Which is the Largest Number That Divides 70 and 125 Leaving Remainders?

Byzhipanyangsheng

I remember a time when I was grappling with a math problem that felt like a real brain teaser. It was one of those questions that pops up in textbooks or math challenges, asking to find a specific number based on division and remainders. The particular problem that sticks with me was: "Which is the largest number that divides 70 and 125 leaving remainder 5 and 8, respectively?" At first glance, it seems straightforward enough, but the dual conditions—one for 70 and another for 125—made it a bit more intricate than a simple Greatest Common Divisor (GCD) problem. It required a deeper dive into how remainders affect the numbers we're working with.

This kind of problem, while perhaps seeming niche, is actually a fantastic illustration of fundamental number theory concepts. It’s not just about crunching numbers; it’s about understanding the relationships between divisors, dividends, and remainders. For me, solving it involved a bit of trial and error, but more importantly, it demanded a structured approach, which is precisely what we'll be exploring here. We’re going to break down how to tackle this specific challenge and, by extension, any similar problems involving multiple divisors and specified remainders.

Understanding the Core Problem: Divisibility with a Twist

So, let's get right to it. The answer to the question, "Which is the largest number that divides 70 and 125 leaving remainder 5 and 8 respectively?" is 9.

Now, you might be wondering how we arrive at that answer. It's not immediately obvious, is it? This isn't as simple as finding the GCD of 70 and 125. The key here lies in understanding what it means for a number to divide another number *leaving a specific remainder*. When a number, let's call it 'd', divides another number, 'N', and leaves a remainder 'r', it means that N can be expressed as: N = q * d + r, where 'q' is the quotient and 'r' is the remainder, and importantly, 0 ≤ r < d.

In our problem, we have two conditions:

  • The largest number (let's call it 'x') divides 70, leaving a remainder of 5.
  • The same largest number 'x' divides 125, leaving a remainder of 8.

This means that if we subtract the remainder from the original number, the resulting number *must* be perfectly divisible by 'x'. Think about it: if 70 divided by 'x' gives a remainder of 5, then 70 - 5 = 65 must be exactly divisible by 'x'. Similarly, if 125 divided by 'x' gives a remainder of 8, then 125 - 8 = 117 must be exactly divisible by 'x'.

This is the crucial insight. We are no longer looking for a number that divides 70 and 125 directly. Instead, we're looking for the largest number that divides the *adjusted* numbers: 65 and 117.

The Strategic Adjustment: Subtracting Remainders

Let's elaborate on this strategic adjustment. When a problem states that a number 'x' divides 'N' and leaves a remainder 'r', it's implying a specific mathematical relationship. As we touched upon, this relationship is formalized by the division algorithm: N = qd + r. The fact that 'x' is the divisor means it plays the role of 'd' in this equation.

So, for our first condition, we have: 70 = q1 * x + 5. If we rearrange this equation, we get 70 - 5 = q1 * x. This clearly shows that (70 - 5), which equals 65, is a multiple of 'x'. In other words, 'x' is a divisor of 65.

Similarly, for our second condition, we have: 125 = q2 * x + 8. Rearranging this gives us 125 - 8 = q2 * x. This tells us that (125 - 8), which equals 117, is also a multiple of 'x'. So, 'x' is a divisor of 117.

Now, the problem asks for the *largest* such number 'x'. This means 'x' must be a common divisor of both 65 and 117. And since we want the largest one, we are specifically looking for the Greatest Common Divisor (GCD) of 65 and 117.

Why Subtracting Remainders is Key

This step is absolutely fundamental and worth reiterating. If we didn't subtract the remainders, we'd be trying to find a number that divides 70 and 125 while *simultaneously* accounting for these leftover amounts. That would be a significantly more complex problem, possibly involving systems of congruences. By subtracting the remainders, we transform the problem into finding a common divisor of two new numbers. This is a standard problem in arithmetic, making the solution path much clearer.

Consider a simpler analogy: If you have a bag of marbles and you want to divide them equally among your friends, but you always have 3 marbles left over, it means that if you removed those 3 marbles first, the remaining amount would be perfectly divisible by the number of friends. The same logic applies here, just with larger numbers and specific remainder conditions.

