# Book of Math Solution of all pages / level Today Touchportal Games released their next free Android App: Book of Math. Now we want to present all solutions for this app.

At the beginning Book of Math contains 30 Levels. Different math puzzles like arithmetic problems, story problems, find the next series numbers, geometry and more funny pages are in this app. Our favorite is the cube puzzle.

## Solution to the math book – all pages / Level

Below you will find an overview of all solutions to the book of math.

• Page 1: 15
• Page 2: 146
• Page 3: 10
• Page 4: 10 (older version: 12)
• Page 5 Solution: 12
• Page 6: 24
• Page 7: 450
• Page 8: 2436
• Page 9 Solution: 300
• Page 10: 429
• Page 11: 180
• Page 12: 4044
• Page 13 Solution: 217972
• Page 14: 9
• Page 15 Solution: 28
• Page 16: 240
• Page 17: 10
• Page 18: 41 (older version: Kevin is 18 years old)
• Page 19: 40
• Page 20: 1
• Page 21: 4100
• Page 22: 59
• Page 23 Solution: 1987
• Page 24: 34 (older version: 32)
• Page 25 Solution: 812520
• Page 26: 4
• Page 27: 130
• Page 28 Solution: 127
• Page 29: 55
• Page 30 Solution: 63
• Page 31: 6
• Page 32: 28
• Page 33: 68
• Page 34: 100
• Page 35: 1882
• Page 36: 5
• Page 37: 5050
• Page 38: 2717
• Page 39: 19180
• Page 40: 1013
• Page 41: 210
• Page 42: 46130
• Page 43: 51170
• Page 44: 72
• Page 45: 235

### Solution of page 1-10

Let us begin with the solution of the page to 1 to 10 of the book of math. If you do not know how to get to this solution, so please get in touch in the comments.

• Page 1: 15
Simply sum 12 and 3 up. There should be no problems.
• Page 2: 146
The number sequence will be continued to resolve. This one always adds the two previous numbers and come to the third number is 56 + 90 = 146
• Page 3: 10
Just flip the Android device. Thus we see 600/60 and the resulting 10
• Page 4: 12
A little trick question, which should have been known. All months have 28 days. Thus, the solution is 12
• Page 5 solution: 12
Here, the triangles are counted. There are 8 small triangles and 4 triangles that result from the combination of three small triangles. The solution is 12
• Page 6: 24
Point bill goes before dashes. Accordingly, we first multiply 2 * 3=6. Now we sum 18 and 6 up. Solution: 24
• Page 7: 450
A small property task. 3 percent of 15,000 is 450. First, divide 15,000 by 100. Now multiply 150 with 3. solution: 450
• Page 8: 2436
On page 8 for the solution of the book Mathematics of numbers must both be determined. In the first sequence of numbers, the distance is increased by 1 each So first of +2, then +3, …. We now add 19+5 = 24 This is our first Number, the second number is obtained from the second series of numbers. This is backward to the squares. So 10 ², 9 ² …. The second number is 36. We do this together now: 2436
• Page 9 solution: 300
To solve the height of the tree must be stated in centimeters. In a right triangle is: a ² + b ² = c ². So we have to determine b. So 5 ² -4 ² = 9 This still extract the root and we get third This is in meters. We should type in in centimeter: 300
• Page 10: 429
On page 10, we have to solve to determine the variables a and b. is to identify a simple, a = 4 Now we need to determine b, this is 29, because 30 = b +1. So b = 29 We give the solution a two variables, ie, 429

### Solution of pages 11 to 20

Here we have the solution to the pages 11 to 20 of the book of mathematics for you:

