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

2
by luca

2014-09-06T18:07:41+05:30
Let,  the unit digit of the number = x
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.

nice
really.
2014-09-07T08:58:33+05:30

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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.