# Problem of the Rabbits

A farmer owns a number of rabbits. If divided by 2, one rabbit remains.If divided by 3, 2 rabbits remain.If divided by 4, 3 rabbits remain.If divided by 5, 4 rabbits remain.If divided by 6, 5 rabbits remain.

How many rabbits does the farmer have ??

P.S : I know the answer but not the solution.

5 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

59.From the conditions we can find out that the answer is $$[2,3,4,5,6] -1$$ which is$$60-1=59$$

- 5 years, 3 months ago

What does [2,3,4,5,6] mean ?

- 5 years, 3 months ago

The LCM of those numbers...that's all.

- 5 years, 3 months ago

answer is 59. Since if it is divided by 2 we get remainder 1, it is odd. Combine this with the fact that when it is divided by 5 remainder 4, its last digit must be 9. After minus 1 out of multiples of 6, the numbers which have 9 as last digit are 29,59,89....And after minus 1 from multiples of 4, the numbers which have 9 as last digit are 19,39,59... Therefore, it is clear that the answer is 59.

- 5 years, 3 months ago

Thanks :)

- 5 years, 3 months ago

first do lcm of 2,3,4,5,6 its 60 as the remainder is 1 less than the divisor so the answer is 60-1=59

- 5 years, 2 months ago

This is simply a system of modular congruence.

- 5 years, 3 months ago