• Page 11: 180
Also a simple task. Tina has a velocity v = 4 km / h For 12 miles they need so 3 hours. We should all specify, in minutes. So 3 hours * 60 = 180 minutes.
• Page 12: 4044
On page 12, we have to determine a and b. So we get for a = 10 * 4 = 40 b result of 46-2 = 44
• Page 13 Solution: 217972
Here we need to solve dislocate something. The Roman numerals us specify the order in which the numbers have to be surrounded. So 63/3 = 21, 16-9 = 7, 9 +9 = 9 * 0 and 12 * 6 = 72 We give the numbers in that order, so 217972nd
• Page 14: 9
The snail is climbing a 10 meter high tree. Every day she creates 2 meters, at night it goes down 1 meter. So we start: Day 1 – 2 feet high, down 1 meter (1 meter), Day 2 – 2 meters …. In the early 9th Day she is so at 8 meters (since Day 8 – 8 meters). This creates on 9 Day 2 meters and has climbed the 10-meter high tree.
• Page 15 Solution: 28
Here you have to count the cubes. This one should be guided primarily by the right. The block front right may otherwise be confused about. Ultimately, 28 blocks are installed.
• Page 16: 240
The number sequence will continue. We have the following scheme: * 3, – 3, +3. So we now proceed. In the 243 we have to subtract 3, so the solution is 240th
• Page 17: 10
Once again, we must again remember that point bill goes before dashes. So 5 * 0 = 0, we add up to 1 +2 +3 +4 = 10
• Page 18: 41 (Older version: Kevin is 18 years old)
• Page 19: 40
At level 19 you have to count squares. First small and then the respectively larger. The resulting 40th
• Page 20: 1
To solve this is only 1, 2 or 3 Now you can either enter a number or just thinking about it. The route is 200 kilometers and 200 kilometers for the outward journey for the return trip. Now you just need to calculate only the speed, how long each individual needs. In the end, Marc is the fastest.

• Marc takes the Inbound 2 hours (200 km – 100 km / h). The same for the return trip. Ie, 4 hours total
• Paul initially runs only 80km / h Need for the outward journey or 2.5 hours. For the return trip, he only needs 1.66 hours. Total So just over four hours, slower than Marc.
• Laura goes first only 50 km / h Alone for the outward journey they need it 4 hours. Now they would have to go with infinite speed in order to arrive at the same time still with Marc. Accordingly, they can also reach 250 km / h and does not create it.

### Book of Mathematics page 21-30

Here we have the solution of 21 to 30 for the book of mathematics for you:

• Page 21: 4100
You just have to add up the numbers. So 1000 +1000 +1000 +1000 = 4000. Now add the still small numbers, that is 40 +30 +10 +20 = 100 So in the 4100th Who should get out here 5000, the math again • Page 22: 59
Again have a series of numbers to continue. It notes that this is some primes in the sequence. The next prime is 59 (Prime numbers are only divisible by 1 and itself)
• Page 23 Solution: 1987
Wanted will be the year when the ships returned to harbor shrink. It is the fourth January. That means we are in the first, the second maximum Calendar week. Each year, more than 51 calendar weeks. The time is irrelevant. Now we must make the least common multiple of 16, 12, 4, and 8. As 8 in the 16 and 4 lies in the 12 (or 8 or 16), we will make the LCM of 12 and 16, ie, 48th In 48 weeks, the ships take so together again in the harbor. So this is still the year 1987. Accordingly, this is also the solution to page 23 of the book of mathematics.
• Page 24: 32
In level 24 of the book of mathematics again blocks must be counted. In this context the different shading. We come finally to 32 blocks.
• Page 25 Solution: 812520
Really complicated, it is on page 25 of the book of mathematics. Here, the variables a, b, c, d are determined. Doing so, the 4 are divided into 4 parts, so that the equation is satisfied. The solution is as follows a = 8, b = 12, c = 5 and d = 20, so we enter 812520:

• 8 +2 = 10
• 12-2 = 10
• 5 × 2 = 10
• 20:2 = 10
• Page 26: 4
On page 26, the equation can be solved to obtain the zero (Equation supposed 0 yield). Here we shall first by 2 and get a ²-8a +16. This is the second Binomial formula. So this means: (a 4) ². We now have a zero at 4, which is also the solution.
• Page 27: 130
Again, the number sequence will continue. This is the power of two, is added with 2 Ie 2 +2 ^ 0 = 3, 2 ^ 2 +2 = 4, 2 ^ 3 +2 = 10, and so on. To get the required number, we expect 2 ^ 7 +2 and the 130th is
• Page 28 Solution: 127
Sought is the number of sheep. We expect just 2 ^ 7-1. Reason: De Shepherd sold the entire 6-times (half and half). So we get the following (at the end of a sheep remains loud task):

• First 127/2-0, 5 = 63
• Second 63/2-0, 5 = 31
• Third 31/2-0, 5 = 15
• 4th 15/2-0, 5 = 7
• 5th 7/2-0, 5 = 3
• 6th 3/2-0, 5 = 1
• Page 29: 55 (older version: solution is 41)
• Page 30 Solution: 63
And finally again blocks count. Orient yourself right again. Ultimately, you get to 63 blocks.

