Natural numbers

Consecutive numbers, even and odd, digits, powers and the number of digits. 5 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution.

Problem 1 Consecutive numbers

a) The sum of four consecutive natural numbers is 74. Find the numbers.

b) The sum of three consecutive odd numbers is 87. Find the numbers.

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a) \(n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 74\), \(n = 17\): 17, 18, 19, 20.

b) \((2k - 1) + (2k + 1) + (2k + 3) = 6k + 3 = 87\), \(k = 14\): 27, 29, 31.

Problem 2 Even and odd

a) Prove that the product of two odd numbers is odd.

b) \(3(4a + b)\) is odd and \(5a + 7\) is even. Is \(a\) even or odd? And \(b\)?

c) \(a\) is odd and \(b\) is even. Is \(3(2a - b) + 5a + 1\) even or odd?

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a) \((2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1\): odd.

b) \(5a + 7\) even \(\Rightarrow\) \(5a\) odd \(\Rightarrow\) \(a\) odd. \(3(4a + b)\) odd \(\Rightarrow\) \(4a + b\) odd \(\Rightarrow\) \(b\) odd (since \(4a\) is even).

c) \(3(2a - b)\) is even, \(5a\) is odd, so even + odd + 1 = even.

Problem 3 Digits

a) \(\overline{ab}\), \(\overline{ba}\) and \(\overline{c3}\) are two-digit numbers with \(\overline{ab} + \overline{ba} = \overline{c3}\). Find \(a + b + c\).

b) When the digits of a two-digit number are reversed, the number increases by 45. The sum of its digits is 9. Find the number.

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a) \(\overline{ab} + \overline{ba} = 11(a + b)\). A two-digit multiple of 11 ending in 3 is \(33\), so \(a + b = 3\) and \(c = 3\): \(a + b + c = 6\).

b) \(\overline{ba} - \overline{ab} = 9(b - a) = 45\), so \(b - a = 5\); with \(a + b = 9\): \(a = 2\), \(b = 7\). The number is 27.

Problem 4 Powers

a) Evaluate \(2^5\cdot 3^2\), \(\dfrac{6^4}{2^3\cdot 3^3}\) and \(\dfrac{4^3\cdot 8^2}{2^{10}}\).

b) \(a = 3^x\). Write \(9^{x + 1}\) and \(27^{x - 1}\) in terms of \(a\).

c) \(m = 2^n\). Write \(8^{n + 1}\) in terms of \(m\).

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a) \(32\cdot 9 = 288\); \(\frac{2^4\cdot 3^4}{2^3\cdot 3^3} = 6\); \(\frac{2^6\cdot 2^6}{2^{10}} = 4\).

b) \(9^{x + 1} = 9\cdot(3^x)^2 = 9a^2\); \(27^{x - 1} = \frac{(3^x)^3}{27} = \frac{a^3}{27}\).

c) \(8^{n + 1} = 8\cdot(2^n)^3 = 8m^3\).

Problem 5 Number of digits

How many digits does each number have? a) \(4^{12}\cdot 5^{20}\) b) \(25^8\cdot 8^6\cdot 3\)

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a) \(2^{24}\cdot 5^{20} = 2^4\cdot 10^{20} = 16\cdot 10^{20}\): 22 digits.

b) \(5^{16}\cdot 2^{18}\cdot 3 = 2^2\cdot 3\cdot 10^{16} = 12\cdot 10^{16}\): 18 digits.