Q: how many natural numbers less than 6000 , exist such as the sum of their digits equal to 8 ? choices: 155 165 164 158


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Question Number 172802 by mnjuly1970 last updated on 01/Jul/22

$Q : how many natural numbers less than$ $6000, exist such as the sum of$ $their digits equal to 8 ?$ $choices : 155 165 164 158$
	Answered by mr W last updated on 02/Jul/22

	$\underset{―}{1 - d i g i t n u m b e r s :}$ $8$ $\Rightarrow 1 n u m b e r$ $\underset{―}{2 - d i g i t n u m b e r s :}$ $17 / 71, 26 / 62, 35 / 53, 44, 80$ $\Rightarrow 8 n u m b e r s$ $\underset{―}{3 - d i g i t n u m b e r s :}$ $a b c w i t h a + b + c = 8 a n d$ $a ⩾ 1 a n d b, c ⩾ 0$ $(x + x^{2} + . . .) {(1 + x + x^{2} + . . .)}^{2} = x \underset{k = 0}{\sum^{\infty}} C_{2}^{k + 2} x^{k}$ $\Rightarrow C_{2}^{7 + 2} = 36 n u m b e r s$ $\underset{―}{4 - d i g i t n u m b e r s :}$ $a b c d w i t h a + b + c + d = 8 a n d$ $1 ⩽ a ⩽ 5 a n d b, c, d ⩾ 0$ $(x + x^{2} + x^{3} + x^{4} + x^{5}) {(1 + x + x^{2} + . . .)}^{3}$ $= x (1 - x^{5}) \underset{k = 0}{\sum^{\infty}} C_{3}^{k + 3} x^{k}$ $\Rightarrow C_{3}^{7 + 3} - C_{3}^{2 + 3} = 110 n u m b e r s$ $\underset{―}{t o t a l l y :}$ $1 + 8 + 36 + 110 = 155 n u m b e r s e x i s t . ✓$ $o r$ $(1 + x + x^{2} + x^{3} + x^{4} + x^{5}) {(1 + x + x^{2} + . . .)}^{3}$ $= (1 - x^{6}) \underset{k = 0}{\sum^{\infty}} C_{3}^{k + 3} x^{k}$ $\Rightarrow C_{3}^{8 + 3} - C_{3}^{2 + 3} = 155 n u m b e r s$
	Commented by Tawa11 last updated on 01/Jul/22

	$Great sir$
	Commented by mr W last updated on 02/Jul/22

	$y o u m e a n t h e a n s w e r i s c o r r e c t M i s s ?$
	Commented by mahdipoor last updated on 02/Jul/22
	$abcd with a+b+c+d=8 and a,b,c,d≥0 ⇒C_(k−1) ^( n+k−1) =C_3 ^( 11) =165 abcd with a+b+c+d=8 and a,b,c,d≥0 and bigger than 6000 : { ((a=6 ⇒ b+c+d=2 ⇒C_2 ^( 4) =6)),((a=7 ⇒ b+c+d=1 ⇒C_2 ^( 3) =3)),((a=8 ⇒ b+c+d=0 ⇒C_2 ^( 2) =1)) :} ⇒6+3+1=10 165−10=155$
	$a b c d w i t h a + b + c + d = 8 a n d a, b, c, d ⩾ 0$ $\Rightarrow C_{k - 1}^{n + k - 1} = C_{3}^{11} = 165$ $a b c d w i t h a + b + c + d = 8 a n d a, b, c, d ⩾ 0$ $a n d b i g g e r t h a n 6000 :$ ${\begin{cases} a = 6 \Rightarrow b + c + d = 2 \Rightarrow C_{2}^{4} = 6 \\ a = 7 \Rightarrow b + c + d = 1 \Rightarrow C_{2}^{3} = 3 \\ a = 8 \Rightarrow b + c + d = 0 \Rightarrow C_{2}^{2} = 1 \end{cases}$ $\Rightarrow 6 + 3 + 1 = 10$ $165 - 10 = 155$
	Commented by peter frank last updated on 02/Jul/22

	$thanks$
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