic. In these contexts, a formula is a set of symbols φ for which it makes sense to ask “Is φ true?” once all the free variables in φ have been instantiated. In formal logic, proofs can be represented by sequences of formulas with certain properties, and the last formula in the sequence is what is proved.

Although the term “formula” can be used for written marks (e.g. on a piece of paper or blackboard), it is more precisely understood as the sequence of symbols being expressed, the marks being a symbolic instance of a formula. Therefore, the same formula can be written more than once, and a formula can in principle be so long that it cannot be written at all within the physical universe.

Formulas themselves are syntactic objects. You get meanings through interpretations. For example, in a propositional formula, each propositional variable can be interpreted as a concrete statement, so that the overall formula expresses a relationship between these statements. However, a formula does not have to be interpreted as just a formula.

Propositional calculus[edit]

The formulas of the propositional calculus, also called propositional formulas,[2] are expressions like ( A ∧ ( B ∨ C ) ) {\displaystyle (A\land (B\lor C))} . Its definition begins with the arbitrary choice of a set V of propositional variables. The alphabet consists of the letters in V together with the symbols for the propositional connectors and brackets “(” and “)”, all of which are assumed not to be in V. The formulas are definite expressions (i.e. character strings) over this alphabet.

The formulas are defined inductively as follows:

Each propositional variable is a formula in itself.

If φ is a formula, then ¬φ is a formula.

If φ and ψ are formulas and • is any binary operation, then ( φ • ψ) is a formula. Here • could be (but is not limited to) the usual operators ∨, ∧, → or ↔.

This definition can also be written as a formal grammar in Backus-Naur form, provided the set of variables is finite:

< alpha set > ::= p | q | r | s | t | you | … (the arbitrary finite set of set variables) < form > ::= < alpha set > | ¬

| ( < form > ∧ < form > ) | ( < form > ∨ < form > ) | ( < form > → < form > ) | ( < form > ↔ < form > )

With this grammar the sequence of symbols

(((p → q) ∧ (r → s)) ∨ (¬q ∧ ¬s))

is a formula because it is grammatically correct. The sequence of symbols

((p → q)→(qq))p))

is not a formula because it does not correspond to the grammar.

A complex formula can be difficult to read, for example because of the proliferation of parentheses. To mitigate this latter phenomenon, rules of precedence (similar to the standard mathematical order of operations) are adopted among the operators, making some operators more binding than others. For example, let’s take the precedence (from highest tie to lowest tie) 1. ¬ 2. → 3. ∧ 4. ∨. Then the formula

(((p → q) ∧ (r → s)) ∨ (¬q ∧ ¬s))

can be abbreviated as

p → q ∧ r → s ∨ ¬q ∧ ¬s

However, this is just a convention used to simplify the written representation of a formula. For example, if precedence is assumed to be left-right associative, in the following order: 1. ¬ 2. ∧ 3. ∨ 4. →, then the same formula above (without parentheses) would be rewritten as

(p → (q ∧ r)) → (s ∨ ((¬q) ∧ (¬s)))

First order logic [ edit ]

The definition of a formula in first order logic is relative to the signature of the present theory. This signature specifies the constant symbols, predicate symbols, and function symbols of the present theory, along with the locations of the function and predicate symbols.

The definition of a formula consists of several parts. First, the set of terms is defined recursively. Terms are informal expressions that represent objects from the realm of discourse.

Each variable is a concept. Each constant symbol from the signature is a term an expression of the form f(t 1 ,…,t n ), where f is an n-ary function symbol and t 1 ,…,t n are terms, is in turn a term.

The next step is to define the atomic formulas.

