Find the value of \(5^3 + 6^3 + 7^3 + ... + 10^3\).
Answer: A
We use the formula for the sum of the cubes of the first n natural numbers: \(S_n = [\frac{n(n+1)}{2}]^2\).
The required sum is (Sum of cubes from 1 to 10) - (Sum of cubes from 1 to 4).
Sum of cubes from 1 to 10 = \([\frac{10(10+1)}{2}]^2 = (\frac{110}{2})^2 = 55^2 = 3025\).
Sum of cubes from 1 to 4 = \([\frac{4(4+1)}{2}]^2 = (\frac{20}{2})^2 = 10^2 = 100\).
The required sum = \(3025 - 100 = 2925\).
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How many numbers between 200 and 400 are divisible by 3, 4, and 5?
Answer: B
A number divisible by 3, 4, and 5 must be divisible by their LCM.
LCM(3, 4, 5) = \(3 \times 4 \times 5 = 60\).
We need to find the multiples of 60 between 200 and 400.
The multiples are: \(60 \times 4 = 240\), \(60 \times 5 = 300\), \(60 \times 6 = 360\).
There are 3 such numbers.
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What is the value of \((32 \times 12) + (17^2) - 95\)?
Answer: A
Follow the BODMAS rule. First, perform multiplication and find the square, then do addition and subtraction.
Step 1: \(32 \times 12 = 384\).
Step 2: \(17^2 = 289\).
Step 3: The expression becomes \(384 + 289 - 95\).
Step 4: \(384 + 289 = 673\).
Step 5: \(673 - 95 = 578\).
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The product of two numbers is 120 and the sum of their squares is 289. The sum of the two numbers is:
Answer: B
Let the numbers be x and y.
Given, xy = 120 and \(x^2 + y^2 = 289\).
We know that \((x+y)^2 = x^2 + y^2 + 2xy\).
Substitute the given values: \((x+y)^2 = 289 + 2(120) = 289 + 240 = 529\).
Therefore, x + y = \(\sqrt{529} = 23\).
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Evaluate \(123 \times 5 + 123 \times 15\).
Answer: A
Using the distributive property \(a \times b + a \times c = a \times (b+c)\).
Here, a = 123, b = 5, c = 15.
The expression is \(123 \times (5 + 15) = 123 \times 20\).
\(123 \times 20 = 2460\).
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The sum of two numbers is 25 and their product is 156. Find the smaller number.
Answer: A
Let the numbers be x and y. x + y = 25 and xy = 156.
We can form a quadratic equation: \(t^2 - (sum)t + (product) = 0\).
\(t^2 - 25t + 156 = 0\).
We need to find two numbers that multiply to 156 and add up to 25. These are 12 and 13.
\((t-12)(t-13) = 0\).
The numbers are 12 and 13. The smaller number is 12.
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How many times does the digit '3' appear in numbers from 1 to 100?
Answer: C
Let's count the occurrences of the digit '3'.
In the units place: 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 (10 times).
In the tens place: 30, 31, 32, 33, 34, 35, 36, 37, 38, 39 (10 times).
The number 33 is counted in both lists. So the total count is 10 + 10 = 20.
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If \(1^2 + 2^2 + 3^2 + ... + 10^2 = 385\), then the value of \(2^2 + 4^2 + 6^2 + ... + 20^2\) is:
Answer: C
Let the required sum be S.
\(S = 2^2 + 4^2 + 6^2 + ... + 20^2\).
\(S = (2 \times 1)^2 + (2 \times 2)^2 + (2 \times 3)^2 + ... + (2 \times 10)^2\).
\(S = 2^2(1^2) + 2^2(2^2) + 2^2(3^2) + ... + 2^2(10^2)\).
\(S = 2^2 (1^2 + 2^2 + 3^2 + ... + 10^2)\).
\(S = 4 \times (\text{given sum})\).
\(S = 4 \times 385 = 1540\).
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If \(x\) and \(y\) are the two digits of the number \(653xy\) such that this number is divisible by 80, then what is the value of \(x+y\)?
Answer: D
A number is divisible by 80 if it is divisible by both 10 and 8.
For the number to be divisible by 10, its last digit must be 0. So, \(y=0\).
The number now is \(653x0\).
For the number to be divisible by 8, the number formed by its last three digits must be divisible by 8. So, \(3x0\) must be divisible by 8.
Let's check values for x:
If x=2, the number is 65320. \(65320 \div 80 = 816.5\) (Not divisible).
If x=6, the number is 65360. \(65360 \div 80 = 817\) (Divisible).
So the number is 65360, which means \(x=6\) and \(y=0\).
The value of \(x+y = 6+0 = 6\).
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What is the unit digit of \(7^{105}\)?
Answer: C
The cyclicity of the unit digit of powers of 7 is 4. The pattern is 7, 9, 3, 1.
Divide the power by 4 and find the remainder.
105 ÷ 4 gives a remainder of 1.
If the remainder is 1, the unit digit is the first in the cycle, which is 7.
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