Question 1209946
<br>
1. The sum of 4 integers is 24 and their product is 945. What are those integers?<br>
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.<br>
ANSWER: 3, 5, 7, and 9<br>
2. Find the sum of all natural numbers between 500 and 1000 which are divisible by 13.<br>
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.<br>
The sum we are to find is 13*(39+40+41+...+75+76)<br>
The sum of an arithmetic sequence is (number of terms)*(average of terms).<br>
Number of terms: (76-39)+1 = 38
Average of terms = average of first and last: (39+76)/2 = 115/2<br>
The sum of the terms is 13(38*115/2) = 13*19*115 = 28405<br>
ANSWER: 28405<br>
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.<br>
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.<br>
4 and 6? 4^2+6^2 = 16+36 = 52, not 58<br>
3 and 7? 3^2+7^2 = 9+49 = 58.  YES!<br>
ANSWER: 3, 5, and 7<br>
NOTE: Unlike the first problem, this problem has a relatively easy formal algebraic solution.<br>
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:<br>
{{{(5-x)^2+(5+x)^2=58}}}
{{{25-10x+x^2+25+10x+x^2=58}}}
{{{50+2x^2=58}}}
{{{2x^2=8}}}
{{{x^2=4}}}
{{{x=2}}}<br>
And then the numbers are 5-x = 5-2 = 3, 5, and 5+x = 5+2 = 7.<br>