There is a number less than 3000 that when divided by 2 leaves a

remainder of 1, when divided by 3 leaves a remainder of 2, when

divided by 4 leaves a remainder of 3, When divided by 5 leaves a

remainder of 4, when divided by 6 leaves a remainder of 5, and so on up to nine.

What is that number?

its l.c.m of (2, 3, 4, 5, 6, 7,8 ,9)k – 1

i found l.c.m = 2520

and k may be any natural no.

and reason for subtracting 1 is that is the common difference between every divisor and its corresponding remainder.

so the ans may be 2519 , 5039 and so on

but its given the no. should be less than 3000

so the final answer is 2519.

its very easy.

2519. This is how it’s done:

2=2

3=3

4=2×2

5=5

6=2×3

7=7

8=2x2x2

9=3×3

number=2x3x2(we already have a 2 for 4)x5x7(we skiped 6, because the number is a multiple of 6 already)x2x3 – 1.

why -1?

because for 2, the smallest number is 2×2-1, for 3 is 3×2-1, for 4 is 4×2-1, and so on.

the condition is that our number is more than, or equals 10×2-1, and it does.

The number is 2519. Found on a TI 84+, using sequence command in the stat are of the calculator.

Great brain teaser!

R