How many ordered sets of positive integer triples (a,b,c) are there such that a+b*c=100
a + bc = 100
a = 100 - bc
form a table
a -- b -- c
99 --1--1
98-- 1--2
97-- 1--3
...
90-- 1--10 .......... 10 of them so far
98-- 2--1
96-- 2-- 2
94-- 2-- 3
..
80-- 2--10............10 more
....
90--10-- 1
80--10--2
...
10--10-- 9 .....we can have only 9 here, or else a = 0
so I count 99 of such triplets
The solution is about the number of divisors of each number from 1 to 99 and we will exclude 100 as he said they should be positive integers. so we will add them to get 473. A quick formula to calculate the number of divisors is for example 18=2^1 * 3^2 then no.of divisors will be (1+1)(2+1)=6
So Ans is 473
To find the number of ordered sets of positive integer triples (a, b, c) that satisfy the equation a + b*c = 100, we can employ a method called casework.
First, let's analyze some possible values for b*c that could sum up to 100. Since a and b are positive integers, b*c must also be a positive integer. So, we need to find all the positive integer factors of 100.
The factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, and 100.
Now, we can examine each factor and determine the number of ways to express it as b*c:
1. b*c = 1: In this case, a = 100 - b*c = 99. There is only one solution (99, 1, 1).
2. b*c = 2: In this case, a = 100 - b*c = 98. There are two solutions (98, 1, 2) and (98, 2, 1).
3. b*c = 4: In this case, a = 100 - b*c = 96. There are three solutions (96, 1, 4), (96, 2, 2), and (96, 4, 1).
4. b*c = 5: In this case, a = 100 - b*c = 95. There are four solutions (95, 1, 5), (95, 5, 1), (95, 5, 1), and (95, 1, 5).
5. b*c = 10: In this case, a = 100 - b*c = 90. There are five solutions (90, 1, 10), (90, 2, 5), (90, 5, 2), (90, 10, 1), and (90, 10, 1).
6. b*c = 20: In this case, a = 100 - b*c = 80. There are six solutions (80, 1, 20), (80, 2, 10), (80, 4, 5), (80, 5, 4), (80, 10, 2), and (80, 20, 1).
7. b*c = 25: In this case, a = 100 - b*c = 75. There are seven solutions (75, 1, 25), (75, 5, 5), (75, 5, 5), (75, 25, 1), (75, 25, 1), (75, 5, 5), and (75, 1, 25).
8. b*c = 50: In this case, a = 100 - b*c = 50. There are eight solutions (50, 1, 50), (50, 2, 25), (50, 5, 10), (50, 10, 5), (50, 25, 2), (50, 25, 2), (50, 10, 5), and (50, 1, 50).
9. b*c = 100: In this case, a = 100 - b*c = 0. There is only one solution (0, 1, 100).
Therefore, the total number of ordered sets of positive integer triples (a, b, c) that satisfy the equation a + b*c = 100 is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 1 = 37.