Someone has x pieces of gold. He wants to distibute them in groups in such a way that each group contain equal number of pieces. When he put two pieces in each group, one piece remained. The same thing happened when he put 3, 4, 5, 6, 7, 8, 9, and 10 pieces in each group: each time, one piece remained. But when he put 11 pieces in each group, no piece is remained. How many iron pieces are there?
x pieces of gold: How many iron pieces are there?
- Thread starter bilalrouf
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Re: multiple of 11
Hello, bilalrouf!
There is an infinite number of anwers.
. . I'll determine the smallest answer.
The LCM of 2, 3, 4, 5, 6, 7, 8, 9, 10 is: \(\displaystyle \,2520\)
. . \(\displaystyle 2520\) is divisible by 2, 3, 4, 5, 6, 7, 8, 9, and 10.
. . Hence, \(\displaystyle x\) is of the form: \(\displaystyle \,x\:=\:2520n\,+\,1\) for some integer \(\displaystyle n.\;\) [1]
Since \(\displaystyle x\) is divisible by 11: \(\displaystyle \:2520n\,+\,1\:=\:11k\) for some integer \(\displaystyle k.\)
. . Solve for \(\displaystyle k:\;\;k\:=\:\frac{2420n\,+\,1}{11}\:=\:229n\,+\,\frac{n\,+\,1}{11}\)
Since \(\displaystyle k\) is an integer, \(\displaystyle \frac{n\,+\,1}{11}\) must be a whole number.
. . This happens first when \(\displaystyle n\,=\,10.\)
Substitute into [1]: \(\displaystyle \:x\:=\:2520(10)\,+\,1\;\;\Rightarrow\;\;\L\fbox{x\,=\,25,201}\)
Hello, bilalrouf!
There is an infinite number of anwers.
. . I'll determine the smallest answer.
The LCM of 2, 3, 4, 5, 6, 7, 8, 9, 10 is: \(\displaystyle \,2520\)
. . \(\displaystyle 2520\) is divisible by 2, 3, 4, 5, 6, 7, 8, 9, and 10.
. . Hence, \(\displaystyle x\) is of the form: \(\displaystyle \,x\:=\:2520n\,+\,1\) for some integer \(\displaystyle n.\;\) [1]
Since \(\displaystyle x\) is divisible by 11: \(\displaystyle \:2520n\,+\,1\:=\:11k\) for some integer \(\displaystyle k.\)
. . Solve for \(\displaystyle k:\;\;k\:=\:\frac{2420n\,+\,1}{11}\:=\:229n\,+\,\frac{n\,+\,1}{11}\)
Since \(\displaystyle k\) is an integer, \(\displaystyle \frac{n\,+\,1}{11}\) must be a whole number.
. . This happens first when \(\displaystyle n\,=\,10.\)
Substitute into [1]: \(\displaystyle \:x\:=\:2520(10)\,+\,1\;\;\Rightarrow\;\;\L\fbox{x\,=\,25,201}\)
Yes
Thanks for helping me. 25201 satisfied the condition. it is solved by Chinese Remander theorem.I havnt leant this theorem in college yet.is there any useful link to learn this theorem easily.
i searched on google.com But the resulting links contain the higher level of knowledge. and was not easily understand-able
Thanks again.
Thanks for helping me. 25201 satisfied the condition. it is solved by Chinese Remander theorem.I havnt leant this theorem in college yet.is there any useful link to learn this theorem easily.
i searched on google.com But the resulting links contain the higher level of knowledge. and was not easily understand-able
Thanks again.
What Soroban done was a simplified form of the Chinese theorem. Without all that 'mod' business. Nice way to do it.
Here's an old classic along the same lines.
An old woman goes to market and a horse steps on her basket and crashes the eggs. The rider offers to pay for the damages and asks her how many eggs she had brought. She does not remember the exact number, but when she had taken them out two at a time, there was one egg left. The same happened when she picked them out three, four, five, and six at a time, but when she took them seven at a time they came out even. What is the smallest number of eggs she could have had?
Use Soroban's method and you get 301 as the smallest number of eggs.
Here's the modular arithmetic method from cut-the-knot:
Solve
q1: x = 1 (mod 2)
q2: x = 1 (mod 3)
q3: x = 1 (mod 4)
q4: x = 1 (mod 5)
q5: x = 1 (mod 6)
q6: x = 0 (mod 7)
With the experience we acquired so far, the combination of q1-q5 is equivalent to
q7: x = 1 (mod 60)
x = 60t + 1. Plugging this into q6 gives 60t = -1 (mod 7). Casting out 7 simplifies this to 4t = 6 (mod 7) and then 2t = 3 (mod 7). From division tables modulo 7, 3/2 = 5 (mod 7). Therefore, t = 7u + 5. Finally, x = 420u + 301. Allowing for an average size farmer, the most likely number of eggs she might expect to be compensated for is 301.
Now, which would you rather use?.
Here's an old classic along the same lines.
An old woman goes to market and a horse steps on her basket and crashes the eggs. The rider offers to pay for the damages and asks her how many eggs she had brought. She does not remember the exact number, but when she had taken them out two at a time, there was one egg left. The same happened when she picked them out three, four, five, and six at a time, but when she took them seven at a time they came out even. What is the smallest number of eggs she could have had?
Use Soroban's method and you get 301 as the smallest number of eggs.
Here's the modular arithmetic method from cut-the-knot:
Solve
q1: x = 1 (mod 2)
q2: x = 1 (mod 3)
q3: x = 1 (mod 4)
q4: x = 1 (mod 5)
q5: x = 1 (mod 6)
q6: x = 0 (mod 7)
With the experience we acquired so far, the combination of q1-q5 is equivalent to
q7: x = 1 (mod 60)
x = 60t + 1. Plugging this into q6 gives 60t = -1 (mod 7). Casting out 7 simplifies this to 4t = 6 (mod 7) and then 2t = 3 (mod 7). From division tables modulo 7, 3/2 = 5 (mod 7). Therefore, t = 7u + 5. Finally, x = 420u + 301. Allowing for an average size farmer, the most likely number of eggs she might expect to be compensated for is 301.
Now, which would you rather use?.