Let a be the first, b be second and c be the third whole number.
Since the sum of these three numbers is 100.
So, [tex] a+b+c=100 [/tex] (equation 1)
Since, The first number is a multiple of 15
Therefore, a = 15n
And, the second number is ten times the third number.
b = 10c
Substituting the values of 'a' and 'b' in equation 1
So, [tex] 15n+10c+c=100 [/tex]
[tex] 15n+11c = 100 [/tex]
[tex] 11c = 100-15n [/tex]
[tex] c=\frac{100-15n}{11} [/tex]
Therefore, 100-15n should be exactly divisible by 11.
So, by taking n= 1 and 2, 100-15n is not divisible by 11
Let n =3
[tex] c=\frac{100-15 \times 3}{11} [/tex]
c= 5
Now, second number (b)= 10c = [tex] 10 \times 5= 50 [/tex]
As, a+b+c=100
a+50+5=100
a+55=100
a=45
Therefore, the three whole numbers are 45,50,5.