We will add this solution of new pages arrived.Today Touchportal Games released their next free Android App: Book of Math. Now we want to present all solutions for this app.

At the beginning Book of Math contains 30 Levels. Different math puzzles like arithmetic problems, story problems, find the next series numbers, geometry and more funny pages are in this app. Our favorite is the cube puzzle.

## Solution to the math book – all pages / Level

Below you will find an overview of all solutions to the book of math.

• Page 1: 15
• Page 2: 146
• Page 3: 10
• Page 4: 12
• Page 5 Solution: 12
• Page 6: 24
• Page 7: 450
• Page 8: 2436
• Page 9 Solution: 300
• Page 10: 429
• Page 11: 180
• Page 12: 4044
• Page 13 Solution: 217972
• Page 14: 9
• Page 15 Solution: 28
• Page 16: 240
• Page 17: 10
• Page 18: 41 (older version: Kevin is 18 years old)
• Page 19: 40
• Page 20: 1
• Page 21: 4100
• Page 22: 59
• Page 23 Solution: 1987
• Page 24: 32
• Page 25 Solution: 812520
• Page 26: 4
• Page 27: 130
• Page 28 Solution: 127
• Page 29: 55
• Page 30 Solution: 63

### Solution of page 1-10

Let us begin with the solution of the page to 1 to 10 of the book of math. If you do not know how to get to this solution, so please get in touch in the comments.

• Page 1: 15
Simply sum 12 and 3 up. There should be no problems.
• Page 2: 146
The number sequence will be continued to resolve. This one always adds the two previous numbers and come to the third number is 56 + 90 = 146
• Page 3: 10
Just flip the Android device. Thus we see 600/60 and the resulting 10
• Page 4: 12
A little trick question, which should have been known. All months have 28 days. Thus, the solution is 12
• Page 5 solution: 12
Here, the triangles are counted. There are 8 small triangles and 4 triangles that result from the combination of three small triangles. The solution is 12
• Page 6: 24
Point bill goes before dashes. Accordingly, we first multiply 2 * 3=6. Now we sum 18 and 6 up. Solution: 24
• Page 7: 450
A small property task. 3 percent of 15,000 is 450. First, divide 15,000 by 100. Now multiply 150 with 3. solution: 450
• Page 8: 2436
On page 8 for the solution of the book Mathematics of numbers must both be determined. In the first sequence of numbers, the distance is increased by 1 each So first of +2, then +3, …. We now add 19+5 = 24 This is our first Number, the second number is obtained from the second series of numbers. This is backward to the squares. So 10 ², 9 ² …. The second number is 36. We do this together now: 2436
• Page 9 solution: 300
To solve the height of the tree must be stated in centimeters. In a right triangle is: a ² + b ² = c ². So we have to determine b. So 5 ² -4 ² = 9 This still extract the root and we get third This is in meters. We should type in in centimeter: 300
• Page 10: 429
On page 10, we have to solve to determine the variables a and b. is to identify a simple, a = 4 Now we need to determine b, this is 29, because 30 = b +1. So b = 29 We give the solution a two variables, ie, 429

### Solution of pages 11 to 20

Here we have the solution to the pages 11 to 20 of the book of mathematics for you:

