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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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 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.
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7
Answer:
27
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