[bonanova]

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.
When I looked for a smaller number with this property I couldn't find one.

Can you find it?

[normdeplume]

My answer is below, is this the answer you were looking for or is there a lower one?

Spoiler for solution:

2519

Just because your voice reaches halfway around the world doesn't mean you are wiser than when it reached only to the end of the bar.

[bonanova]

That's it.

Here's my method: [care to share yours?]

Spoiler for solution:

The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.

Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.

Still think of this as kind of brute force.
Maybe there is no elegant solution.

[normdeplume]

I am afraid, it was pretty much the 'brute force' method.

As you say 59 and 119 looked promising, both working upto and including 6, but 7 was the dificulty, so I tried each increment of 60 (179, 239, 299 ...), until I got to 419, which worked for 7 as well, (hooray) but not 9 or 10 (boo). Then using the same logic as before (59 being the first number that worked up the number 6, Increments of 60 always worked for the first 6 numbers. So 419 being the first that works for numbers up to 7 I incremented by 420.) So 419, 839, 1259, 1679, 2099, 2519.

Not a very elogant solution, but it passed a little time.

I assume, some mathemarical genius is going to post an easier way

Just because your voice reaches halfway around the world doesn't mean you are wiser than when it reached only to the end of the bar

[ghyspran]

If we're looking for n where n divided by x (x being each number 2-10) leaves a remainder of x-1, n+1 divided by each number 2-10 leaves no remainder. To find n+1, we must simply multiply all prime factors of 2-10, excluding any duplicates. This is equal to 2*2*2*3*3*5*7 = 2520, then just subtract one to get the answer of 2519.

[brhan]

Quote

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.
When I looked for a smaller number with this property I couldn't find one.

Can you find it?

Let X be the number. From the above properties of X, one can see that,
"X+1" is a multiple of the numbers from 2 to 10. As Bonanova is asking for the smallest number with such properties, "X+1" must be the LCM (least common multiple) of the numbers from 2 to 10 -- which is 2520.

And hence, X = 2520-1 = 2519.

[brhan]

Another variation of the above problem is as follows:

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 1.
When I divide it by 4, the remainder is 1.
When I divide it by 5, the remainder is 1.
When I divide it by 6, the remainder is 1.
When I divide it by 7, the remainder is 1.
When I divide it by 8, the remainder is 1.
When I divide it by 9, the remainder is 1.
When I divide it by 10, the remainder is 1.

But when I divide it by 11, the remainder is 0 (just to make the procedure a little bit different).

Can you find it?

[brhan]

Not sure if people are checking the variation of the above question. I think it is better to post it as a new question.

[vivek_t]

All we need to do is to take the LCM of all these numbers and subtract 1 from that..
LCM of 1 to 10 is 5x8x9x7=2520
-1 is 2519

[PtGrill]

brhan, on Oct 3 2007, 08:57 AM, said:

Another variation of the above problem is as follows:

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 1.
When I divide it by 4, the remainder is 1.
When I divide it by 5, the remainder is 1.
When I divide it by 6, the remainder is 1.
When I divide it by 7, the remainder is 1.
When I divide it by 8, the remainder is 1.
When I divide it by 9, the remainder is 1.
When I divide it by 10, the remainder is 1.

But when I divide it by 11, the remainder is 0 (just to make the procedure a little bit different).

Can you find it?

A number that satisfies all of these properties is:

Spoiler for above problem:

25201