If t 1 and t 2 are terms, then t 1 = t 2 is an atomic formula. If R is an n-ary predicate symbol and t 1 ,…,t n are terms, then R(t 1 ,…,t n ) is an atomic formula

Finally, the set of formulas is defined as the smallest set that contains the set of atomic formulas, so:

¬ ϕ {\displaystyle

eg \phi } ϕ {\displaystyle \phi } ( ϕ ∧ ψ ) {\displaystyle (\phi \land \psi )} ( ϕ ∨ ψ ) {\displaystyle (\phi \lor \psi )} ϕ {\displaystyle \ phi } ψ {\displaystyle \psi } ∃ x ϕ {\displaystyle \exists x\,\phi } x {\displaystyle x} ϕ {\displaystyle \phi } ∀ x ϕ {\displaystyle \forall x\,\phi } x {\displaystyle x} ϕ {\displaystyle \phi } ∀ x ϕ {\displaystyle \for all x\,\phi } ¬ ∃ x ¬ ϕ {\displaystyle

eg \exists x\,

eg \phi }

If a formula has no occurrence of ∃ x {\displaystyle \exists x} or ∀ x {\displaystyle \forall x} for any variable x {\displaystyle x}, it is said to be quantifier-free. An existential formula is a formula that begins with a sequence of existential quantification, followed by a quantifier-free formula.

Atomic and open formulas[ edit ]

An atomic formula is a formula that contains no logical operators or quantifiers, or equivalently a formula that has no strict subformulae. The exact form of atomic formulas depends on the formal system under consideration; for propositional logic, for example, the atomic formulas are the propositional variables. For first-order logic, the atoms along with their arguments are predicate symbols, with each argument being a term.

According to some terminology, an open formula is formed by combining atomic formulas only with logical connectors, excluding quantifiers.[3] This is not to be confused with a formula that is not complete.

Closed formulas [ edit ]

A closed formula, also known as a basic formula or theorem, is a formula in which there are no free occurrences of any variable. If A is a formula of a first-order language in which the variables v 1 , …, v n have free occurrences, then A preceded by ∀v 1 ⋯ ∀v n is a closure of A.

Properties applicable to formulas[ edit ]

A formula A in a language Q {\displaystyle {\mathcal {Q}}} is valid if it holds for every interpretation of Q {\displaystyle {\mathcal {Q}}} .

in a language if it holds for every interpretation of A Formula A in a language Q {\displaystyle {\mathcal {Q}}} satisfiable if it holds for some interpretations of Q {\displaystyle {\mathcal {Q}}}

in a language if it holds for any interpretation of A Formula A of the language of arithmetic is decidable if it represents a decidable set, i.e. if there is an effective method which, given a substitution of the free variables of A, says both the resulting instance of A is provable or its negation is provable.

Use of terminology[edit]

In earlier works on mathematical logic (e.g. by Church[4]), formulas referred to arbitrary strings of symbols, and among these strings, well-formed formulas were the strings that followed the formation rules of (correct) formulas.

Several authors simply say formula.[5][6][7][8] Modern usages (particularly in the context of computer science with mathematical software such as model checkers, automated theorem provers, interactive theorem provers) tend to retain only the algebraic concept of the formula term and raise the question of well-formedness, i.e. the concrete string representation of formulas (with this or that symbol for connectors and quantifiers, with this or that bracket convention, with Polish or infix notation, etc.) as a pure notation problem.

While the expression well-formed formula is still used, [9][10][11] these authors do not necessarily use it [weasel words] in contrast to the old formula sense, which is no longer common in mathematical logic. [citation required]

The expression “Well-Formed Formulas” (WFF) has also crept into popular culture. WFF is part of an esoteric pun used in the name of the academic game WFF ‘N PROOF: The Game of Modern Logic by Layman Allen[12] which was developed while he was at Yale Law School (he was later Professor at Yale Law School). the University of Michigan). The game series was designed to teach children the principles of symbolic logic (in Polish notation).[13] His name is an echo of whiffenpoof, a nonsense word used as a cheer at Yale University and made popular in The Whiffenpoof Song and The Whiffenpoofs.

See also[edit]

Notes [edit]

type of puzzle

Mathematical puzzles are a staple of recreational math. They have specific rules but usually do not involve competition between two or more players. Instead, to solve such a puzzle, the solver must find a solution that satisfies the given conditions. Mathematical puzzles require mathematics to solve them. Logic puzzles are a common type of math puzzle.

The car has to travel a path given by a mathematical function.

Conway’s Game of Life and fractals, to name two examples, can also be viewed as mathematical puzzles, although the solver only interacts with them initially by providing a set of initial conditions. After these conditions are established, the rules of the puzzle dictate all subsequent changes and moves. Many of the puzzles are well known, having been discussed by Martin Gardner in his Mathematical Games column in Scientific American. Mathematical puzzles are sometimes used to motivate students when teaching math problem-solving techniques in elementary school.[1] Creative thinking (thinking outside the box) often helps to find the solution.

