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Question Number 186864 by mathlove last updated on 11/Feb/23

i f A B, B A a n d B B t h r e e t o w d i g i t s n a t u r a l

n u m b e r s i f \frac{(A B + B A)}{B B} = 4

t h e n f i n d t h e m a x i m u m v o l u e o f

(A + B) = ?

Answered by mr W last updated on 11/Feb/23

\frac{(A B + B A)}{B B} = \frac{11 (a + b)}{11 b} = \frac{a + b}{b} = 4

a = 3 b

b_{m a x} = 3

a_{m a x} = 9

\Rightarrow {(a + b)}_{m a x} = 12

Commented by mathlove last updated on 11/Feb/23

h o w i s B A = B B ?

Commented by mr W last updated on 11/Feb/23

w h o s a i d B A = B B ?

Commented by mathlove last updated on 11/Feb/23

y o u s a i d B A = b a n d B B = b

Commented by mathlove last updated on 11/Feb/23

p l e a s i a l i t t e l m o r e d e s c r i b y o u s o l u t i o n

Commented by mr W last updated on 11/Feb/23

i {d i d n}^{'} t s a y B A = b, B B = b .

s a y t h e t w o d i g i t s a r e a a n d b .

(i p r e f e r u s i n g l o w e r c a s e, a = A,

b = B)

A B = \underset{―}{a b} = 10 a + b

B A = \underset{―}{b a} = 10 b + a

A B + B A = 10 a + b + 10 b + a = 11 (a + b)

B B = \underset{―}{b b} = 10 b + b = 11 b

\frac{A B + B A}{B B} = \frac{11 (a + b)}{11 b} = \frac{a + b}{b} = 4

\Rightarrow a = 3 b

s i n c e a i s a d i g i t, i t c a n b e a t m o s t 9,

t h e r e f o r e b i s a t m o s t 3, a a t m o s t 9 .

b_{m a x} = 3, a_{m a x} = 9 .

Commented by mathlove last updated on 12/Feb/23

t h a n k s m r n o w u n d e r s t a n d

Answered by manxsol last updated on 12/Feb/23

A B = 10 A + B

B A = 10 B + A

B B = 10 B + B = 11 B

\frac{A B + B A}{B B} = \frac{10 A + B + 10 B + A}{11 B}

\frac{11 A + 11 B m}{11 B} = 4

\frac{A}{B} + 1 = 4

\frac{A}{B} = 3

A = 9 m a x

B = 3 m a x

A + B = 12

c o n v e n t i o n

n u m b e r t w o c i f r a s

\overset{―}{a b} = 10 a + b

h o w s c r i p t

a b [s e l e c t] [\leftarrow] [\leftarrow] [\overset{}{}] [-]

\overset{―}{a b}

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