Two-Digit Numbers: Sum Of Remainders When Divided By 7

Hey guys! Today, let's dive into a fun math problem: figuring out the sum of all the remainders when we divide two-digit numbers by 7. Sounds like a mouthful, but trust me, it’s totally doable and kinda cool once you get the hang of it. We’ll break it down step by step so you can ace it like a math whiz.

Understanding the Problem

First off, what exactly are we trying to do? We need to take every two-digit number (that's from 10 to 99), divide each one by 7, and then add up all the remainders we get. So, for example, if we divide 10 by 7, we get a remainder of 3. If we divide 11 by 7, we get a remainder of 4, and so on. Our mission is to do this for every number up to 99 and then sum all those remainders. Sounds tedious, right? Well, we're going to find a smarter way to tackle this!

Breaking It Down

Let's start by identifying the first and last two-digit numbers that give us a remainder when divided by 7. The smallest two-digit number is 10. When you divide 10 by 7, you get a remainder of 3. The largest two-digit number is 99. When you divide 99 by 7, you get a remainder of 1. So, we know our remainders will fall somewhere between these values. The key here is to recognize that the remainders will cycle through a pattern.

Why a pattern? Because after every multiple of 7, the remainders start over. Think about it: If 14 divided by 7 has a remainder of 0, then 15 divided by 7 has a remainder of 1, 16 has a remainder of 2, and so on. This pattern repeats itself, which is super helpful for us!

Strategy

Okay, so here’s the plan. Instead of dividing every single number and adding the remainders individually, we're going to use some clever observations to make our lives easier. We'll look for the repeating pattern of remainders and then use that to calculate the sum more efficiently.

Finding the Pattern

Let's list out the remainders for a few two-digit numbers when divided by 7 to see the pattern more clearly:

• 10 ÷ 7 = 1 remainder 3
• 11 ÷ 7 = 1 remainder 4
• 12 ÷ 7 = 1 remainder 5
• 13 ÷ 7 = 1 remainder 6
• 14 ÷ 7 = 2 remainder 0
• 15 ÷ 7 = 2 remainder 1
• 16 ÷ 7 = 2 remainder 2

See the pattern? The remainders go from 3 to 6, then 0 to 2. They cycle through 3, 4, 5, 6, 0, 1, 2. This cycle has a length of 7, which makes sense because we're dividing by 7.

How Many Full Cycles?

Now we need to figure out how many complete cycles of these remainders we have between 10 and 99. First, let's find how many numbers are in the range from 10 to 99, inclusive. That's 99 - 10 + 1 = 90 numbers. Since each cycle has 7 numbers, we can divide 90 by 7 to find out how many full cycles we have: 90 ÷ 7 = 12 with a remainder of 6. So we have 12 full cycles and then 6 extra numbers.

Each full cycle contains the remainders 0, 1, 2, 3, 4, 5, and 6. The sum of these remainders is 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21. Since we have 12 full cycles, the sum of the remainders from these cycles is 12 * 21 = 252.

Handling the Leftovers

We still have those 6 extra numbers to deal with. We know that the remainders cycle through 3, 4, 5, 6, 0, 1, 2. So the first 6 remainders in our cycle are 3, 4, 5, 6, 0, and 1. The sum of these remainders is 3 + 4 + 5 + 6 + 0 + 1 = 19.

Calculating the Total Sum

Now we just add the sum of the remainders from the full cycles and the sum of the remainders from the extra numbers:

Total sum = (Sum of remainders from full cycles) + (Sum of remainders from extra numbers)

Total sum = 252 + 19 = 271

So, the sum of all the remainders when two-digit numbers are divided by 7 is 271. Woo-hoo! That's how it’s done, guys!

An Alternative Approach: Arithmetic Progression

Alright, mathletes, let's tackle this problem from a slightly different angle! We can use the concept of arithmetic progression to simplify our calculations. This method is super handy if you're comfortable with sequences and series.

Setting Up the Sequence

First, we need to identify the smallest and largest two-digit numbers that give us remainders when divided by 7. As we already found out, 10 divided by 7 gives a remainder of 3, and 99 divided by 7 gives a remainder of 1. Now, let's list the remainders in a sequence. The sequence will consist of the remainders we get when dividing each number from 10 to 99 by 7.

