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Posted

My answer is 17. It is neither a multiple of 3 or 10, nor is it a sum of a multiple of 3 and a multiple of 10. 

 

To eliminate all larger numbers, note the numbers that ARE multiples of 3 or 10, or sums of a multiple of 3 / multiple of 10, first from 10 to 20:

10 12 13 15 16 18 19 20

 

If you take the last three numbers, 18, 19, and 20, and add 3 to each of them, you get 21, 22, and 23. The first is a multiple of 3, the second can be represented by 10 + 12, and the third can be represented by 20 + 3.

 

Keep adding 3 and you get 24, 25, 26. Multiple of 3, represented by 10 + 15, and represented by 20 + 6. 

 

One last time for 27, 28, 29. Once again, multiple of 3, represented by 10 + 18, and represented by 20 + 9.

 

Adding 3 to a multiple of three simply makes another multiple of 3, adding 3 to (10 + 3x) gets (10 + 3 (x+1)), which is still a sum of a multiple of 10 and a multiple of 3, and adding 3 to (20 + 3x) gets (20 + 3 (x+1)). From this, it can be inferred that all numbers larger than 17 can be represented by either 3x, 10 + 3x, or 20 + 3x. Therefore, all numbers larger than 17 are disqualified.

  • Upvote 3
Posted

^Correct

But my solution involved subtracting 10 from multiple of 3, 3 up to 27, since any number ending in 0-9 results from these numbers plus a multiple of 10. Biggest=3x9-10=17. But it's still that.

x/y= 4/5+ 4/50+ 4/500+ 4/5000.....

x+y=?

Posted

Just saw this thread 'n' the most recent problem, so I wanted to give it a quick try.

 

Is the answer 17?

 

Here's the explanation:

4/5=0.8

4/50=0.08

4/500=0.008

.

.

.

So, the final result will be 0.888...

From that, we can say that x<y

I've just started searching for the first two numbers that would meet these two conditions, 'n' I ended up with 8 'n' 9.

  • Upvote 1
Posted

I don't really have any problems to post, except for some random problems that I've solved while on the University. But that'll just kill your will to live. So, if anyone has an interestin' problem, feel free to post it in my stead.

Posted

Simple problem: x is the smallest nonnegative number that satisfies the ff:

 

1) x-1, when divided by a number y wherin 2<=y<=10, the remainder is y-1; and

2) x+1, when divided by a number y wherin 2<=y<=10, the remainder is always 1

 

What is x?

Posted

If x+1 has a remainder of 1, x has a remainder of 0. So x is divisible by y. Becuase it's the least possible number, y must be 2. And so x is either 2 or 1 (factors of 2) If x is one, then x+1=2, which does not fulfill the conditions. If x is 2, x-1=1, which has a remainder of 2-1=1 when divided by 2, and x+1=3, which has a remainder of 1 when divided by 3/

My answer: 2


 

But if y is less than or equal to 2 and also greater than or equal to 10, y does not exist.

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