How-many-natural-numbers-less-than-1000-have-the-sum-of-their-digits-equal-to-5-

Question Number 111541 by Aina Samuel Temidayo last updated on 04/Sep/20

How many natural numbers less than

1000 have the sum of their digits equal

to 5 ?

Answered by nimnim last updated on 04/Sep/20

21 .

Answered by mr W last updated on 04/Sep/20

o n e d i g i t n u m b e r s :

1 n u m b e r

t w o d i g i t n u m b e r s :

a + b = 5

a \in [1, 9]

b \in [0, 9]

(x + x^{2} + x^{3} + \dots) (1 + x + x^{2} + x^{3} + \dots)

= \frac{x}{{(1 - x)}^{2}}

= x \underset{k = 0}{\sum^{\infty}} C_{1}^{k + 1} x^{k}

c o e f . o f x^{5} i s C_{1}^{5} = 5

\Rightarrow 5 n u m b e r s

t h r e e d i g i t n u m b e r s :

(x + x^{2} + x^{3} + \dots) {(1 + x + x^{2} + x^{3} + \dots)}^{2}

\frac{x}{{(1 - x)}^{3}} = x \underset{k = 0}{\sum^{\infty}} C_{2}^{k + 2} x^{k}

c o e f . o f x^{5} i s C_{2}^{6} = 15

\Rightarrow 15 n u m b e r s

\Rightarrow t o t a l 1 + 5 + 15 = 21 n u m b e r s

Commented by mr W last updated on 04/Sep/20

s e e a l s o Q 102816

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

For the two digit numbers and

three digit numbers, please how did you get

(x + x^{2} + x^{3} + \dots) (1 + x^{2} + x^{3} + \dots) and

(x + x^{2} + x^{3} + \dots) {(1 + x^{2} + x^{3} + \dots)}^{2} respectively ?

Commented by mr W last updated on 04/Sep/20

Answered by 1549442205PVT last updated on 04/Sep/20

The numbers have three digits :

5 = 5 + 0 + 0 = 4 + 1 + 0 = 3 + 2 + 0 = 3 + 1 + 1

= 2 + 2 + 1

(5, 0, 0) : one number : 500

(4, 1, 0) : four numbers : 410, 401, 104, 140

(3, 2, 0) : four numbers : 320, 302, 203, 230

(3, 1, 1) : three nimbers : 311, 131, 113

(2, 2, 1) : three numbers : 212, 221, 122

The numbers have two digits :

50, 41, 14, 32, 23

The numbers have one digits : 5

Thus, all has 21 numbers smaller than

1000 sum their digits whose equal to 5