Euler's Eyes
2024-07-10
Exercises for Adding Natural Numbers
Warm-up with figures
Solved Exercises on Adding Natural Numbers - Elementary School
Example 1: Simple sum
Statement: Maria has 3 candy and John has 4 candy. How many bullets do they have together?
Resolution:
Maria has 3 candy and John has 4 candy. To find out how many bullets they have together, we add:
$$ 3 + 4 = 7 $$Therefore, Mary and John have 7 candies together.
Example 2: Sum of several numbers
Statement: Pedro bought 2 apples, 3 bananas and 5 oranges. How many fruits did he buy in total?
Resolution:
Pedro bought 2 apples, 3 bananas and 5 oranges. To find out the total number of fruits, we add:
$$ 2 + 3 + 5 = 10 $$Therefore, Peter bought 10 fruit in total.
Example 3: Sum with zero
Statement: How many balls does Maria have if she has 7 balls and then receives 0 balls as a gift?
Resolution:
Maria has 7 balls and doesn’t get any extra balls. Therefore, the sum is:
$$ 7 + 0 = 7 $$Maria continues with 7 balls.
Example 4: Sum of larger numbers
Statement: Lucas has 12 stickers and his friend gives him 8 stickers. How many stickers does Lucas have now?
Resolution:
Lucas had 12 stickers and got more 8 stickers. The sum is:
$$ 12 + 8 = 20 $$Now Lucas has 20 stickers.
Example 5: Sum of two large numbers
Statement: What is the sum of 25 and 17?
Resolution:
To find the sum of 25 and 17, we do:
$$ 25 + 17 = 42 $$Therefore, the sum of 25 and 17 is 42.
Example 6: Sum on a timeline
Statement: John starts with 10 toys. He gets more 5 from his parents and more 3 from his friends. How many toys does João have now?
Resolution:
João starts with 10 toys, gets more 5 from his parents and then 3 from his friends. The sum is:
$$ 10 + 5 + 3 = 18 $$Therefore, John now has 18 toys.
Example 7: Addition challenge
Statement: What is the sum of 15, 20 and 10?
Resolution:
We add the three numbers:
$$ 15 + 20 + 10 = 45 $$Therefore, the sum of 15, 20 and 10 is 45.
Solved Exercises on Adding Natural Numbers - Elementary School II
Example 1: Sum of larger numbers
Statement: Calculate the sum of $56$ and $79$.
Resolution:
To add $56$ and $79$, we line up the numbers and add:
$$ 56 + 79 = 135 $$Therefore, the sum of $56$ and $79$ is $135$.
Example 2: Sum of multiple natural numbers
Statement: Calculate the sum of $25, 48$ and $37$.
Resolution:
We add $25 + 48 + 37$ as follows:
$$ 25 + 48 = 73 $$$$ 73 + 37 = 110 $$
Therefore, the sum of $25, 48$ and $37$ is $110$.
Example 3: Sum of a sequence of consecutive natural numbers
Statement: What is the sum of consecutive natural numbers from $1$ to $10$?
Resolution:
The sum of the natural numbers from $1$ to $10$ can be done by adding directly:
$$ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 $$Therefore, the sum of the numbers from $1$ to $10$ is $55$.
Example 4: Sum of large numbers
Statement: Calculate the sum of $245$ and $678$.
Resolution:
Adding the numbers $245$ and $678$:
$$ 245 + 678 = 923 $$Therefore, the sum of $245$ and $678$ is $923$.
Example 5: Sum using a regular sequence
Statement: Calculate the sum of even numbers from $2$ to $20$.
Resolution:
The even numbers from $2$ to $20$ are: $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$.
Now we add up all the numbers:
$$ 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110 $$Therefore, the sum of even numbers from $2$ to $20$ is $110$.
Example 6: Sum in context problems
Statement: Ana is saving money. In the first month, she saved $50$ reais. In the second month, he saved $65$ reais. In the third month, he saved $85$ reais. How much did Ana save in total?
Resolution:
We add up the amounts saved over the three months:
$$ 50 + 65 + 85 = 200 $$Therefore, Ana saved $200$ reais in total.
Example 7: Sum of an arithmetic progression
Statement: What is the sum of the numbers $5, 10, 15, 20, 25$?
Resolution:
The given numbers form an arithmetic progression with ratio $5$. We add:
$$ 5 + 10 + 15 + 20 + 25 = 75 $$Therefore, the sum of the numbers $5, 10, 15, 20, 25$ is $75$.
Example 8: Sum of natural numbers with properties
Statement: Calculate the sum of the multiples of 5 between $1$ and $30$.
Resolution:
The multiples of 5 between $1$ and $30$ are: $5, 10, 15, 20, 25, 30$. We add these numbers together:
$$ 5 + 10 + 15 + 20 + 25 + 30 = 105 $$Therefore, the sum of multiples of $5$ between $1$ and $30$ is $105$.
