# There is a number which is very peculiar this number is three times the sum of its digits .can you find the number

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by luca

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by luca

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And 10th digit of the number = y

therefore number (N) = 10y+x

or, N/x= 10(y/x)+1 ---------(1)

A/q,

3(x+y) = 10y+x -----------------(2)

or, 3x+3y = 10y+x

or, 7y = 2x

or, y/x = 2/7

therefore from (1)

N/x = 10(2/7)+1

or, N/x = 27/7

therefore number(N) will be 27.

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It is not given whether the number is one-digit or 2-digit or 3 -digit number.

let us say the number consists**only 1 digit:** a. Its value is a. a cannot be 3 times itself. So the given number is not one digit number.

Let us say the number has** 3 digits.** Let it be a b c. Its value is 100a+10b+c.

Three times sum of digits is 3 (a+b+c). We know maximum value of a, b, or c is 9. So maximum value of sum of digits is = 3 (9+9+9) = 81

But the minimum value of a b c = 100. So they cannot be equal.

Given number is not a 3 digit number or 4 -digit or higher.

It is probably a** two digit number: say a b.**

Hence 3(a+b) = 10 a + b

2 b = 7 a

**a = 2 b / 7 ** and ** b = 7a / 2**

2 and 7 are prime. If a is an integer, b has to be multiple of 7. There is only one multiple of 7 between 0 and 9. b cannot be 0 because then a is also 0.

So b = 7 a = 2

*the number is 27.*

there is also__ a trivial solution : a =0 and b=0__. that is number is 0. it is equal to three times sum of digits.

let us say the number consists

Let us say the number has

Three times sum of digits is 3 (a+b+c). We know maximum value of a, b, or c is 9. So maximum value of sum of digits is = 3 (9+9+9) = 81

But the minimum value of a b c = 100. So they cannot be equal.

Given number is not a 3 digit number or 4 -digit or higher.

It is probably a

Hence 3(a+b) = 10 a + b

2 b = 7 a

2 and 7 are prime. If a is an integer, b has to be multiple of 7. There is only one multiple of 7 between 0 and 9. b cannot be 0 because then a is also 0.

So b = 7 a = 2

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