Number Sums and Series

The digit 6 appears 20 times in numbers from 1 to 100 (10 times in the units place and 10 times in the tens place). Replacing each 6 with a 9 increases its value by 3. Thus, the sum increases by $10 \times 3 + 10 \times 30 = 330$.

Sum of squares of first n natural numbers is n(n+1)(2n+1)/6. For n=9: 9*10*19/6 = 285. Sum of first n is n(n+1)/2. For n=9: 9*10/2 = 45. 285 / 45 = 19/3.

Let the terms be a/r, a, ar. Product: a^3 = 1728, so a = 12. Sum: 12/r + 12 + 12r = 38. Solving 12/r + 12r = 26 gives 6r^2 - 13r + 6 = 0. Roots are 3/2 and 2/3. The terms are 8, 12, 18. Smallest is 8.

This is the classic Sum and Product Puzzle. The unique solution for the sum being less than 100 and the conversation holding true is $m = 4$ and $n = 13$, where their sum is 17 and product is 52. Other options fail the logical constraints of the mathematicians' statements.

Let the numbers be x and y. x - y = 5 and xy = 336. Using the identity (x + y)^2 = (x - y)^2 + 4xy, we get (x + y)^2 = 5^2 + 4(336) = 25 + 1344 = 1369. Taking the square root of 1369 gives 37. The numbers are 21 and 16.

Use the arithmetic series sum formula: n(n+1)/2. For n=26: 26(27)/2 = 351. This elegant formula (attributed to Gauss) pairs numbers from opposite ends: (1+26)+(2+25)+... = 27, with 13 such pairs. 27 × 13 = 351.

To find the sum of the numbers from 1 to 36, there is a formula that can be used. This formula is known as the arithmetic series formula:

$$S = \frac{n}{2}(a + l)$$

Where:

• S is the sum of the series
• n is the number of terms in the series
• a is the first term of the series
• l is the last term of the series

In this case, we have:

• n = 36 (as we are adding the numbers from 1 to 36)
• a = 1 (the first term of the series)
• l = 36 (the last term of the series)

Substituting these values into the formula:

$$S = \frac{36}{2}(1 + 36)$$

$$S = 18(37)$$

$$S = 666$$

Hence, the sum of the numbers from 1 to 36 is 666.

Therefore, the correct answer is: C. 666

Let the numbers be a and b. We have ab = 120 and a² + b² = 289. Then (a+b)² = a² + 2ab + b² = 289 + 2(120) = 289 + 240 = 529. Therefore, a + b = √529 = 23. Option B is correct.

Let the consecutive even numbers be x and x+2. The difference of squares is (x+2)² - x² = 4x + 4 = 84. Solving: 4x = 80, so x = 20. The numbers are 20 and 22. Their sum = 20 + 22 = 42. Option C is correct.

Let first number = x ≤ 8, second number = y. We have 2x + 3y < 24. To maximize y, minimize x. When x = 1: 2 + 3y < 24 → 3y < 22 → y ≤ 7 (max integer). Verify: 2(1) + 3(7) = 23 < 24 ✓

Let consecutive odd numbers be n and n+2. Then (n+2)²-n²=48, giving 4n+4=48, so n=11. The numbers are 11 and 13, with sum 24. This uses the algebraic identity (x+y)²-(x-y)²=4xy.

For 5 consecutive odd numbers, they can be represented as x, x+2, x+4, x+6, x+8. Sum = 5x + 20 = 1075, so 5x = 1055 and x = 211. The numbers are 211, 213, 215, 217, 219. The largest is 219. Alternatively, for 5 consecutive numbers with same difference, the middle number (third) equals average, so middle = 1075/5 = 215, making the largest 215+4 = 219.

Let numbers be x and y with x - y = 5 and xy = 336. Then (x + y)² = (x - y)² + 4xy = 25 + 1344 = 1369, so x + y = √1369 = 37. Options A (21), B (51), and C (28) don't satisfy this relationship. This uses the identity (x+y)² = (x-y)² + 4xy.

The SUM function in SAS with the 'of' syntax sums a range of variables. SUM(of X1-X3) sums variables X1, X2, and X3. With X1=9, X2=4, X3=1, the sum is 9+4+1=14. The SUM function also handles missing values gracefully (treating them as 0). Option C is correct.

SUM(OF X1-X3) sums the variables X1, X2, X3 (values 9, 4, 1). 9+4+1 = 14. The SUM function handles missing values differently than simple addition, though here all values are present.