Lets have fun
This calculation will tell you who you really are?
Pick any integral number between 1 to 9 inclusive.
Now multiply it by 3 .
Add 3 .
Now again multiply it by 3 .
You will get two digits, now add them. now this number will tell who you really are
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Kind Gentle Honest Caring Reliable Adventurous Romantic Faithful Somewhat Stupid Protective
You have to enter the sum of digits obtained. No offense.
The answer is 9.
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14 solutions
Moderator note:
Cute one, Tanishq Varshney!
BTW only 72% people got it right
Hahahaha Thumbs up @Tanishq Varshney
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My friend gave me this problem and the qualities mentioned were different. U know what i mean , but it was a hilarious encounter.
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Hai its crct
Haha...yup!! Good one, though:D
This was rigged, was it?
oooooooooo
your head !!!!!!!!!!!!
But how would you prove (for completeness?) that the sum of the two digits must add up to nine? Though i do see that for n ranging from 1 to 9 produces multiples of 9 and hence the sum adds up to nine (intuition, as the tens digit increases by 1 the units digit decreases by 1 and hence sum is constant) i cannot think of a way to prove this, would you help?
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This is the "divisibility by 9 " rule, a general proof of which is as follows:
Suppose we have a ( k + 1 ) -digit number N = a k a k − 1 a k − 2 . . . . a 1 a 0 where 1 ≤ a k ≤ 9 and 0 ≤ a i ≤ 9 for 0 ≤ i ≤ k − 1 .
We can then write the numerical value of N as
N = a k ∗ 1 0 k + a k − 1 ∗ 1 0 k − 1 + . . . . + a 1 ∗ 1 0 1 + a 0 ∗ 1 0 0 .
Now look at N ( m o d 9 ) . Since 1 0 ≡ 1 ( m o d 9 ) , we know that 1 0 j ≡ 1 j ≡ 1 ( m o d 9 ) , and so
N ( m o d 9 ) ≡ a k ∗ 1 + a k − 1 ∗ 1 + . . . . + a 1 ∗ 1 + a 0 ∗ 1 ( m o d 9 ) ,
which is precisely the digit sum of N modulo 9 . Thus if any integer N is divisible by 9 , so too will its digit sum be divisible by 9 , (and visa versa).
Wait, so if I choose 3 4 then I am honest?
pretty clever, BTW i edited to take only integers between 1 and 9.
Haha... honest people play by the rules. Picking 4/3 would be very contradictory, as it would prove that the system says you're honest only when you're dishonest. Actually, it would be no more contradictory than if you picked an integer 1 to 9 because you would be intelligent for solving the problem, but the system would say you are stupid.
Integral number
Alex, the question asks you to choose an integer.
No it has to be a whole number
Let the number you choose be x , 3 x + 3 is a multiple of three. So, 3 ( 3 x + 3 ) is a multiple of 3 ∗ 3 .So, the sum of its digit is always
9
Couldnt have said it better myself. Its all about the number play. More specifically, any time you add the digits of a multiple of nine, like what is achieved when you multiply 3 by itself, the digits will always add up to 9. Rule of 9s
That's not very nice but true at the same time..
So when we choose a number between 0-9, the given operations tend to make the answer 9 and ...... as we get the equation 3(3n+3)=9n+9=9(n+1) which tends to make it a multiple of 9 and hence the answer always is 9....
look at the "No offense" part at the bottom, see the most offensive #, and click.
Haha... I am STUPID??.. cool! :D
"Relaible" is spelled wrong. Who's stupid now?
The only important part of the problem is the last sentence. This implies that the problem will end up being some sort of insult. Since only 9 can be considered a insult, 9 is the correct answer.
An intelligent person always knows that he is stupid :-b cute question...
you will always get 9 ;)
Let n be the chosen integer such that 1 ≤ n ≤ 9 . Then the given operations leave us with the value 3 ( 3 n + 3 ) = 9 ( n + 1 ) , which, given the possible values of n , will be a 2-digit positive integer divisible by 9 . But the sum of the digits of any such integer is 9 , and so anyone who answers this question is .....
Nice one, @Tanishq Varshney .:D