A two-digit number is seven times its units digit. The number is:
Answer: D
Write the number as \(10t + u = 7u\), which gives \(10t = 6u\) and so \(5t = 3u\). The only single digits that fit are \(t = 3\) and \(u = 5\), giving 35. Check: \(7 \times 5 = 35\). Testing each option against its own units digit is just as quick here.
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The digits of a two-digit number add up to 12, and the number formed by reversing them is 18 greater. The number is:
Answer: A
Let the number be \( 10a + b \) with \( a + b = 12 \).
Reversing adds 18:
\[ (10b + a) - (10a + b) = 9(b - a) = 18 \quad\Longrightarrow\quad b - a = 2 \]
\[ a = 5, \quad b = 7 \quad\Longrightarrow\quad 57 \]
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The product of the digits of a two-digit number is 24, and reversing the digits increases the number by 18. The number is:
Answer: B
Reversing adds 18, so \( 9(b-a) = 18 \) and \( b - a = 2 \); the digits also multiply to 24.
\[ a(a+2) = 24 \quad\Longrightarrow\quad a = 4, \ b = 6 \]
\[ \text{number} = 46 \]
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In a three-digit number the digits add to 16, the hundreds digit is three times the units digit, and the tens digit is one more than the units digit. The number is:
Answer: C
Let the units digit be \(u\). Then the hundreds digit is \(3u\) and the tens digit is \(u + 1\). Their sum is \(3u + (u+1) + u = 5u + 1 = 16\), so \(u = 3\), the tens digit is 4 and the hundreds digit is 9. The number is 943.
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The digits of a two-digit number add up to 8. Subtracting 18 from the number reverses its digits. The number is:
Answer: D
Subtracting 18 reverses the digits, so \( 9(a-b) = 18 \) and \( a - b = 2 \), while \( a + b = 8 \).
\[ a = 5, \quad b = 3 \quad\Longrightarrow\quad 53 \]
Checking: \( 53 - 18 = 35 \).
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The difference between a two-digit number and the number formed by reversing its digits is 45. The difference between its digits is:
Answer: A
\[ (10a + b) - (10b + a) = 9(a - b) = 45 \]
\[ a - b = 5 \]
The difference of a two-digit number and its reversal is always nine times the difference of the digits.
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The digits of a two-digit number have product 18 and sum 9. The larger of the two numbers that satisfy this is:
Answer: B
Two digits with sum 9 and product 18 are 3 and 6, since \(3 + 6 = 9\) and \(3 \times 6 = 18\). They can be arranged as 36 or 63, and the larger is 63. Treating the digits as roots of \(x^2 - 9x + 18 = 0\) finds them directly.
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The sum of a positive number and its reciprocal is \(\dfrac{10}{3}\). The greater of the possible values is:
Answer: C
From \(x + \dfrac{1}{x} = \dfrac{10}{3}\), multiplying through by \(3x\) gives \(3x^2 - 10x + 3 = 0\). Factoring, \((3x - 1)(x - 3) = 0\), so \(x = 3\) or \(x = \dfrac{1}{3}\). The two roots are always reciprocals of each other, and the greater is 3.
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A positive number decreased by 8 equals 128 times its reciprocal. The number is:
Answer: D
The condition is \(x - 8 = \dfrac{128}{x}\). Multiplying by \(x\) gives \(x^2 - 8x - 128 = 0\), which factors as \((x - 16)(x + 8) = 0\). The negative root is rejected because the number is positive, leaving \(x = 16\). Check: \(16 - 8 = 8\) and \(128 \div 16 = 8\).
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The product of two consecutive positive integers is 156. The smaller integer is:
Answer: A
Solve \(n(n+1) = 156\), so \(n^2 + n - 156 = 0\) and \((n + 13)(n - 12) = 0\), giving \(n = 12\). The integers are 12 and 13. A quicker route is to note that \(\sqrt{156} \approx 12.5\), so the pair must straddle that value.
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