• Page 11: 180
Also a simple task. Tina has a velocity v = 4 km / h For 12 miles they need so 3 hours. We should all specify, in minutes. So 3 hours * 60 = 180 minutes.
• Page 12: 4044
On page 12, we have to determine a and b. So we get for a = 10 * 4 = 40 b result of 46-2 = 44
• Page 13 Solution: 217972
Here we need to solve dislocate something. The Roman numerals us specify the order in which the numbers have to be surrounded. So 63/3 = 21, 16-9 = 7, 9 +9 = 9 * 0 and 12 * 6 = 72 We give the numbers in that order, so 217972nd
• Page 14: 9
The snail is climbing a 10 meter high tree. Every day she creates 2 meters, at night it goes down 1 meter. So we start: Day 1 – 2 feet high, down 1 meter (1 meter), Day 2 – 2 meters …. In the early 9th Day she is so at 8 meters (since Day 8 – 8 meters). This creates on 9 Day 2 meters and has climbed the 10-meter high tree.
• Page 15 Solution: 28
Here you have to count the cubes. This one should be guided primarily by the right. The block front right may otherwise be confused about. Ultimately, 28 blocks are installed.
• Page 16: 240
The number sequence will continue. We have the following scheme: * 3, – 3, +3. So we now proceed. In the 243 we have to subtract 3, so the solution is 240th
• Page 17: 10
Once again, we must again remember that point bill goes before dashes. So 5 * 0 = 0, we add up to 1 +2 +3 +4 = 10
• Page 18: 41 (Older version: Kevin is 18 years old)
• Page 19: 40
At level 19 you have to count squares. First small and then the respectively larger. The resulting 40th
• Page 20: 1
To solve this is only 1, 2 or 3 Now you can either enter a number or just thinking about it. The route is 200 kilometers and 200 kilometers for the outward journey for the return trip. Now you just need to calculate only the speed, how long each individual needs. In the end, Marc is the fastest.

• Marc takes the Inbound 2 hours (200 km – 100 km / h). The same for the return trip. Ie, 4 hours total
• Paul initially runs only 80km / h Need for the outward journey or 2.5 hours. For the return trip, he only needs 1.66 hours. Total So just over four hours, slower than Marc.
• Laura goes first only 50 km / h Alone for the outward journey they need it 4 hours. Now they would have to go with infinite speed in order to arrive at the same time still with Marc. Accordingly, they can also reach 250 km / h and does not create it.

### Book of Mathematics page 21-30

Here we have the solution of 21 to 30 for the book of mathematics for you:

• Page 21: 4100
You just have to add up the numbers. So 1000 +1000 +1000 +1000 = 4000. Now add the still small numbers, that is 40 +30 +10 +20 = 100 So in the 4100th Who should get out here 5000, the math again • Page 22: 59
Again have a series of numbers to continue. It notes that this is some primes in the sequence. The next prime is 59 (Prime numbers are only divisible by 1 and itself)
• Page 23 Solution: 1987
Wanted will be the year when the ships returned to harbor shrink. It is the fourth January. That means we are in the first, the second maximum Calendar week. Each year, more than 51 calendar weeks. The time is irrelevant. Now we must make the least common multiple of 16, 12, 4, and 8. As 8 in the 16 and 4 lies in the 12 (or 8 or 16), we will make the LCM of 12 and 16, ie, 48th In 48 weeks, the ships take so together again in the harbor. So this is still the year 1987. Accordingly, this is also the solution to page 23 of the book of mathematics.
• Page 24: 32
In level 24 of the book of mathematics again blocks must be counted. In this context the different shading. We come finally to 32 blocks.
• Page 25 Solution: 812520
Really complicated, it is on page 25 of the book of mathematics. Here, the variables a, b, c, d are determined. Doing so, the 4 are divided into 4 parts, so that the equation is satisfied. The solution is as follows a = 8, b = 12, c = 5 and d = 20, so we enter 812520:

• 8 +2 = 10
• 12-2 = 10
• 5 × 2 = 10
• 20:2 = 10
• Page 26: 4
On page 26, the equation can be solved to obtain the zero (Equation supposed 0 yield). Here we shall first by 2 and get a ²-8a +16. This is the second Binomial formula. So this means: (a 4) ². We now have a zero at 4, which is also the solution.
• Page 27: 130
Again, the number sequence will continue. This is the power of two, is added with 2 Ie 2 +2 ^ 0 = 3, 2 ^ 2 +2 = 4, 2 ^ 3 +2 = 10, and so on. To get the required number, we expect 2 ^ 7 +2 and the 130th is
• Page 28 Solution: 127
Sought is the number of sheep. We expect just 2 ^ 7-1. Reason: De Shepherd sold the entire 6-times (half and half). So we get the following (at the end of a sheep remains loud task):

• First 127/2-0, 5 = 63
• Second 63/2-0, 5 = 31
• Third 31/2-0, 5 = 15
• 4th 15/2-0, 5 = 7
• 5th 7/2-0, 5 = 3
• 6th 3/2-0, 5 = 1
• Page 29: 55 (older version: solution is 41)
• Page 30 Solution: 63
And finally again blocks count. Orient yourself right again. Ultimately, you get to 63 blocks.