List of math puzzles[edit]

This list is incomplete.

Numbers, arithmetic, and algebra[ edit ]

Combinatorial[ edit ]

Analytical or differential[ edit ]

Probability [edit]

Tiling, packing and dissecting[ edit ]

Includes a board[ edit ]

Checkerboard problems[edit]

Topology, nodes, graph theory [ edit ]

The fields of knot theory and topology, particularly their non-intuitive conclusions, are often viewed as part of recreational mathematics.

Mechanical [ edit ]

0-player puzzles [ edit ]

$\begingroup$

$\wedge$ is exactly ‘and’ in this context. $\vee$ means ‘or’. You can see the similarity in both form and meaning to $\cap$ and $\cup$ from set theory.

In differential geometry, $\omega_1\wedge\omega_2$ also means the wedge product of two differential forms.

Divide each side by 4 to get #x=6#.

As used in physics, the term “exact” generally refers to a solution that captures all of the physics and mathematics of a problem, as opposed to a solution that is approximate, buggy, etc. Therefore, exact solutions do not have to have a closed form.

video transcript

We’re asked to simplify 8 plus 5 times 4 minus, and then in parentheses 6 plus 10 divided by 2 plus 44. Whenever you see a crazy expression like this where you have parentheses and addition and subtraction and division, you always die want to keep track of the order of operations. Let me write them down over here. So when performing the order of operations, or really when evaluating an expression, keep in mind that parentheses have the highest precedence. And those are those little brackets over here, or whatever you want to call them. Those are the brackets right there. That gets top priority. After that, you want to take care of exponents. There are no exponents in this expression, but I’m just writing it down for future reference: Exponents. One way I like to think about it is that parentheses are always top priority, but after that we go in descending order, or I think we should say, in– well, yeah, descending order, how fast this calculation is. If I say quickly how fast it grows. If I take something to an exponent, if I take something to a power, it grows really fast. Then it grows a little slower or shrinks a little slower when I multiply or divide, so next comes: multiply or divide. Next comes multiplication and division, and last comes addition and subtraction. So these are the slowest operations. That’s a little faster. This is the fastest process. And then, no matter what, the brackets always take precedence. So let’s apply it here. Let me rewrite that whole expression. So it’s 8 plus 5 times 4 minus, in parentheses, 6 plus 10 divided by 2 plus 44. So let’s do the parentheses first. We have brackets here and there. Well, these brackets are pretty simple. Well, inside the parentheses is already evaluated, so we could really just think of this as 5 times 4. So let’s evaluate that right from the start. So that’s going to be 8 plus — and really, if you evaluate the parentheses, if you evaluate those parentheses, you literally only get 5, and when you evaluate those parentheses, you literally only get 4, and then they’re next to each other, so you multiply them So 5 times 4 is 20 minus – let me be consistent with the colors. Now let me write the next bracket right there, and then we would evaluate those first. Let me close the bracket right there. And then we have plus 44. So what is this thing, this thing in the brackets? Well, you might be tempted to say, well, just let me go left to right. 6 plus 10 is 16 and then divided by 2 and you would get 8. But remember: order of operations. Division takes precedence over addition, so you actually want to do division first, and we could actually write it that way here. You could imagine putting a few more brackets. let me do it in the same purple You could imagine putting more parentheses here to emphasize the fact that you will be dividing first. So 10 divided by 2 is 5, so that’s 6, plus 10 divided by 2, is 5. 6 plus 5. Well, we still have to evaluate those brackets, so that’s – what’s 6 plus 5? Well, that’s 11. So that leaves us with 20 – let me rewrite it all. That leaves 8 plus 20 minus 6 plus 5, which is 11 plus 44. And now that we have everything at this level of operation, we can just go left to right. So 8 plus 20 is 28, so you can look at that as 28 minus 11 plus 44. 28 minus 11 – 28 minus 10 would be 18, so that will be 17. It will be 17 plus 44. And then 17 plus 44– I’ll scroll down a little. 7 plus 44 would be 51, so this will be 61. So this will equal 61. And we’re done!