Solved Exercises on Adding Natural Numbers - High School
Example 1: Sum of large numbers
Statement: Calculate the sum of $3789$ and $6453$.
Resolution:
We add the numbers directly:
$$ 3789 + 6453 = 10242 $$Therefore, the sum of $3789$ and $6453$ is $10242$.
Example 2: Sum of a sequence of consecutive natural numbers
Statement: What is the sum of the natural numbers from $1$ to $100$?
Resolution:
We can use the formula for the sum of the first $n$ natural numbers:
$$ S = \frac{n(n+1)}{2} $$We replace $n = 100$:
$$ S = \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 5050 $$Therefore, the sum of natural numbers from $1$ to $100$ is $5050$.
Example 3: Sum of an arithmetic progression (PA)
Statement: Find the sum of the arithmetic progression $3, 7, 11, 15, \dots, 55$.
Resolution:
First, we identify the BP parameters. The first term is $a_1 = 3$, the ratio is $r = 4$, and the last term is $a_n = 55$.
The number of terms $n$ is given by:
$$ a_n = a_1 + (n - 1)r $$$$ 55 = 3 + (n - 1) \times 4 $$
$$ 55 - 3 = (n - 1) \times 4 $$
$$ 52 = (n - 1) \times 4 \implies n - 1 = \frac{52}{4} = 13 $$
$$ n = 14 $$
Now, we use the formula for the sum of a PA:
$$ S_n = \frac{n(a_1 + a_n)}{2} $$$$ S_{14} = \frac{14(3 + 55)}{2} = \frac{14 \times 58}{2} = 406 $$
Therefore, the sum of PA $3, 7, 11, 15, \dots, 55$ is $406$.
Example 4: Sum of multiples of a number
Statement: Calculate the sum of the multiples of $7$ between $1$ and $70$.
Resolution:
The multiples of $7$ between $1$ and $70$ are: $7, 14, 21, 28, 35, 42, 49, 56, 63, 70$.
We use the formula for the sum of a PA, where the first term $a_1 = 7$, the last term $a_n = 70$ and the ratio $r = 7$.
The number of terms $n$ is given by:
$$ a_n = a_1 + (n - 1)r $$$$ 70 = 7 + (n - 1) \times 7 $$
$$ 70 - 7 = (n - 1) \times 7 $$
$$ 63 = (n - 1) \times 7 \implies n - 1 = \frac{63}{7} = 9 $$
$$ n = 10 $$
Now, we calculate the sum:
$$ S_n = \frac{n(a_1 + a_n)}{2} $$$$ S_{10} = \frac{10(7 + 70)}{2} = \frac{10 \times 77}{2} = 385 $$
Therefore, the sum of multiples of $7$ between $1$ and $70$ is $385$.
Example 5: Sum of a power series
Statement: Calculate the sum of $1^2 + 2^2 + 3^2 + \dots + 10^2$.
Resolution:
The sum of the squares of the first $n$ natural numbers can be calculated with the formula:
$$ S = \frac{n(n+1)(2n+1)}{6} $$Replacing $n = 10$:
$$ S = \frac{10(10+1)(2 \times 10+1)}{6} = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385 $$Therefore, the sum $1^2 + 2^2 + 3^2 + \dots + 10^2$ is $385$.
Example 6: Sum of natural numbers in financial contexts
Statement: João decided to save money progressively. In the first month, he saves $50$ reais, in the second month, $60$ reais, and so on, until the sixth month. How much will he have saved at the end of six months?
Resolution:
We observe that the saved values form an arithmetic progression with $a_1 = 50$, $r = 10$, and $n = 6$.
The last term $a_n$ is:
$$ a_n = a_1 + (n - 1)r = 50 + (6 - 1) \times 10 = 50 + 50 = 100 $$Now, we use the formula for the sum of a PA:
$$ S_n = \frac{n(a_1 + a_n)}{2} $$$$ S_6 = \frac{6(50 + 100)}{2} = \frac{6 \times 150}{2} = 450 $$
Therefore, at the end of six months, João will have saved $450$ reais.
Example 7: Sum of the digits of a number
Statement: Calculate the sum of the digits of the number $9352$.
Resolution:
The number $9352$ has the following digits: $9, 3, 5, 2$.
We add the digits:
$$ 9 + 3 + 5 + 2 = 19 $$Therefore, the sum of the digits of $9352$ is $19$.
Solved Exercises on Adding Natural Numbers - Entrance Exam
Example 1: Sum of large consecutive numbers
Statement: Calculate the sum of the natural numbers from $1$ to $200$.
Resolution:
The sum of the first $n$ natural numbers is given by the formula:
$$ S = \frac{n(n+1)}{2} $$We replace $n = 200$:
$$ S = \frac{200(200+1)}{2} = \frac{200 \times 201}{2} = 20100 $$Therefore, the sum of natural numbers from $1$ to $200$ is $20100$.
