Mathematical tricks
Found this on Quora:
Give a person a calculator to use. Tell them to type any three digits (example 839). Tell them to repeat the digits so they get a six figure number (839839). Then you show them how incredibly quick you are! Just glance at the number and say, That number will divide exactly by 13?. Tell them to do it. Glance at this answer and say, That number now will divide exactly by 11. Tell them to do it. Finally say, Your answer will now divide exactly by 7. Tell them to do it and to their surprise, the final answer will be the first three digits they thought of!!! (ie 839) It is a good exercise to get people to prove this always works.
@jgill What is the trick here? Why does this work? It must have something to do with how prime numbers function, yes? If you divide any number by 2 for example, the answer will always divide evenly or with a remainder of 1. Is it that and the fact that there are 6 digits involved?
My favourite maths trick was always the missing pound one:
Three men decide to pool together to buy a TV. They find a second hand one at £30. They each handed £10 to the salesman and left with the TV. The boss of the store tells the salesman that he charged them £5 too much. The salesman gave the £5 in £1 notes to an assistant, to run after the three men and return the money. The assistant decides to keep £2 and give each man £1 back. Each man paid £10 and they have had £1 returned so each man had paid £9. 3x9 is £27. Add the £2 the assistant stole and the total is £29. What happened to the other £1?
At least I understand the trick in that one!!
Anyone know any other maths tricks?
Give a person a calculator to use. Tell them to type any three digits (example 839). Tell them to repeat the digits so they get a six figure number (839839). Then you show them how incredibly quick you are! Just glance at the number and say, That number will divide exactly by 13?. Tell them to do it. Glance at this answer and say, That number now will divide exactly by 11. Tell them to do it. Finally say, Your answer will now divide exactly by 7. Tell them to do it and to their surprise, the final answer will be the first three digits they thought of!!! (ie 839) It is a good exercise to get people to prove this always works.
@jgill What is the trick here? Why does this work? It must have something to do with how prime numbers function, yes? If you divide any number by 2 for example, the answer will always divide evenly or with a remainder of 1. Is it that and the fact that there are 6 digits involved?
My favourite maths trick was always the missing pound one:
Three men decide to pool together to buy a TV. They find a second hand one at £30. They each handed £10 to the salesman and left with the TV. The boss of the store tells the salesman that he charged them £5 too much. The salesman gave the £5 in £1 notes to an assistant, to run after the three men and return the money. The assistant decides to keep £2 and give each man £1 back. Each man paid £10 and they have had £1 returned so each man had paid £9. 3x9 is £27. Add the £2 the assistant stole and the total is £29. What happened to the other £1?
At least I understand the trick in that one!!
Anyone know any other maths tricks?
Comments (7)
[math]\frac{{{10}^{3}}\left( a\cdot {{10}^{2}}+b\cdot 10+c \right)+\left( a\cdot {{10}^{2}}+b\cdot 10+c \right)}{1001}=abc[/math]
I didn't respond to your example of a good maths trick prof, as it whizzed right over my head.
I did not get the trick, or fully understand the equation, what is the . for, for example? Does that signify multiplication? I was also hoping you would explain the 6 digit trick, divided by the three primes, as to why it worked. Sorry my maths skills are perhaps a bit rudimentary in your eyes.
That one expression I presented shows it all. The dots are multiplication. When you write out abcabc, like 869869, what you are really expressing is [a(10^5)+b(10^4)+c(10^3)+a(10^2)+b(10^1)+c] , and dividing by 13, 11, and 7 is the same as dividing once by their product, 1001. (Since the product divides evenly, so will the factors of the product: 13, 11,7)
Factor out the expression in my original equation that is enclosed in parentheses and you are left with (10^3 + 1)/1001 = 1.
Sorry, I have neither the time nor interest to delve into this. I think number theory may be full of stuff like this, but I never even took a course in the subject. My general area was complex analysis, dealing with the limit concept in the complex plane.
Ok, thanks for your equation, which I now understand, if your . notation means multiply, but when I multiply out the top line and divide by 1001. I am left with 100a+10b+c, not abc?
Addition: The penny just dropped! 100a+10b+c IS abc, on a base 10 number system, sorry, I get it now.