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What are the solutions to these math problems concerning divisibility, number properties, and basic arithmetic?

Mathematics

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Here are the solutions to your questions:

  1. a) The number 2,135 is divisible by 5. b) Reason: A number is divisible by 5 if its last digit is 0 or 5. The last digit of 2,135 is 5.

  2. Step 1: Identify the greatest 8-digit number. The greatest 8-digit number is 99,999,999.

  1. Identify the smallest 9-digit number.

    The smallest 9-digit number is 100,000,000.
    
  2. Calculate the difference.

    $$
    100,000,000 - 99,999,999 = 1
    $$
    The difference is $\boxed{1}$.
    

    3. Given a=6a=6 and b=1b=1.

  3. Substitute the values of aa and bb into the expression.

    $$
    (a+b) \times 6 = (6+1) \times 6
    $$
    
  4. Perform the addition inside the parentheses.

    $$
    7 \times 6
    $$
    
  5. Perform the multiplication.

    $$
    42
    $$
    The value is $\boxed{42}$.
    

    4. a) Seven hundred and forty-five million, two hundred and three thousand, six hundred and twelve is written as 745,203,612\boxed{745,203,612}. b) Eighty-nine million, four thousand and seventy is written as 89,004,070\boxed{89,004,070}.

    1. a) 508,076,314 is written as Five hundred eight million, seventy-six thousand, three hundred fourteen. b) 300,000,009 is written as Three hundred million, nine.

    2. a) In 927,534,218, the digit 7 is in the millions place. Its total value is 7×1,000,000=7,000,0007 \times 1,000,000 = \boxed{7,000,000}. b) In 240,815,963, the digit 4 is in the ten millions place. Its total value is 4×10,000,000=40,000,0004 \times 10,000,000 = \boxed{40,000,000}.

    3. a) Round 489,736,214 to the nearest hundred million. The hundred millions digit is 4. The digit to its right is 8. Since 8≥58 \ge 5, we round up the 4. The rounded number is 500,000,000\boxed{500,000,000}. b) Round 72,361,983 to the nearest ten million. The ten millions digit is 7. The digit to its right is 2. Since 2<52 < 5, we keep the 7 as it is. The rounded number is 70,000,000\boxed{70,000,000}.

    4. a) The sequence is 2, 6, 18, .... The common ratio is 6÷2=36 \div 2 = 3 and 18÷6=318 \div 6 = 3. The next three numbers are: 18×3=5418 \times 3 = 54 54×3=16254 \times 3 = 162 162×3=486162 \times 3 = 486 The next three numbers are 54,162,486\boxed{54, 162, 486}. b) The sequence is 105, 95, 85, .... The common difference is 95−105=−1095 - 105 = -10 and 85−95=−1085 - 95 = -10. The next three numbers are: 85−10=7585 - 10 = 75 75−10=6575 - 10 = 65 65−10=5565 - 10 = 55 The next three numbers are 75,65,55\boxed{75, 65, 55}.

    5. The numbers are 308,419,621 and 381,429,216. Comparing the numbers, we look at the digits from left to right. The first digit (hundred millions place) is 3 for both. The second digit (ten millions place) is 0 for the first number and 8 for the second. Since 0<80 < 8, the first number is smaller. Arranged in increasing order: 308,419,621;381,429,216\boxed{308,419,621; 381,429,216}.

    6. a) Test 3456 by 2. A number is divisible by 2 if its last digit is an even number. The last digit of 3456 is 6, which is an even number. Therefore, 3456 is divisible by 2. b) Test 936 by 11. A number is divisible by 11 if the alternating sum of its digits is divisible by 11. For 936: 6−3+9=126 - 3 + 9 = 12. Since 12 is not divisible by 11, 936 is not divisible by 11.

    7. a) Express 84 as a product of prime factors.

  6. Divide 84 by the smallest prime number, 2.

    $$
    84 = 2 \times 42
    $$
    
  7. Divide 42 by 2.

    $$
    42 = 2 \times 21
    $$
    
  8. Divide 21 by the next smallest prime number, 3.

    $$
    21 = 3 \times 7
    $$
    
  9. Combine the prime factors.

    $$
    84 = 2 \times 2 \times 3 \times 7 = \boxed{2^2 \times 3 \times 7}
    $$
    
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