The remainders repeat in a cycle of 7: 3, 4, 5, 6, 0, 1, 2. We know there are 90 two-digit numbers (from 10 to 99). We want to find the sum of the remainders when each of these numbers is divided by 7.

Identifying the Terms

Let's denote the remainders as ${r_1, r_2, r_3, \ldots, r_{90}}$. We already know that ${r_1 = 3}$ (the remainder when 10 is divided by 7) and that the remainders repeat every 7 numbers. To find the sum of these remainders, we can break it down into cycles.

As we determined earlier, there are 12 full cycles of remainders and 6 additional remainders. Each full cycle has the remainders 0, 1, 2, 3, 4, 5, 6. The sum of the numbers in each cycle is 21.

Sum of the Cycles

Since we have 12 full cycles, the sum of the remainders from these cycles is ${12 \times 21 = 252}$.

For the additional 6 remainders, we have the sequence 3, 4, 5, 6, 0, 1. The sum of these remainders is ${3 + 4 + 5 + 6 + 0 + 1 = 19}$.

Total Sum

Now, we add the sum of the remainders from the full cycles and the sum of the additional remainders:

Total sum = (Sum of remainders from full cycles) + (Sum of additional remainders)

Total sum = ${252 + 19 = 271}$

Wrapping Up the Arithmetic Approach

So, using the arithmetic progression approach, we've once again found that the sum of all the remainders when two-digit numbers are divided by 7 is 271. This method provides a structured way to approach the problem, making it easier to manage the calculations.

Tips and Tricks for Remainder Problems

Alright, let's arm you with some extra tips and tricks to handle remainder problems like a pro. These strategies can make solving similar questions much easier and faster.

Recognizing Patterns

• Look for Cycles: As we saw in our main problem, remainders often repeat in cycles. Identifying these cycles is crucial for simplifying the problem. Divide the range of numbers by the divisor to find out how many full cycles you have, and then deal with the leftovers.

• Start Small: If the problem seems overwhelming, start by listing out the remainders for the first few numbers. This can help you spot the pattern more easily.

Using Modular Arithmetic

• Understand the Basics: Modular arithmetic is all about working with remainders. The expression ${a \equiv b \pmod{m}}$ means that ${a}$ and ${b}$ have the same remainder when divided by ${m}$. This can be super useful for simplifying calculations.

• Apply Properties: Remember that ${(a + b) \pmod{m} \equiv (a \pmod{m} + b \pmod{m}) \pmod{m}}$ and ${(a \times b) \pmod{m} \equiv (a \pmod{m} \times b \pmod{m}) \pmod{m}}$. These properties allow you to break down complex expressions into simpler parts.

Efficient Calculation Techniques

• Group Numbers: If possible, group numbers that give you easy-to-calculate remainders. For example, numbers that are close to multiples of the divisor can simplify your work.

• Use Arithmetic Progressions: As demonstrated earlier, if the remainders form an arithmetic progression, you can use the formulas for the sum of an arithmetic series to find the total sum quickly.

Example Scenario

Let's say you want to find the sum of the remainders when dividing the numbers from 50 to 100 by 5.

1. Identify the Range: We're looking at numbers from 50 to 100, which is a total of 51 numbers.
2. Find the Cycle: When dividing by 5, the remainders are 0, 1, 2, 3, 4. This is a cycle of 5.
3. Determine Full Cycles: Divide the number of values in the range by the cycle length: ${51 \div 5 = 10}$ with a remainder of 1. So we have 10 full cycles and 1 extra number.
4. Calculate the Sum of One Cycle: ${0 + 1 + 2 + 3 + 4 = 10}$.
5. Sum of Full Cycles: ${10 \times 10 = 100}$.
6. Remainder: The extra number is 50, which gives a remainder of 0 when divided by 5.
7. Total Sum: ${100 + 0 = 100}$.

So, the sum of the remainders when dividing the numbers from 50 to 100 by 5 is 100.

By keeping these tips and tricks in mind, you'll be well-equipped to tackle all sorts of remainder problems with confidence. Happy calculating, guys!

