|
Question 1012027: I am a multiple of 8 of you subtract one from me you get a multiple of 5 and 11 at the same time-who am I
Answer by ValorousDawn(53)   (Show Source):
You can put this solution on YOUR website! A quick way to do it is not to check through the multiples of 8, as that requires you to go through 12.5% of all numbers, but rather the multiples of both 5 and 11: 55, which is only 1.81% of all numbers. So lets go through the multiples of 55, and add one, and test if it's a multiple of 8. If this fails, we multiply by EITHER 5, 11, or BOTH. This way the multiplicative property holds.
5*11=55; 55+1=56. 56/8=7. The number is 56.
If you're a fan of number theory, you may want to do it an analytical way, or find all of the numbers that satisfy this property.
We can do this with modular arithmetic. You want a number x such that x=-1 (mod 8) as well as 0 (mod 5) and 0 (mod 11).
If it is BOTH 0 (mod 5) and (mod 11), it is the same as saying 0 (mod 55).
55 (mod 8) is 7, which is really -1 (mod 8). You might be inclined to think that every multiple of 55 satisfies this property, but this is false. 55*2=120, and 121 is not a multiple of 8. Due to the parity (even/oddness), every OTHER multiple of 55 thus satisfies this property.
But you might not always be multiplying by 55. You can also multiply by 5 and 11 and still get this property. So, we do the same process.
5 (mod 8) = 5.
11 (mod 8) = 3.
5*3 (mod 8)=-1
-1*5 (mod 8)=3
Oh. Interesting, multiplying by 5 gives the same result as multiplying by 11. One more time.
3*5 (mod 8) = -1.
We are back to where we started, so the same parity point applies. Every other multiple of 5 and 11 multiplied to 55 gives this property.
From this we can determine that the all numbers which satisfy this property have the form 5^2m+1*11^2n+1, with m and n whole numbers. This basically means an odd number of 5 and 11 in the prime factorization.
|
|
|
| |
