Question

the sum of the digits of a three digit number is 10 the tens digit in three times the hundred digit and the ones digit is reversed the number is increased by 495 find the number​

Answer 1

Answer:

Let hundred 's digit is a, team's digit is b and unit's digit is c.

a+b+c=12 ………(1)

The number is 100a+10b+c

Reversed number =100c+10b+a

100c+10b+a-495=100a+10b+c

Or, 99c-99a =495 , c-a =5…….(2)

Again another reversed number = 100a+10c+b

Or, 100a+10c+b-36=100a+10b+c

Or, 9c-9b =36, c-b =4……….(3)

From(2) a=c-5, from (3) b=c-4

From(1) a+b+c=12,

So, c-5+c-4+c =12,→ 3c=21,→c=7

So a=c-5,→a=7–5, → a=2

And, b=c-4,→b=7–4,→ b=3.

The number is 237 ans

Step-by-step explanation:

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Answer 2

Answer:

237 is the correct answer