We will add this solution of new pages arrived.Today Touchportal Games released their next free Android App: Book of Math. Now we want to present all solutions for this app.

At the beginning Book of Math contains 30 Levels. Different math puzzles like arithmetic problems, story problems, find the next series numbers, geometry and more funny pages are in this app. Our favorite is the cube puzzle.

## Solution to the math book – all pages / Level

Below you will find an overview of all solutions to the book of math.

• Page 1: 15
• Page 2: 146
• Page 3: 10
• Page 4: 12
• Page 5 Solution: 12
• Page 6: 24
• Page 7: 450
• Page 8: 2436
• Page 9 Solution: 300
• Page 10: 429
• Page 11: 180
• Page 12: 4044
• Page 13 Solution: 217972
• Page 14: 9
• Page 15 Solution: 28
• Page 16: 240
• Page 17: 10
• Page 18: 41 (older version: Kevin is 18 years old)
• Page 19: 40
• Page 20: 1
• Page 21: 4100
• Page 22: 59
• Page 23 Solution: 1987
• Page 24: 32
• Page 25 Solution: 812520
• Page 26: 4
• Page 27: 130
• Page 28 Solution: 127
• Page 29: 55
• Page 30 Solution: 63

### Solution of page 1-10

Let us begin with the solution of the page to 1 to 10 of the book of math. If you do not know how to get to this solution, so please get in touch in the comments.

• Page 1: 15
Simply sum 12 and 3 up. There should be no problems.
• Page 2: 146
The number sequence will be continued to resolve. This one always adds the two previous numbers and come to the third number is 56 + 90 = 146
• Page 3: 10
Just flip the Android device. Thus we see 600/60 and the resulting 10
• Page 4: 12
A little trick question, which should have been known. All months have 28 days. Thus, the solution is 12
• Page 5 solution: 12
Here, the triangles are counted. There are 8 small triangles and 4 triangles that result from the combination of three small triangles. The solution is 12
• Page 6: 24
Point bill goes before dashes. Accordingly, we first multiply 2 * 3=6. Now we sum 18 and 6 up. Solution: 24
• Page 7: 450
A small property task. 3 percent of 15,000 is 450. First, divide 15,000 by 100. Now multiply 150 with 3. solution: 450
• Page 8: 2436
On page 8 for the solution of the book Mathematics of numbers must both be determined. In the first sequence of numbers, the distance is increased by 1 each So first of +2, then +3, …. We now add 19+5 = 24 This is our first Number, the second number is obtained from the second series of numbers. This is backward to the squares. So 10 ², 9 ² …. The second number is 36. We do this together now: 2436
• Page 9 solution: 300
To solve the height of the tree must be stated in centimeters. In a right triangle is: a ² + b ² = c ². So we have to determine b. So 5 ² -4 ² = 9 This still extract the root and we get third This is in meters. We should type in in centimeter: 300
• Page 10: 429
On page 10, we have to solve to determine the variables a and b. is to identify a simple, a = 4 Now we need to determine b, this is 29, because 30 = b +1. So b = 29 We give the solution a two variables, ie, 429

### Solution of pages 11 to 20

Here we have the solution to the pages 11 to 20 of the book of mathematics for you:

