Question

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B – A is perfectly divisible by 7, then how many possible values A can have ?

Answer 1

Step-by-step explanation:

Given The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B – A is perfectly divisible by 7, then how many possible values A can have ?  

• So let A = 100 m + 10 n + o
•       So B = 100 o + 10 n + m
• Now according to the question
• B – A is perfectly divisible by 7
• So B – A = 99 (o – m) / 7
• So (o – m) should also be divisible by 7.
• So o and m will have the values 8 and 1 or 9 and 2
• So 8 – 1 = 7 and 9 – 2 = 7, So y will have the values from 0 to 9
• Therefore lowest possible value of A will be 108 and the highest possible value of A is 299

Reference link will be

https://brainly.in/question/14281217

Answer 2

Lowest possible value of A will be 108 and the highest possible value of A is 299.

Step-by-step explanation:

Given: The digits of a three-digit number A are written in the reverse order to form another three-digit number B. B > A and B – A is perfectly divisible by 7.

Find: Possible values A can have.

Solution:

Three-digit number A can be written as = 100x + 10y + z where x, y and z are the 3 digits respectively.

B = 100z + 10y + x

Given B – A is perfectly divisible by 7.

B – A = 99x - 99z / 7 = 99 (x-z) / 7

 x - z should also be divisible by 7.

So x and z will have values 8 and 1 or 9 and 2, such that the difference is 7.

y can have any values from 0 to 9.

So the lowest possible value of A will be 108 and the highest possible value of A is 299.

The total combination can be 40 different numbers.