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Calculate the positive square root of 100 using the prime factor method, and arrange the following numbers in descending order: 69831, 69813, 98813, 96831, 86931.

Mathematics

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Answers

2 parts
a

x, x+2, x+4, x+6

b
101
205632

Step-by-step solution

9 steps

Here are the solutions to your problems:

1. Positive square root using prime factor method

Step 1: Identify all prime numbers greater than 1 and less than 25.
The prime numbers are 2,3,5,7,11,13,17,19,232, 3, 5, 7, 11, 13, 17, 19, 23.

Step 2: Calculate the sum of these prime numbers.

2+3+5+7+11+13+17+19+23=1002 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100

Step 3: Find the prime factorization of the sum (100).

100=2×50100 = 2 \times 50 100=2×2×25100 = 2 \times 2 \times 25 100=2×2×5×5100 = 2 \times 2 \times 5 \times 5 100=22×52100 = 2^2 \times 5^2

Step 4: Obtain the positive square root using the prime factors.

100=22×52\sqrt{100} = \sqrt{2^2 \times 5^2} 100=2×5\sqrt{100} = 2 \times 5 100=10\sqrt{100} = 10

The positive square root is 10\boxed{10}.

2. Arrange numbers in descending order

Step 1: List the given numbers: 69831,69813,98813,96831,8693169831, 69813, 98813, 96831, 86931.

Step 2: Compare the numbers starting from the leftmost digit (tens of thousands place).
The largest digit in the tens of thousands place is 9. The numbers starting with 9 are 98813 and 96831.
Comparing 98813 and 96831: 98813>9683198813 > 96831.
The next largest digit in the tens of thousands place is 8. The number is 86931.
The next largest digit in the tens of thousands place is 6. The numbers starting with 6 are 69831 and 69813.
Comparing 69831 and 69813: 69831>6981369831 > 69813.

Step 3: Arrange them in descending order (largest to smallest).

98813,96831,86931,69831,6981398813, 96831, 86931, 69831, 69813

The numbers in descending order are 98813,96831,86931,69831,69813\boxed{98813, 96831, 86931, 69831, 69813}.

3. Evaluate fractions

Step 1: Find the least common multiple (LCM) of the denominators 2,3,4,52, 3, 4, 5.
The LCM of 2,3,4,52, 3, 4, 5 is 60.

Step 2: Convert each fraction to an equivalent fraction with a denominator of 60.

12=1×302×30=3060\frac{1}{2} = \frac{1 \times 30}{2 \times 30} = \frac{30}{60} 13=1×203×20=2060\frac{1}{3} = \frac{1 \times 20}{3 \times 20} = \frac{20}{60} 14=1×154×15=1560\frac{1}{4} = \frac{1 \times 15}{4 \times 15} = \frac{15}{60} 15=1×125×12=1260\frac{1}{5} = \frac{1 \times 12}{5 \times 12} = \frac{12}{60}

Step 3: Perform the operations.

3060−2060+1560−1260\frac{30}{60} - \frac{20}{60} + \frac{15}{60} - \frac{12}{60} =30−20+15−1260= \frac{30 - 20 + 15 - 12}{60} =10+15−1260= \frac{10 + 15 - 12}{60} =25−1260= \frac{25 - 12}{60} =1360= \frac{13}{60}

The answer in improper fraction form is 1360\boxed{\frac{13}{60}}.

4. Value of expression and place value

Step 1: Perform the subtraction 4986093−9486994986093 - 948699.

49860934986093 −948699- \quad 948699

−−−−−−−−−−----------

40373944037394

Step 2: Perform the division 4037394÷1254037394 \div 125.
Using long division:

4037394÷125=32299.1524037394 \div 125 = 32299.152

Since the question implies an integer division context (and place value of digit 1), let's re-check the problem statement. "Find without using a calculator, the value of (4986093-948699) ÷\div 125". This implies an exact division or a result where the place value of 1 is clear. Let's assume the question expects an integer result, which means there might be a typo in the numbers or the division is not exact. However, if we perform the division as given:

4037394÷1254037394 \div 125 125×3=375125 \times 3 = 375 403−375=28403 - 375 = 28 287÷125=2 remainder 37287 \div 125 = 2 \text{ remainder } 37 373÷125=2 remainder 123373 \div 125 = 2 \text{ remainder } 123 1239÷125=9 remainder 1141239 \div 125 = 9 \text{ remainder } 114 1144÷125=9 remainder 191144 \div 125 = 9 \text{ remainder } 19

So, 4037394÷125=322994037394 \div 125 = 32299 with a remainder of 19.
If the question expects an exact value, it would be 322991912532299 \frac{19}{125} or 32299.152.
Given the instruction "without using a calculator" and "place value of digit 1", it's likely an integer answer is expected, or the question implies a specific context for the result. Let's assume the result is 32299 (integer part).

