Question

$ba \times b3 = 57a$
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Answer 1

Question:

In the product BA×B3=57A, what are the respective positional values of B and A?

Answer:

BA×B3=57A

as B is in 10′ s place

⇒(10×B+A)×(10B+3)=(57×10)+A

⇒100B² +10(A×B)+30B+3A=570+A

⇒100B

2

+10(A×B)+30B+2A=570

⇒2A=570−100B² −10(A×B)−30B

⇒10/2A as 10 divides RHS

⇒A=5

or A=0

for A=0, 100B² +30B=570

⇒10B² +3B=50+7

sno B satisfies this as

3B=7

for A=5,100B² +80B+10=570

⇒10B² +8B=56

⇒5B² +4B=28=20+8

⇒B=2

∴B=2,A=5