Finding the Greatest Common Divisor (GCD) of 65 and 117

With our problem transformed into finding the GCD of 65 and 117, we can now employ established methods. There are a couple of common approaches:

  • Prime Factorization Method: This involves breaking down each number into its prime factors and then identifying the common prime factors raised to their lowest powers.
  • Euclidean Algorithm: This is a more efficient method, especially for larger numbers, and it's based on the principle that the GCD of two numbers does not change if the larger number is replaced by its difference with the smaller number. It can be further optimized by using the remainder of the division.

Let's try the prime factorization method first, as it's often more intuitive for understanding the concept of common divisors.

Prime Factorization of 65 and 117

For 65:

We start by looking for the smallest prime number that divides 65. It's not divisible by 2 (since it's odd). It's not divisible by 3 (sum of digits 6+5=11, not divisible by 3). It *is* divisible by 5.

65 ÷ 5 = 13

Now, 13 is a prime number. So, the prime factorization of 65 is 5 × 13.

For 117:

Let's check for divisibility by small primes.

Not divisible by 2 (odd).

Sum of digits = 1 + 1 + 7 = 9. Since 9 is divisible by 3, 117 is divisible by 3.

117 ÷ 3 = 39

Now we factor 39. It's divisible by 3 again.

39 ÷ 3 = 13

And 13 is a prime number.

So, the prime factorization of 117 is 3 × 3 × 13, or 3² × 13.

Identifying Common Factors:

Prime factors of 65: 5, 13

Prime factors of 117: 3, 3, 13

The only common prime factor is 13.

Therefore, the GCD(65, 117) = 13.

Wait a minute! This is where I need to pause and double-check. The problem asks for the largest number that divides 70 leaving remainder 5 and 125 leaving remainder 8. If our GCD is 13, let's test it.

  • Does 13 divide 70 leaving a remainder of 5? 70 ÷ 13 = 5 with a remainder of 5. (70 = 5 * 13 + 5). Yes!
  • Does 13 divide 125 leaving a remainder of 8? 125 ÷ 13 = 9 with a remainder of 8. (125 = 9 * 13 + 8). Yes!

So, 13 is a common divisor that satisfies the conditions. But is it the *largest*? My initial answer was 9. What did I miss?

Ah, I see the mistake in my initial calculation or recall. The common factor is indeed 13. Let me retrace my steps and ensure I haven't misapplied any logic. The process of subtracting remainders and finding the GCD of the resulting numbers is correct. The prime factorization method should yield the correct GCD. Let me re-evaluate the prime factorization or consider the Euclidean Algorithm, as sometimes subtle errors can creep in.

Revisiting the Calculation: The Euclidean Algorithm

Let's use the Euclidean Algorithm to find the GCD of 65 and 117. This method is systematic and less prone to calculation errors for me.

The principle is: GCD(a, b) = GCD(b, a mod b), where 'a mod b' is the remainder when 'a' is divided by 'b'. We continue this process until the remainder is 0. The last non-zero remainder is the GCD.

  1. Divide the larger number (117) by the smaller number (65):

    2. 117 = 1 * 65 + 52

    (The remainder is 52)

  2. Now, replace the larger number (117) with the smaller number (65), and the smaller number with the remainder (52):

    3. GCD(117, 65) = GCD(65, 52)

    Divide 65 by 52:

    65 = 1 * 52 + 13

    (The remainder is 13)

  3. Again, replace the numbers:

    4. GCD(65, 52) = GCD(52, 13)

    Divide 52 by 13:

    52 = 4 * 13 + 0

    (The remainder is 0)

The last non-zero remainder is 13. Therefore, the GCD(65, 117) is indeed 13.

My apologies for the initial misstatement of the answer. The largest number that divides 70 leaving a remainder of 5 and 125 leaving a remainder of 8 is 13.

This highlights the importance of careful calculation and verification, even when the underlying logic is sound. It’s a good reminder for all of us who work with numbers!

Important Consideration: The Remainder Must Be Smaller Than the Divisor

There's a crucial condition in division with remainders that we must always keep in mind: the remainder must always be strictly less than the divisor. In our problem, this means:

  • The divisor 'x' must be greater than the remainder 5. So, x > 5.
  • The divisor 'x' must be greater than the remainder 8. So, x > 8.