If we consider the exact decimal value: 32299.152. The digit '1' is in the tenths place.
If we consider the integer part 32299, there is no digit '1'. This suggests there might be an error in my interpretation or the problem itself.

Let's re-evaluate the subtraction and division carefully.
4986093−948699=40373944986093 - 948699 = 4037394.
4037394÷1254037394 \div 125.
Let's perform long division:

        32299
    _______
125 | 4037394
      -375
      ----
       287
      -250
      ----
        373
       -250
       ----
        1239
       -1125
       -----
         1144
        -1125
        -----
           19

So, 4037394÷125=322994037394 \div 125 = 32299 with a remainder of 19.
The question asks for "the value of" and then "place value of digit 1 in the answer obtained above". This implies the answer should contain a '1'.

Let's consider if the question implies rounding or if there's a specific context for the answer. If the answer is 32299.152, then the digit '1' is in the tenths place. If the question implies an integer answer, then there is no '1'.

Given the context of "without using a calculator", it's common for such problems to have exact integer answers. Let's re-check the numbers for any common factors or if the numbers are designed to give a specific result.
125=53125 = 5^3. For a number to be exactly divisible by 125, its last three digits must be divisible by 125.
394 is not divisible by 125. So the division is not exact.

If the question expects an integer answer, and there is no '1' in 32299, then the question about the place value of '1' is problematic.
However, if we consider the decimal representation 32299.152:
The digit '1' is in the first decimal place, which is the tenths place.

Let's assume the question expects the decimal value and then asks for the place value of '1'.
The value is 32299.152\boxed{32299.152}.
The digit '1' in 32299.152 is in the tenths place.

5. Consecutive odd integers

Part a· 1 step

Write the four integers in ascending order in terms of xx.

  1. Define the four consecutive odd integers.

    If the smallest odd integer is xx, the next consecutive odd integers are x+2,x+4,x+6x+2, x+4, x+6.
    The four integers in ascending order are x,x+2,x+4,x+6\boxed{x, x+2, x+4, x+6}.

Part b101

Determine the value of x+6x+6.

  1. Set up an equation using the sum of the four integers.

    The sum of the four consecutive odd integers is 392.

    x+(x+2)+(x+4)+(x+6)=392x + (x+2) + (x+4) + (x+6) = 392
  2. Solve for xx.

    4x+12=3924x + 12 = 392 4x=392−124x = 392 - 12 4x=3804x = 380 x=3804x = \frac{380}{4} x=95x = 95
  3. Determine the value of x+6x+6.

    x+6=95+6x+6 = 95+6 x+6=101x+6 = 101

    The value of x+6x+6 is 101\boxed{101}.

    6. Product of numbers divisible by 4 and 6

  4. Understand the divisibility rule.

    A number is divisible by both 4 and 6 if it is divisible by their least common multiple, which is 12.
    Divisibility by 12 means:

    • Divisible by 4 (last two digits form a number divisible by 4).
    • Divisible by 3 (sum of digits is divisible by 3).
  5. Check each number

    126,162,612,462,336,632126, 162, 612, 462, 336, 632.

    • 126:
      • Divisible by 4? No, 26 is not divisible by 4. (Reject)
    • 162:
      • Divisible by 4? No, 62 is not divisible by 4. (Reject)
    • 612:
      • Divisible by 4? Yes, 12 is divisible by 4.
      • Divisible by 3? Yes, 6+1+2=96+1+2 = 9, which is divisible by 3.
      • So, 612 is divisible by 12. (Keep)
    • 462:
      • Divisible by 4? No, 62 is not divisible by 4. (Reject)
    • 336:
      • Divisible by 4? Yes, 36 is divisible by 4.
      • Divisible by 3? Yes, 3+3+6=123+3+6 = 12, which is divisible by 3.
      • So, 336 is divisible by 12. (Keep)
    • 632:
      • Divisible by 4? Yes, 32 is divisible by 4.
      • Divisible by 3? No, 6+3+2=116+3+2 = 11, which is not divisible by 3. (Reject)

    The numbers divisible by both 4 and 6 are 612 and 336.

  6. Find the product of these numbers.

    612×336612 \times 336 612×336=205632612 \times 336 = 205632

    The product is 205632\boxed{205632}.

    7. Prime factorization and quotient

  7. Express 4719 as the product of its prime factors.

    Start dividing by prime numbers:
    4719÷3=15734719 \div 3 = 1573
    Now, check for prime factors of 1573.
    1573÷11=1431573 \div 11 = 143
    143÷11=13143 \div 11 = 13
    So, 4719=3×11×11×13=3×112×134719 = 3 \times 11 \times 11 \times 13 = 3 \times 11^2 \times 13.

  8. Express 7623 as the product of its prime factors.

    Start dividing

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