Number Sums and Series

Let the numbers be $x$ and $x+4$. The equation $1/x + 1/(x+4) = 10/21$ simplifies to $10x^2 + 40x = 42x + 84$, or $5x^2 - x - 42 = 0$. Solving gives $x=3$. Thus the numbers are 3 and 7. Option 20964 is correct.

Let the numbers be x-1, x, x+1. Sum: 3x = 15, so x = 5. The middle number is 5, and its square is 25.

Let the numbers be x and x+7. 6x + (x+7) = 77 => 7x = 70 => x = 10. The numbers are 10 and 17.

Even numbers between 1 and n are 2, 4, 6, ..., (n-1). Sum = 2+4+6+...+(n-1) = 2(1+2+3+...+(n-1)/2). Number of terms = (n-1)/2. Sum = 2 × [(n-1)/2 × ((n-1)/2 + 1)] / 2 = (n-1)/2 × (n+1)/2 = (n²-1)/4. Given (n²-1)/4 = 79×80, so n²-1 = 31680, n² = 31681, n = 177. But wait - let's recalculate: sum of first k even numbers = k(k+1). Here k = (n-1)/2, so sum = (n-1)/2 × (n+1)/2 = 79×80. Solving: (n²-1)/4 = 6320, n²-1 = 25280, n² = 25281, n = 159.

If the product of two integers is odd, then both integers must be odd (since even × anything = even). The sum of two odd numbers is always even: odd + odd = even. For example: 3 × 5 = 15 (odd), and 3 + 5 = 8 (even). The other options are not necessarily true.

To solve this question, the user needs to know that an odd number can only be obtained by adding an odd number of odd numbers. Therefore, the sum of two integers whose product is odd must be an even number.

Now, let's go through each option and explain why it is right or wrong:

A. Odd number: This option is incorrect because the sum of two integers whose product is odd cannot be odd. Adding two odd numbers together results in an even number.

B. Even number: This option is correct. The sum of two integers whose product is odd must be an even number. This is because the product of two odd numbers is odd, and odd + odd = even.

C. Prime number: This option is incorrect because the sum of two integers whose product is odd has nothing to do with prime numbers. Prime numbers are numbers that are only divisible by 1 and themselves.

D. Divisible by difference of two numbers: This option is incorrect because there is no relation between the sum of two integers whose product is odd and the difference of the two numbers.

E. Perfect square: This option is incorrect because there is no relation between the sum of two integers whose product is odd and perfect squares.

Therefore, the answer is:

The Answer is: B. Even number

Let the integers be n, n+1, n+2. Their sum is 3n+3. For 3(n+1) to be odd, n+1 must be odd, meaning n must be even. If n is even, then n+2 is also even. Thus, the first and last integers must both be even.

The series is of squares: 1, 4, 9... n^2. The sum of the first n squares is [n(n+1)(2n+1)]/6. For n=25: (25 * 26 * 51) / 6 = 25 * 13 * 17 = 5525. Since 17 is a prime factor in the formula (2n+1 = 51, which is 3*17), 17 is a factor.

96 = 2^5 × 3, so even divisors are 2, 4, 6, 8, 12, 16, 24, 32, 48, 96 (sum = 248). 3600 = 2^4 × 3^2 × 5^2, so odd factors are powers of 3 × 5: (1+3+9) × (1+5+25) = 13 × 31 = 403. Total = 248 + 403 = 651.

The sum of the first n positive integers is given by the formula n(n+1)/2. For n=9: 9(10)/2 = 90/2 = 45. This is a standard arithmetic series formula.

Let the 5 numbers in AP be a-2d, a-d, a, a+d, a+2d. Their sum is 5a = 30, giving a = 6. The third term (middle term) is a = 6. Verifying with sum of squares: (a-2d)² + (a-d)² + a² + (a+d)² + (a+2d)² = 5a² + 10d² = 180 + 10d² = 220, so d² = 4. The third term remains 6 regardless of d.

Let the four consecutive numbers be x, x+1, x+2, x+3. Their sum is 4x + 6 = 154, so 4x = 148 and x = 37. Checking: 37 + 38 + 39 + 40 = 154. This confirms 37 is the smallest number.

Basic addition: 1 + 1 = 2. The other options like 11 are results of string concatenation rather than mathematical addition.

Basic addition: 2 + 2 = 4. The other options like 22 or 44 are either concatenations or incorrect multiplications.