• Page 11: 180
Also a simple task. Tina has a velocity v = 4 km / h For 12 miles they need so 3 hours. We should all specify, in minutes. So 3 hours * 60 = 180 minutes.
• Page 12: 4044
On page 12, we have to determine a and b. So we get for a = 10 * 4 = 40 b result of 46-2 = 44
• Page 13 Solution: 217972
Here we need to solve dislocate something. The Roman numerals us specify the order in which the numbers have to be surrounded. So 63/3 = 21, 16-9 = 7, 9 +9 = 9 * 0 and 12 * 6 = 72 We give the numbers in that order, so 217972nd
• Page 14: 9
The snail is climbing a 10 meter high tree. Every day she creates 2 meters, at night it goes down 1 meter. So we start: Day 1 – 2 feet high, down 1 meter (1 meter), Day 2 – 2 meters …. In the early 9th Day she is so at 8 meters (since Day 8 – 8 meters). This creates on 9 Day 2 meters and has climbed the 10-meter high tree.
• Page 15 Solution: 28
Here you have to count the cubes. This one should be guided primarily by the right. The block front right may otherwise be confused about. Ultimately, 28 blocks are installed.
• Page 16: 240
The number sequence will continue. We have the following scheme: * 3, – 3, +3. So we now proceed. In the 243 we have to subtract 3, so the solution is 240th
• Page 17: 10
Once again, we must again remember that point bill goes before dashes. So 5 * 0 = 0, we add up to 1 +2 +3 +4 = 10
• Page 18: 41 (Older version: Kevin is 18 years old)
• Page 19: 40
At level 19 you have to count squares. First small and then the respectively larger. The resulting 40th
• Page 20: 1
To solve this is only 1, 2 or 3 Now you can either enter a number or just thinking about it. The route is 200 kilometers and 200 kilometers for the outward journey for the return trip. Now you just need to calculate only the speed, how long each individual needs. In the end, Marc is the fastest.

• Marc takes the Inbound 2 hours (200 km – 100 km / h). The same for the return trip. Ie, 4 hours total
• Paul initially runs only 80km / h Need for the outward journey or 2.5 hours. For the return trip, he only needs 1.66 hours. Total So just over four hours, slower than Marc.
• Laura goes first only 50 km / h Alone for the outward journey they need it 4 hours. Now they would have to go with infinite speed in order to arrive at the same time still with Marc. Accordingly, they can also reach 250 km / h and does not create it.

### Book of Mathematics page 21-30

Here we have the solution of 21 to 30 for the book of mathematics for you:

• Page 21: 4100
You just have to add up the numbers. So 1000 +1000 +1000 +1000 = 4000. Now add the still small numbers, that is 40 +30 +10 +20 = 100 So in the 4100th Who should get out here 5000, the math again • Page 22: 59
Again have a series of numbers to continue. It notes that this is some primes in the sequence. The next prime is 59 (Prime numbers are only divisible by 1 and itself)
• Page 23 Solution: 1987
Wanted will be the year when the ships returned to harbor shrink. It is the fourth January. That means we are in the first, the second maximum Calendar week. Each year, more than 51 calendar weeks. The time is irrelevant. Now we must make the least common multiple of 16, 12, 4, and 8. As 8 in the 16 and 4 lies in the 12 (or 8 or 16), we will make the LCM of 12 and 16, ie, 48th In 48 weeks, the ships take so together again in the harbor. So this is still the year 1987. Accordingly, this is also the solution to page 23 of the book of mathematics.
• Page 24: 32
In level 24 of the book of mathematics again blocks must be counted. In this context the different shading. We come finally to 32 blocks.
• Page 25 Solution: 812520
Really complicated, it is on page 25 of the book of mathematics. Here, the variables a, b, c, d are determined. Doing so, the 4 are divided into 4 parts, so that the equation is satisfied. The solution is as follows a = 8, b = 12, c = 5 and d = 20, so we enter 812520:

• 8 +2 = 10
• 12-2 = 10
• 5 × 2 = 10
• 20:2 = 10
• Page 26: 4
On page 26, the equation can be solved to obtain the zero (Equation supposed 0 yield). Here we shall first by 2 and get a ²-8a +16. This is the second Binomial formula. So this means: (a 4) ². We now have a zero at 4, which is also the solution.
• Page 27: 130
Again, the number sequence will continue. This is the power of two, is added with 2 Ie 2 +2 ^ 0 = 3, 2 ^ 2 +2 = 4, 2 ^ 3 +2 = 10, and so on. To get the required number, we expect 2 ^ 7 +2 and the 130th is
• Page 28 Solution: 127
Sought is the number of sheep. We expect just 2 ^ 7-1. Reason: De Shepherd sold the entire 6-times (half and half). So we get the following (at the end of a sheep remains loud task):

• First 127/2-0, 5 = 63
• Second 63/2-0, 5 = 31
• Third 31/2-0, 5 = 15
• 4th 15/2-0, 5 = 7
• 5th 7/2-0, 5 = 3
• 6th 3/2-0, 5 = 1
• Page 29: 55 (older version: solution is 41)
• Page 30 Solution: 63
And finally again blocks count. Orient yourself right again. Ultimately, you get to 63 blocks.