Combining these, the divisor 'x' must be greater than 8 (since if it's greater than 8, it's automatically greater than 5). So, we are looking for the GCD of 65 and 117, with the additional constraint that our answer must be greater than 8.

Let's check our GCD, which is 13.

  • Is 13 > 8? Yes.

So, 13 is a valid candidate. Since the GCD is the *greatest* common divisor, and 13 satisfies the condition of being greater than the remainders, 13 is indeed the largest number that meets all the criteria.

This constraint is vital. If, for example, the remainders were 5 and 2, and we found the GCD of the adjusted numbers to be, say, 3, then 3 wouldn't be a valid answer because it's not greater than the remainder 5. In such a hypothetical case, we would have to look at the *next* largest common divisor of the adjusted numbers and see if that satisfies the remainder condition.

Illustrative Example: A Similar Problem

To solidify our understanding, let's consider a similar problem and work through it systematically.

Problem:

Find the largest number that divides 50 and 75, leaving remainders of 2 and 3, respectively.

Step-by-Step Solution:

  1. Adjust the numbers by subtracting the remainders:
    • For 50, the remainder is 2. Adjusted number = 50 - 2 = 48.
    • For 75, the remainder is 3. Adjusted number = 75 - 3 = 72.
  2. Find the GCD of the adjusted numbers (48 and 72):

    3. We can use the Euclidean Algorithm:

    • 117 = 1 * 65 + 52 (Mistake here, should be 72 = 1 * 48 + 24)
    • Divide 72 by 48:

      • 72 = 1 * 48 + 24

      (Remainder is 24)

    • Now find GCD(48, 24):

      • 48 = 2 * 24 + 0

      (Remainder is 0)

    The last non-zero remainder is 24. So, GCD(48, 72) = 24.

  3. Check the remainder condition:

    4. The divisor must be greater than both remainders. Here, the remainders are 2 and 3. Our GCD is 24.

    • Is 24 > 2? Yes.
    • Is 24 > 3? Yes.

    Since 24 is greater than both remainders, it is a valid divisor.

  4. Conclusion:

    5. The largest number that divides 50 leaving a remainder of 2 and 75 leaving a remainder of 3 is 24.

Let's verify this:

  • 50 ÷ 24 = 2 with a remainder of 2. (50 = 2 * 24 + 2). Correct.
  • 75 ÷ 24 = 3 with a remainder of 3. (75 = 3 * 24 + 3). Correct.

This example further solidifies the method. The process is consistent and reliable.

When the GCD is Not the Answer: A Deeper Dive

What happens if the GCD we calculate isn't greater than the remainders? This is a critical edge case that requires careful consideration. Let's imagine a hypothetical scenario to illustrate this.

Hypothetical Problem:

Find the largest number that divides 30 and 45, leaving remainders of 4 and 7, respectively.

Working Through the Hypothetical:

  1. Adjust the numbers:
    • 30 - 4 = 26
    • 45 - 7 = 38
  2. Find the GCD of 26 and 38:

    3. Using the Euclidean Algorithm:

    • 38 = 1 * 26 + 12
    • 26 = 2 * 12 + 2
    • 12 = 6 * 2 + 0

    The GCD(26, 38) = 2.

  3. Check the remainder condition:

    4. The remainders are 4 and 7. The divisor must be strictly greater than both.

    • Is 2 > 4? No.
    • Is 2 > 7? No.

    Since 2 is not greater than either 4 or 7, it cannot be the correct answer. This means that no number satisfying these conditions can exist as stated.

Why does this happen?

The condition that the remainder 'r' must be less than the divisor 'd' (r < d) is a fundamental part of the definition of division with remainder. If our calculated GCD is not greater than one of the remainders, it implies that a number satisfying both divisibility conditions *and* the remainder constraints cannot exist.

Let's think about it mathematically. If 'x' divides N with remainder r, then N = qx + r, which implies N - r = qx. This means 'x' must be a divisor of N - r. Furthermore, for the remainder to be 'r', we must have r < x.

In our hypothetical case, we need a number 'x' such that:

  • x divides 26
  • x divides 38
  • x > 4
  • x > 7

Combining the last two, we need x > 7.

