How-many-3-digited-numbers-which-are-divisible-by-1-3-2-4-3-5-4-6-5-7-6-8-7-9-with-repetetion-of-digits-is-NOT-allowed-one-problem-process-

Question Number 177309 by SLVR last updated on 03/Oct/22

H o w m a n y 3 d i g i t e d n u m b e r s

w h i c h a r e d i v i s i b l e b y 1) 3

2) 4 3) 5 4) 6 5) 7 6) 8 7) 9

w i t h r e p e t e t i o n o f d i g i t s i s

N O T a l l o w e d \dots o n e p r o b l e m p r o c e s s

Commented by SLVR last updated on 03/Oct/22

i n e e d f o r d i v i s i b l e b y 7

Answered by mahdipoor last updated on 04/Oct/22

N = (a b c) = 100 a + 10 b + c

0 ⩽ a, b, c \in N ⩽ 9 a \neq 0

\Rightarrow N \overset{7}{\equiv} 0 a n d r e p e t e t i o n o f d i g i t s

\Rightarrow 100 a + 10 b + c \overset{7}{\equiv} 2 a + 3 b + c \overset{7}{\equiv} 0

i) a = b \Rightarrow c \overset{7}{\equiv} 2 a \Rightarrow 13 n u m b e r s

i i) a = c \Rightarrow b \overset{7}{\equiv} 6 a \Rightarrow 12

i i i) b = c \Rightarrow a \overset{7}{\equiv} 5 b \Rightarrow 12

i v) a = b = c \Rightarrow 6 a \overset{7}{\equiv} 0 \Rightarrow 1

\Rightarrow a l l n u m b e r s = 13 + 12 + 12 - 2 \times 1 = 35

3 d i g i t n u m b e r a r e d i v i s i b l e b y 7 :

[\frac{999}{7}] - [\frac{99}{7}] = 128

\Rightarrow 128 - 35 = 93

Commented by SLVR last updated on 04/Oct/22

T h a n k s f o r y o u r c o n c e r n s i r

b u t h o w 4 t h l i n e 2 a + 3 b + c \equiv 0 m o d 7

a n d h o w 13 n u m b e r s i f a = b

Commented by SLVR last updated on 04/Oct/22

k i n d l y e x p l a i n w h a t t h o s e 34

n u m b e r s b e d e d u c t e d . . p l e a s e

Commented by mr W last updated on 04/Oct/22

i i) a = c

2 a + 3 b + c = 3 (a + b) \overset{7}{\equiv} 0

a + b = 7 \Rightarrow 7

a + b = 14 \Rightarrow 5

t o t a l 12 n o t 11 ?

i c o u n t e d t o t a l l y 93 v a l i d n u m b e r s .

Commented by mr W last updated on 04/Oct/22

Commented by mahdipoor last updated on 04/Oct/22

a n s w e r t o S L V R :

i >

100 a + 10 b + c \overset{7}{\equiv} (98 a + 7 b) + (2 a + 3 b + c) \overset{7}{\equiv}

2 a + 3 b + c 7 ∣ 98 a + 7 b = 7 (14 a + b)

i i > f o r e x a m p l e :

a = c \Rightarrow 3 (a + b) \overset{7}{\equiv} (a + b) \overset{7}{\equiv} 0 \Rightarrow

b \overset{7}{\equiv} - a + 7 a \Rightarrow b \overset{7}{\equiv} 6 a 0 ⩽ a, b \in N ⩽ 9 a \neq 0

(a, b) = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (5, 9),

(6, 1), (6, 8), (7, 0), (7, 7), (8, 6), (9, 5)

\Rightarrow 12 n u m b e r

a n s w e r t o M r . W :

t h x, i e d i t e d .