We will add this solution of new pages arrived.Today Touchportal Games released their next free Android App: Book of Math. Now we want to present all solutions for this app.

At the beginning Book of Math contains 30 Levels. Different math puzzles like arithmetic problems, story problems, find the next series numbers, geometry and more funny pages are in this app. Our favorite is the cube puzzle.

## Solution to the math book – all pages / Level

Below you will find an overview of all solutions to the book of math.

• Page 1: 15
• Page 2: 146
• Page 3: 10
• Page 4: 12
• Page 5 Solution: 12
• Page 6: 24
• Page 7: 450
• Page 8: 2436
• Page 9 Solution: 300
• Page 10: 429
• Page 11: 180
• Page 12: 4044
• Page 13 Solution: 217972
• Page 14: 9
• Page 15 Solution: 28
• Page 16: 240
• Page 17: 10
• Page 18: 41 (older version: Kevin is 18 years old)
• Page 19: 40
• Page 20: 1
• Page 21: 4100
• Page 22: 59
• Page 23 Solution: 1987
• Page 24: 32
• Page 25 Solution: 812520
• Page 26: 4
• Page 27: 130
• Page 28 Solution: 127
• Page 29: 55
• Page 30 Solution: 63

### Solution of page 1-10

Let us begin with the solution of the page to 1 to 10 of the book of math. If you do not know how to get to this solution, so please get in touch in the comments.

• Page 1: 15
Simply sum 12 and 3 up. There should be no problems.
• Page 2: 146
The number sequence will be continued to resolve. This one always adds the two previous numbers and come to the third number is 56 + 90 = 146
• Page 3: 10
Just flip the Android device. Thus we see 600/60 and the resulting 10
• Page 4: 12
A little trick question, which should have been known. All months have 28 days. Thus, the solution is 12
• Page 5 solution: 12
Here, the triangles are counted. There are 8 small triangles and 4 triangles that result from the combination of three small triangles. The solution is 12
• Page 6: 24
Point bill goes before dashes. Accordingly, we first multiply 2 * 3=6. Now we sum 18 and 6 up. Solution: 24
• Page 7: 450
A small property task. 3 percent of 15,000 is 450. First, divide 15,000 by 100. Now multiply 150 with 3. solution: 450
• Page 8: 2436
On page 8 for the solution of the book Mathematics of numbers must both be determined. In the first sequence of numbers, the distance is increased by 1 each So first of +2, then +3, …. We now add 19+5 = 24 This is our first Number, the second number is obtained from the second series of numbers. This is backward to the squares. So 10 ², 9 ² …. The second number is 36. We do this together now: 2436
• Page 9 solution: 300
To solve the height of the tree must be stated in centimeters. In a right triangle is: a ² + b ² = c ². So we have to determine b. So 5 ² -4 ² = 9 This still extract the root and we get third This is in meters. We should type in in centimeter: 300
• Page 10: 429
On page 10, we have to solve to determine the variables a and b. is to identify a simple, a = 4 Now we need to determine b, this is 29, because 30 = b +1. So b = 29 We give the solution a two variables, ie, 429

### Solution of pages 11 to 20

Here we have the solution to the pages 11 to 20 of the book of mathematics for you:

• Page 11: 180
Also a simple task. Tina has a velocity v = 4 km / h For 12 miles they need so 3 hours. We should all specify, in minutes. So 3 hours * 60 = 180 minutes.
• Page 12: 4044
On page 12, we have to determine a and b. So we get for a = 10 * 4 = 40 b result of 46-2 = 44
• Page 13 Solution: 217972
Here we need to solve dislocate something. The Roman numerals us specify the order in which the numbers have to be surrounded. So 63/3 = 21, 16-9 = 7, 9 +9 = 9 * 0 and 12 * 6 = 72 We give the numbers in that order, so 217972nd
• Page 14: 9
The snail is climbing a 10 meter high tree. Every day she creates 2 meters, at night it goes down 1 meter. So we start: Day 1 – 2 feet high, down 1 meter (1 meter), Day 2 – 2 meters …. In the early 9th Day she is so at 8 meters (since Day 8 – 8 meters). This creates on 9 Day 2 meters and has climbed the 10-meter high tree.
• Page 15 Solution: 28
Here you have to count the cubes. This one should be guided primarily by the right. The block front right may otherwise be confused about. Ultimately, 28 blocks are installed.
• Page 16: 240
The number sequence will continue. We have the following scheme: * 3, – 3, +3. So we now proceed. In the 243 we have to subtract 3, so the solution is 240th
• Page 17: 10
Once again, we must again remember that point bill goes before dashes. So 5 * 0 = 0, we add up to 1 +2 +3 +4 = 10
• Page 18: 41 (Older version: Kevin is 18 years old)
• Page 19: 40
At level 19 you have to count squares. First small and then the respectively larger. The resulting 40th
• Page 20: 1
To solve this is only 1, 2 or 3 Now you can either enter a number or just thinking about it. The route is 200 kilometers and 200 kilometers for the outward journey for the return trip. Now you just need to calculate only the speed, how long each individual needs. In the end, Marc is the fastest.