Example 2: Sum of multiples of a number
Statement: Calculate the sum of the multiples of $5$ between $1$ and $1000$.
Resolution:
The multiples of $5$ between $1$ and $1000$ are $5, 10, 15, 20, \dots, 1000$, forming an arithmetic progression (AP) with $a_1 = 5$, $r = 5$, and $a_n = 1000$.
To find the number of terms $n$, we use the $n$-th position formula of a PA:
$$ a_n = a_1 + (n - 1)r $$$$ 1000 = 5 + (n - 1) \times 5 $$
$$ 1000 - 5 = (n - 1) \times 5 \implies 995 = (n - 1) \times 5 $$
$$ n - 1 = \frac{995}{5} = 199 \implies n = 200 $$
Now, we calculate the sum of PA:
$$ S_n = \frac{n(a_1 + a_n)}{2} $$$$ S_{200} = \frac{200(5 + 1000)}{2} = \frac{200 \times 1005}{2} = 100500 $$
Therefore, the sum of multiples of $5$ between $1$ and $1000$ is $100500$.
Example 3: Sum of an arithmetic progression
Statement: Find the sum of the terms of the arithmetic progression $12, 15, 18, 21, \dots, 60$.
Resolution:
This is an arithmetic progression with first term $a_1 = 12$, ratio $r = 3$, and last term $a_n = 60$.
To find the number of terms $n$, we use the formula:
$$ a_n = a_1 + (n - 1)r $$$$ 60 = 12 + (n - 1) \times 3 $$
$$ 60 - 12 = (n - 1) \times 3 \implies 48 = (n - 1) \times 3 $$
$$ n - 1 = \frac{48}{3} = 16 \implies n = 17 $$
Now, we use the formula for the sum of a PA:
$$ S_n = \frac{n(a_1 + a_n)}{2} $$$$ S_{17} = \frac{17(12 + 60)}{2} = \frac{17 \times 72}{2} = 612 $$
Therefore, the sum of the PA terms $12, 15, 18, 21, \dots, 60$ is $612$.
Example 4: Sum of the squares of the first $n$ natural numbers
Statement: Calculate the sum of the squares of the natural numbers from $1$ to $20$.
Resolution:
The sum of the squares of the first $n$ natural numbers is given by the formula:
$$ S = \frac{n(n+1)(2n+1)}{6} $$Replacing $n = 20$:
$$ S = \frac{20(20+1)(2 \times 20 + 1)}{6} = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870 $$Therefore, the sum of the squares of the numbers $1$ to $20$ is $2870$.
Example 5: Sum of the digits of a large number
Statement: Calculate the sum of the digits of the number $123456789$.
Resolution:
The digits of $123456789$ are: $1, 2, 3, 4, 5, 6, 7, 8, 9$.
We add the digits:
$$ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 $$Therefore, the sum of the digits of $123456789$ is $45$.
Example 6: Sum in financial problems
Statement: Carlos saved $200$ reais in the first month, $250$ reais in the second month, $300$ reais in the third month, and so on, until the sixth month. How much did he save at the end of six months?
Resolution:
We observe that the saved values form a PA with $a_1 = 200$, ratio $r = 50$, and $n = 6$.
The last term $a_n$ is:
$$ a_n = a_1 + (n - 1)r = 200 + (6 - 1) \times 50 = 200 + 250 = 450 $$Now, we calculate the sum:
$$ S_n = \frac{n(a_1 + a_n)}{2} $$$$ S_6 = \frac{6(200 + 450)}{2} = \frac{6 \times 650}{2} = 1950 $$
Therefore, at the end of six months, Carlos will have saved $1950$ reais.
Example 7: Sum of a geometric series
Statement: Calculate the sum of the series $1 + 2 + 4 + 8 + \dots + 512$.
Resolution:
This is a geometric progression with $a_1 = 1$, ratio $r = 2$, and the last term $a_n = 512$.
To find the number of terms $n$, we use the $n$-th position formula of a GP:
$$ a_n = a_1 \cdot r^{(n-1)} $$$$ 512 = 1 \cdot 2^{(n-1)} \implies 512 = 2^{(n-1)} $$
We know that $2^9 = 512$, then $n - 1 = 9 \implies n = 10$.
The sum of a PG is given by:
$$ S_n = \frac{a_1(r^n - 1)}{r - 1} $$$$ S_{10} = \frac{1(2^{10} - 1)}{2 - 1} = 2^{10} - 1 = 1024 - 1 = 1023 $$
Therefore, the sum of the series $1 + 2 + 4 + 8 + \dots + 512$ is $1023$.
Example 8: Sum of two prime numbers
Statement: Find two prime numbers whose sum is $52$.
Resolution:
We know that $52 = 29 + 23$, and both are prime numbers.
Therefore, the prime numbers whose sum is $52$ are $29$ and $23$.
Training
Time to practice! Answer and see how many stars you get.