Practice Problems

Okay, math enthusiasts, it’s time to put your knowledge to the test! Here are a few practice problems to help you master the art of finding sums of remainders. Work through these, and you'll be a pro in no time!

1. Problem 1: Sum of Remainders When Dividing by 4

   Find the sum of the remainders when all two-digit numbers are divided by 4.

   Hint: Identify the cycle of remainders and how many full cycles you have within the two-digit numbers.

2. Problem 2: Sum of Remainders When Dividing by 9

   Calculate the sum of the remainders when the numbers from 20 to 50 are divided by 9.

   Hint: Use modular arithmetic and look for patterns to simplify your calculations.

3. Problem 3: Sum of Remainders with Three-Digit Numbers

   Determine the sum of the remainders when all three-digit numbers are divided by 6.

   Hint: Three-digit numbers range from 100 to 999. Break down the problem into manageable cycles.

4. Problem 4: Sum of Remainders in a Specific Range

   Find the sum of the remainders when the numbers from 15 to 45 are divided by 7.

   Hint: Pay close attention to the start and end points of the range and how they affect the remainder cycle.

Solutions

1. Solution to Problem 1

   • Two-digit numbers are from 10 to 99, totaling 90 numbers.
   • When dividing by 4, the cycle of remainders is 0, 1, 2, 3 (cycle length is 4).
   • Number of full cycles: ${90 \div 4 = 22}$ with a remainder of 2.
   • Sum of one cycle: ${0 + 1 + 2 + 3 = 6}$.
   • Sum of full cycles: ${22 \times 6 = 132}$.
   • Remainders for the extra 2 numbers: 0, 1.
   • Sum of extra remainders: ${0 + 1 = 1}$.
   • Total sum: ${132 + 1 = 133}$.

   Therefore, the sum of the remainders when all two-digit numbers are divided by 4 is 133.

2. Solution to Problem 2

   • Numbers from 20 to 50, totaling 31 numbers.
   • When dividing by 9, the cycle of remainders is 0, 1, 2, 3, 4, 5, 6, 7, 8 (cycle length is 9).
   • Number of full cycles: ${31 \div 9 = 3}$ with a remainder of 4.
   • Sum of one cycle: ${0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36}$.
   • Sum of full cycles: ${3 \times 36 = 108}$.
   • Remainders for the extra 4 numbers: 2, 3, 4, 5 (since 20 divided by 9 gives a remainder of 2).
   • Sum of extra remainders: ${2 + 3 + 4 + 5 = 14}$.
   • Total sum: ${108 + 14 = 122}$.

   Therefore, the sum of the remainders when the numbers from 20 to 50 are divided by 9 is 122.

3. Solution to Problem 3

   • Three-digit numbers are from 100 to 999, totaling 900 numbers.
   • When dividing by 6, the cycle of remainders is 0, 1, 2, 3, 4, 5 (cycle length is 6).
   • Number of full cycles: ${900 \div 6 = 150}$ with no remainder.
   • Sum of one cycle: ${0 + 1 + 2 + 3 + 4 + 5 = 15}$.
   • Total sum: ${150 \times 15 = 2250}$.

   Therefore, the sum of the remainders when all three-digit numbers are divided by 6 is 2250.

4. Solution to Problem 4

   • Numbers from 15 to 45, totaling 31 numbers.
   • When dividing by 7, the cycle of remainders is 0, 1, 2, 3, 4, 5, 6 (cycle length is 7).
   • Number of full cycles: ${31 \div 7 = 4}$ with a remainder of 3.
   • Sum of one cycle: ${0 + 1 + 2 + 3 + 4 + 5 + 6 = 21}$.
   • Sum of full cycles: ${4 \times 21 = 84}$.
   • Remainders for the extra 3 numbers: 1, 2, 3 (since 15 divided by 7 gives a remainder of 1).
   • Sum of extra remainders: ${1 + 2 + 3 = 6}$.
   • Total sum: ${84 + 6 = 90}$.

   Therefore, the sum of the remainders when the numbers from 15 to 45 are divided by 7 is 90.

Keep practicing, and you’ll become a master at solving remainder problems. You got this, guys!