• Marc takes the Inbound 2 hours (200 km – 100 km / h). The same for the return trip. Ie, 4 hours total
• Paul initially runs only 80km / h Need for the outward journey or 2.5 hours. For the return trip, he only needs 1.66 hours. Total So just over four hours, slower than Marc.
• Laura goes first only 50 km / h Alone for the outward journey they need it 4 hours. Now they would have to go with infinite speed in order to arrive at the same time still with Marc. Accordingly, they can also reach 250 km / h and does not create it.

### Book of Mathematics page 21-30

Here we have the solution of 21 to 30 for the book of mathematics for you:

• Page 21: 4100
You just have to add up the numbers. So 1000 +1000 +1000 +1000 = 4000. Now add the still small numbers, that is 40 +30 +10 +20 = 100 So in the 4100th Who should get out here 5000, the math again • Page 22: 59
Again have a series of numbers to continue. It notes that this is some primes in the sequence. The next prime is 59 (Prime numbers are only divisible by 1 and itself)
• Page 23 Solution: 1987
Wanted will be the year when the ships returned to harbor shrink. It is the fourth January. That means we are in the first, the second maximum Calendar week. Each year, more than 51 calendar weeks. The time is irrelevant. Now we must make the least common multiple of 16, 12, 4, and 8. As 8 in the 16 and 4 lies in the 12 (or 8 or 16), we will make the LCM of 12 and 16, ie, 48th In 48 weeks, the ships take so together again in the harbor. So this is still the year 1987. Accordingly, this is also the solution to page 23 of the book of mathematics.
• Page 24: 32
In level 24 of the book of mathematics again blocks must be counted. In this context the different shading. We come finally to 32 blocks.
• Page 25 Solution: 812520
Really complicated, it is on page 25 of the book of mathematics. Here, the variables a, b, c, d are determined. Doing so, the 4 are divided into 4 parts, so that the equation is satisfied. The solution is as follows a = 8, b = 12, c = 5 and d = 20, so we enter 812520:

• 8 +2 = 10
• 12-2 = 10
• 5 × 2 = 10
• 20:2 = 10
• Page 26: 4
On page 26, the equation can be solved to obtain the zero (Equation supposed 0 yield). Here we shall first by 2 and get a ²-8a +16. This is the second Binomial formula. So this means: (a 4) ². We now have a zero at 4, which is also the solution.
• Page 27: 130
Again, the number sequence will continue. This is the power of two, is added with 2 Ie 2 +2 ^ 0 = 3, 2 ^ 2 +2 = 4, 2 ^ 3 +2 = 10, and so on. To get the required number, we expect 2 ^ 7 +2 and the 130th is
• Page 28 Solution: 127
Sought is the number of sheep. We expect just 2 ^ 7-1. Reason: De Shepherd sold the entire 6-times (half and half). So we get the following (at the end of a sheep remains loud task):

• First 127/2-0, 5 = 63
• Second 63/2-0, 5 = 31
• Third 31/2-0, 5 = 15
• 4th 15/2-0, 5 = 7
• 5th 7/2-0, 5 = 3
• 6th 3/2-0, 5 = 1
• Page 29: 55 (older version: solution is 41)
• Page 30 Solution: 63
And finally again blocks count. Orient yourself right again. Ultimately, you get to 63 blocks.

We will add this solution of new pages arrived.

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