The common divisors of 26 and 38 are the divisors of their GCD, which is 2. The divisors of 2 are 1 and 2.

Neither 1 nor 2 is greater than 7.

Therefore, in this hypothetical situation, there is no such number that satisfies all the conditions. The problem statement itself would contain incompatible conditions.

What if the GCD is not the *only* common divisor?

Consider another hypothetical:

Find the largest number that divides 30 and 40, leaving remainders of 3 and 5, respectively.

  1. Adjust numbers:
    • 30 - 3 = 27
    • 40 - 5 = 35
  2. Find GCD(27, 35):
    • 35 = 1 * 27 + 8
    • 27 = 3 * 8 + 3
    • 8 = 2 * 3 + 2
    • 3 = 1 * 2 + 1
    • 2 = 2 * 1 + 0

    GCD(27, 35) = 1.

  3. Check remainder condition:

    4. Remainders are 3 and 5. Divisor must be > 5.

    Our GCD is 1, which is not > 5.

    The common divisors of 27 and 35 are the divisors of their GCD, which is 1. The only common divisor is 1. Since 1 is not greater than 5, there is no such number.

Okay, let's try to construct a scenario where the GCD isn't the answer, but a smaller common divisor might be, if the GCD itself fails the remainder test.

Constructed Hypothetical Problem:

Find the largest number that divides 100 and 150, leaving remainders of 10 and 10, respectively.

  1. Adjust numbers:
    • 100 - 10 = 90
    • 150 - 10 = 140
  2. Find GCD(90, 140):
    • 140 = 1 * 90 + 50
    • 90 = 1 * 50 + 40
    • 50 = 1 * 40 + 10
    • 40 = 4 * 10 + 0

    GCD(90, 140) = 10.

  3. Check remainder condition:

    4. The remainder is 10. The divisor must be strictly greater than the remainder.

    • Is 10 > 10? No.

    So, 10 is not a valid answer because the remainder (10) cannot be equal to or greater than the divisor (10).

  4. Consider other common divisors:

    5. The common divisors of 90 and 140 are the divisors of their GCD, which is 10. The divisors of 10 are 1, 2, 5, and 10.

    We need a divisor that is strictly greater than the remainder (10).

    Let's check each common divisor against the condition x > 10:

    • 1 > 10? No.
    • 2 > 10? No.
    • 5 > 10? No.
    • 10 > 10? No.

    In this constructed scenario, there is no number that satisfies the conditions. The problem statement is again inherently flawed because the remainder specified is not less than any of the common divisors of the adjusted numbers.

This illustrates why the initial problem statement is so critical. The remainders given *must* be smaller than the number we are looking for. If they aren't, then no such number exists.

Let's consider a case where the GCD works, but the remainder condition needs careful application.

Another Scenario:

Find the largest number that divides 80 and 100, leaving remainders of 6 and 6, respectively.

  1. Adjust numbers:
    • 80 - 6 = 74
    • 100 - 6 = 94
  2. Find GCD(74, 94):
    • 94 = 1 * 74 + 20
    • 74 = 3 * 20 + 14
    • 20 = 1 * 14 + 6
    • 14 = 2 * 6 + 2
    • 6 = 3 * 2 + 0

    GCD(74, 94) = 2.

  3. Check remainder condition:

    4. The remainder is 6. The divisor must be strictly greater than 6 (x > 6).

    Our GCD is 2. Is 2 > 6? No.

  4. Consider other common divisors:

    5. The common divisors of 74 and 94 are the divisors of 2: which are 1 and 2.

    We need a divisor 'x' such that x > 6.

    • 1 > 6? No.
    • 2 > 6? No.

    Again, in this case, no such number exists. The problem statement implies a remainder that is larger than any possible common divisor that would satisfy the divisibility rule after adjustment.

It appears my initial answer of 9 for the original problem was likely from a miscalculation or a different problem entirely. The consistent application of the method clearly points to 13. It's a good lesson in meticulousness!

The Number Theory Behind It All

The problem we've been dissecting is a classic application of number theory, specifically focusing on the concepts of divisibility, remainders, and the Greatest Common Divisor (GCD). Let's delve a bit deeper into why these concepts are so fundamental and how they interact.

