Question 1209946: 1. The sum of 4 integers is 24 and their product is 945. What are those integers?
2. Find the sum of all natural numbers between 500 and 1000 which are divisible by 13.
3. If the sum of three consecutive numbers of an AP is 15 and the sum of the squares of its 1st and 3rd terms is 58, find the numbers.
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve each problem step-by-step.
**1. Finding the 4 Integers**
* Let the 4 integers be $a, b, c, d$.
* We are given:
* $a + b + c + d = 24$
* $a \cdot b \cdot c \cdot d = 945$
* First, find the prime factorization of 945:
* $945 = 3^3 \cdot 5 \cdot 7 = 3 \cdot 3 \cdot 3 \cdot 5 \cdot 7$
* We need to find 4 integers that multiply to 945 and add to 24.
* Let's try different combinations:
* If we take 3, 5, 7, the remaining factor is 9. 3+5+7+9 = 24. 3*5*7*9 = 945.
* Thus, the integers are 3, 5, 7, and 9.
**2. Sum of Natural Numbers Divisible by 13**
* We need to find the sum of natural numbers between 500 and 1000 that are divisible by 13.
* The first number divisible by 13 greater than 500 is:
* $500 / 13 \approx 38.46$, so the first number is $39 \cdot 13 = 507$.
* The last number divisible by 13 less than 1000 is:
* $1000 / 13 \approx 76.92$, so the last number is $76 \cdot 13 = 988$.
* We have an arithmetic progression (AP) with:
* First term ($a_1$) = 507
* Common difference ($d$) = 13
* Last term ($a_n$) = 988
* To find the number of terms ($n$):
* $a_n = a_1 + (n - 1)d$
* $988 = 507 + (n - 1)13$
* $481 = (n - 1)13$
* $n - 1 = 481 / 13 = 37$
* $n = 38$
* To find the sum of the AP ($S_n$):
* $S_n = \frac{n}{2}(a_1 + a_n)$
* $S_{38} = \frac{38}{2}(507 + 988)$
* $S_{38} = 19(1495) = 28405$
**3. Consecutive Numbers in AP**
* Let the three consecutive numbers in AP be $a - d$, $a$, and $a + d$.
* We are given:
* $(a - d) + a + (a + d) = 15$
* $(a - d)^2 + (a + d)^2 = 58$
* From the first equation:
* $3a = 15$
* $a = 5$
* Substitute $a = 5$ into the second equation:
* $(5 - d)^2 + (5 + d)^2 = 58$
* $25 - 10d + d^2 + 25 + 10d + d^2 = 58$
* $50 + 2d^2 = 58$
* $2d^2 = 8$
* $d^2 = 4$
* $d = \pm 2$
* If $d = 2$, the numbers are $5 - 2$, $5$, $5 + 2$, which are 3, 5, 7.
* If $d = -2$, the numbers are $5 - (-2)$, $5$, $5 + (-2)$, which are 7, 5, 3.
**Summary**
1. The integers are 3, 5, 7, and 9.
2. The sum is 28405.
3. The numbers are 3, 5, 7.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
1. The sum of 4 integers is 24 and their product is 945. What are those integers?
Find that the prime factorization of 945 is 3*3*3*5*7. That's 5 factors; we need to combine 2 of them to express 945 as the product of 4 factors. Play with the numbers to find 3*5*7*9 = 945 and 3+5+7+9 = 24.
ANSWER: 3, 5, 7, and 9
2. Find the sum of all natural numbers between 500 and 1000 which are divisible by 13.
500/13 = 38.46... so 13*39 = 507 is the first number we are looking for.
1000/13 = 76.93... so 13*76 = 988 is the last one.
The sum we are to find is 13*(39+40+41+...+75+76)
The sum of an arithmetic sequence is (number of terms)*(average of terms).
Number of terms: (76-39)+1 = 38
Average of terms = average of first and last: (39+76)/2 = 115/2
The sum of the terms is 13(38*115/2) = 13*19*115 = 28405
ANSWER: 28405
3. If the sum of three consecutive numbers of an AP is 15 and the sum of the squares of its 1st and 3rd terms is 58, find the numbers.
Since the sum of 3 consecutive terms of an AP is 15, the middle number is 5. Do some quick calculations to find the other two numbers.
4 and 6? 4^2+6^2 = 16+36 = 52, not 58
3 and 7? 3^2+7^2 = 9+49 = 58. YES!
ANSWER: 3, 5, and 7
NOTE: Unlike the first problem, this problem has a relatively easy formal algebraic solution.
Again starting from the fact that the middle number is 5, let the other two numbers be 5-x and 5+x. The sum of the squares of those two numbers is 58:
And then the numbers are 5-x = 5-2 = 3, 5, and 5+x = 5+2 = 7.
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