Help! A remainder is chasing me

Number Theory Level 2

I just found a positive integer 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.

Find the smallest number with such property.


The answer is 2519.

Payal Sheth by Payal Sheth , 22, India

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2 solutions

We just need to find l c m ( 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 ) lcm(2,3,4,5,6,7,8,9,10) and then subtract 1 1 from it to find the least value with the desired property.

Now the lcm is 2 3 ∗ 3 2 ∗ 5 ∗ 7 = 2520 2^{3} * 3^{2} * 5 * 7 = 2520 , so the desired value is 2519 \boxed{2519} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Exactly... @brian charlesworth

Yagnesh Baria - 6 years, 10 months ago
Payal Sheth
Jul 9, 2014

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.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

That's not a correct solution. Suppose the numbers to divide was extended to say 31, your method would not work. Brian Charlesworth's solution is the accurate one and can be used for any n n numbers.

Kunal Verma - 6 years ago

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