Divisibility and the Division Algorithm

At its heart, division is about partitioning a quantity into equal groups. The Division Algorithm is the formal mathematical statement that for any integer 'a' (the dividend) and any positive integer 'd' (the divisor), there exist unique integers 'q' (the quotient) and 'r' (the remainder) such that:

a = qd + r

and 0 ≤ r < d.

This inequality, 0 ≤ r < d, is paramount. It states that the remainder must be non-negative and strictly less than the divisor. This is why, in our problem, the number we are looking for must be strictly greater than the given remainders.

Greatest Common Divisor (GCD)

The GCD of two or more integers is the largest positive integer that divides each of the integers without leaving a remainder. For instance, the divisors of 12 are 1, 2, 3, 4, 6, and 12. The divisors of 18 are 1, 2, 3, 6, 9, and 18. The common divisors are 1, 2, 3, and 6. The greatest among these is 6, so GCD(12, 18) = 6.

The Euclidean Algorithm is an efficient method for computing the GCD. It relies on the property that the GCD of two numbers does not change if the larger number is replaced by its remainder when divided by the smaller number. This iterative process eventually leads to a remainder of zero, and the last non-zero remainder is the GCD.

Connecting the Dots: How GCD Solves Remainder Problems

The critical step in solving these remainder problems is the transformation: if a number 'x' divides 'N' leaving a remainder 'r', then 'x' must divide 'N - r'.

Let's formalize this. We are given:

70 = q1 * x + 5 (Equation 1)

125 = q2 * x + 8 (Equation 2)

From Equation 1, rearranging gives: 70 - 5 = q1 * x. This means 'x' is a divisor of 65.

From Equation 2, rearranging gives: 125 - 8 = q2 * x. This means 'x' is a divisor of 117.

Since 'x' must be a divisor of *both* 65 and 117, it must be a *common divisor* of 65 and 117. The problem asks for the *largest* such number, which by definition is the Greatest Common Divisor (GCD) of 65 and 117.

Additionally, from the definition of remainder, we know:

5 < x (from Equation 1, as remainder must be less than divisor)

8 < x (from Equation 2, as remainder must be less than divisor)

Therefore, 'x' must be the GCD(65, 117), with the additional constraint that x > 8.

As we calculated using the Euclidean Algorithm:

GCD(65, 117) = 13.

Now we check the condition: Is 13 > 8? Yes.

Therefore, 13 is the largest number that satisfies all conditions.

Common Pitfalls and How to Avoid Them

Based on my own experience and observing others, there are a few common traps when dealing with these types of problems:

  • Forgetting to subtract the remainders: This is probably the most frequent error. People might try to find the GCD of 70 and 125 directly, which would give a different result and not account for the specified remainders.
  • Ignoring the remainder constraint (r < d): As we saw in the hypothetical examples, simply finding the GCD of the adjusted numbers isn't enough. The resulting GCD *must* be greater than the specified remainders. If it isn't, and there are no other common divisors that satisfy the condition, then no such number exists.
  • Calculation errors in GCD: Whether using prime factorization or the Euclidean Algorithm, a simple arithmetic mistake can lead to the wrong GCD. Double-checking calculations is always a good practice.
  • Misinterpreting "largest number": The phrasing "largest number" unambiguously points to the GCD. If the question were phrased differently, for instance, asking for "a number" or "all numbers," the approach might change slightly to include other common divisors.

To avoid these pitfalls:

  1. Always start by adjusting the numbers: Subtract the remainder from the dividend.
  2. Systematically find the GCD: Use a reliable method like the Euclidean Algorithm and double-check your steps.
  3. Crucially, verify the remainder condition: Ensure that your GCD is strictly greater than *all* the specified remainders. If not, analyze the divisors of the GCD to see if any satisfy the condition. If none do, state that no such number exists.

Frequently Asked Questions (FAQs)

How do I know if a number leaves a specific remainder when divided?

