What number should be added to 1385 to make it divisible by 25?
Answer: C
A number is divisible by 25 if its last two digits form a number divisible by 25 (i.e., 00, 25, 50, or 75).
The last two digits of 1385 are 85.
The next multiple of 25 after 85 is 100. To get from 85 to 100, we need to add 15.
So, we must add 15 to the number (1385 + 15 = 1400).
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The sum of four consecutive odd numbers is 112. The smallest number is:
Answer: A
Four consecutive odd numbers are symmetric about their average.
\[ \frac{112}{4} = 28 \]
The four numbers straddle 28 in steps of 2:
\[ 25,\ 27,\ 29,\ 31 \]
So the smallest is 25. Setting up \( x + (x+2) + (x+4) + (x+6) = 112 \) gives \( 4x + 12 = 112 \) and the same answer.
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Find the value of \(0.125 + 0.027\) / \(0.25 - 0.15 + 0.09\).
Answer: D
The numerator is in the form \(a^3+b^3\) and the denominator is in the form \(a^2-ab+b^2\). Let a=0.5 and b=0.3. Then \(a^3=0.125\) and \(b^3=0.027\). Denominator is \(a^2-ab+b^2 = 0.5^2 - (0.5)(0.3) + 0.3^2 = 0.25 - 0.15 + 0.09\). This matches.
We know that \(a^3+b^3 = (a+b)(a^2-ab+b^2)\).
So the expression simplifies to \(a+b\).
Value = 0.5 + 0.3 = 0.8.
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Find the total number of prime factors in the expression \(4^{11} \times 7^5 \times 11^2\).
Answer: A
First, express all bases as prime numbers.
\(4^{11} = (2^2)^{11} = 2^{22}\).
The expression is \(2^{22} \times 7^5 \times 11^2\).
The total number of prime factors is the sum of the exponents of the prime bases.
Total prime factors = 22 + 5 + 2 = 29.
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