Math, asked by niveditatuli1941, 6 months ago

The integers a, b and c are all positive.
Given that a + b + c = 14 and 156a + 13 b + c = 873, find 100a + 10b + c.

Answers

Answered by abhi569
6

Answer:

572

Step-by-step explanation:

As we know, each 3 digit number(in form of abc) can be written as 100a + 10b + c.

So assuming that we have to find the same and a, b, c are real positive integers.

156a + 13b + c = 873, 156 & 13 are multiples of 13, but c is not. It means the RHS had to be a multiple of 13 but due to c it got increased by + c.

Nearest(to 873) multiple of 13 is 871. So we can say,

=> 156a + 13b + c = 871 + 2,

=> 13(12a + b) + c = 13(67) + 2

=> c = 2

Now, compare 12a + b & 67:

=> 12a + b = 67, RHS had to be a multiple of 12 but due to b it got increased by + b. Nearest(to 67) multiple of 12 is 60. So we can say,

=> 12a + b = 60 + 7

=> 12a = 60 & b = 7

It means, a = 60/12 = 5 & b = 7 & c = 2.

It means 100a + 10b + c= 100(5) + 10(7) + 2

= 572

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