A number 'N' leaves a remainder 'r' when divided by a number 'd' if, when you perform the division, the amount left over after making as many full groups of 'd' as possible is exactly 'r'. Mathematically, this is expressed by the division algorithm: N = qd + r, where 'q' is the quotient and 'r' is the remainder, with the condition that 0 ≤ r < d. So, if you divide N by d and get a quotient 'q' and a remainder 'r', and this remainder satisfies 0 ≤ r < d, then yes, N leaves a remainder of 'r' when divided by 'd'.

For instance, to check if 70 leaves a remainder of 5 when divided by some number 'x', you would perform the division. If 70 divided by 'x' gives a quotient 'q' and a remainder of 5, and importantly, if 5 is less than 'x', then the condition is met. This leads us to the core technique used in these problems: if N leaves a remainder r when divided by x, then N - r must be perfectly divisible by x.

Why do we subtract the remainders in these problems?

We subtract the remainders because the problem statement implies a relationship that can be simplified. When we say a number 'x' divides 'N' leaving a remainder 'r', we are essentially saying that 'N' is 'r' more than some multiple of 'x'.

So, if N = qx + r (where q is the quotient and x is the divisor), we can rearrange this equation to get N - r = qx. This new equation, N - r = qx, tells us that the number (N - r) is now perfectly divisible by 'x'. By subtracting the remainders from the original numbers (70 and 125 in our case), we transform the problem from one with remainders into a standard Greatest Common Divisor (GCD) problem. We are looking for a number 'x' that divides both (70 - 5) and (125 - 8).

What if the GCD I find is not larger than the remainders?

This is a crucial point that often trips people up. The definition of a remainder in division states that the remainder 'r' must always be strictly less than the divisor 'd' (i.e., r < d). If you find that the GCD of the adjusted numbers is not strictly greater than the remainders specified in the problem, then that GCD cannot be your answer.

In such a scenario, you must consider the other common divisors of the adjusted numbers. These common divisors are simply the divisors of the GCD you found. You would then check each of these smaller common divisors to see if any of them are strictly greater than *both* remainders. If you find one that satisfies this condition, that would be your answer (the largest among those that satisfy it).

However, if *none* of the common divisors (including the GCD itself) are strictly greater than both remainders, then it means that no such number exists that can satisfy all the conditions of the problem simultaneously. This indicates that the problem statement itself might contain contradictory conditions, where the specified remainders are too large for any common divisor to fulfill the requirement.

Can there be multiple numbers that divide with the given remainders?

Yes, there can be multiple numbers that divide the original numbers leaving the specified remainders. For example, if we found that the common divisors of the adjusted numbers were, say, 1, 2, 4, 8, 16, and the remainders were 3 and 5, then we would need a divisor greater than 5. In this list, only 8 and 16 would be valid candidates. The question usually asks for the *largest* such number, which is why we focus on the GCD and its divisors.

The set of all numbers that divide N leaving a remainder r are the divisors of N-r. If we have two conditions, the numbers must be common divisors of (N1-r1) and (N2-r2). The set of common divisors is precisely the set of divisors of the GCD of (N1-r1) and (N2-r2). So, there are as many such numbers as there are divisors of the GCD, provided these divisors are greater than the respective remainders.

How can I be sure my answer is correct?

The best way to be sure your answer is correct is to perform a final verification. Once you have identified a candidate number (which should be the GCD of the adjusted numbers, provided it meets the remainder condition), plug it back into the original problem statement:

  1. Divide the first original number (e.g., 70) by your candidate divisor. Check if the remainder matches the first specified remainder (e.g., 5).
  2. Divide the second original number (e.g., 125) by your candidate divisor. Check if the remainder matches the second specified remainder (e.g., 8).
  3. Confirm that your candidate divisor is indeed the largest possible. This is guaranteed if you followed the steps correctly: you found the GCD of the adjusted numbers and ensured it's greater than the remainders.

For our original problem, the candidate is 13:

  • 70 ÷ 13 = 5 with a remainder of 5. (Correct)
  • 125 ÷ 13 = 9 with a remainder of 8. (Correct)

Since 13 is the GCD of 65 and 117 and is greater than both 5 and 8, it is indeed the largest such number.

This systematic approach—adjusting the numbers, finding the GCD, checking the remainder condition, and verifying—ensures accuracy and confidence in the solution.

Which is the largest number that divides 70 and 125 leaving remainder 5 and 8

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