Think of any number. Multiply it by 6 and add 3. Then divide it by 3 and add 8. Lastly, subtract twice the original number. I know what number you have left. What must it be?
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Take any number (except from 1 to 8) find the sum of it's digits and then subtract it from the original number the resultant number will always be divisible by ( 3 & 9) Ex - consider the number 245 the addition of digits comes out to be 11 , 245-11 = 234 is divisible by 3 & 9
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If you add the sum of ANY number, it will be divisible by 3. Simple divisibility.
it should be subtract it from!!!!
I couldn't have put it better myself.
Initially I solved by following the instructions given. But you have posted an appropriate solution. Thanks for it.
What if we take a fraction, its not written that it must be an integer.
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Very true fact, but Im pretty sure the same result will become of it. Eg. 1/9 + 9 -9 = 1/9. The thought is dismissed😊
ohhh good laugh
but i dint understood that when we do twice of 2a it will be 4a then how your answer is coming 2a
Suppose that your number is x . Multiply by 6 : 6 x Add 3 : 6 x + 3 Divide by 3 : 2 x + 1 Add 8: 2 x + 9 Subtract twice the original number (subtract 2 x ): 9 The output is 9, irrespective of the original number. EDIT: Of course, you could just follow the steps with some number, but that takes all the fun out of it...xD
Thanks a lot this made it much easier
You see the first two numbers 6&3 and your brain automatically thinks of 2 since 6÷3=2
How does that solve the problem?
My thought was 10. Please rephrase YOUR brain to MY brain.
I thought of 7, so no.
You can think of any number and that is your choice.😁
Let the number is 3 which you think that Then multiple by 6 is = 18 then add 3 is = 21 , now divide by 3 is = 7 then add 8 is = 15 .... Finally subtracting twice the original number which you think that is 3 So , 15 - 3 - 3 = 9
Hence we will get [9]
This is one of the many reasons why 9 is a mysterious number.
How I did it: 7 x 6 = 42 + 3 = 45 - 3 = 15 + 8 = 23 - 14 = 9. This is how I got 9 .
Let the number be x.
Multiply x by 6: 6x And add 3: 6x+3 Divide by 3: 2x+1 And add 8: 2x+9 Lastly, subtract twice the original number:9
(6x + 3)/3 + 8 - 2x
= (2x + 1) + 8 - 2x
= 2x + 1 + 8 - 2x
= 9
the answer will always 9
Suppose x is the number.
3 6 x + 3 + 8 − 2 x = 2 x + 1 + 8 − 2 x = 1 + 8 = 9
Try the numbers one two and three
You should already be able to guess that the input number is arbitrary, so just make it 0.
0 * 6 = 0
0 + 3 = 3
3 / 3 = 1
1 - (0 * 2) = 0
1 + 8 = 9
Only two statements actually matter, and they don't use the arbitrary number.
let the no. be 'x' ATQ multiply the no. with 6 and add 3.so, the no be comes '6x+3' divide by 3 and add 9.so the no becomes 3 6 x + 3 +9=2x+9 at last the subtract the original no from it will give 9
ANS=9
Since we know that the chosen number is irrelevant, choose 0 so the arithmetic is easy.
This also works with 0.
It works with negatives too
Yes, because of algebra
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Let the number be a . On multiplying with 6 we have 6 a and on adding 3 we have 6 a + 3 .
Dividing by 3 would give 2 a + 1 and then adding 8 would make it 2 a + 9 .
Finally subtracting twice the original number we get 2 a + 9 − 2 a = 9
Hence we'll always get 9