How many ordered sets of positive integer triples (a,b,c) are there such that a+b*c=100

Answer 1

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

Answer 2

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