Meenu had some toffees to be distributed equally among some children. if she put 5 toffees each in a packet, she found only 3 toffees were left. Then she tried putting 6,7 and 8 toffees in a packet and found that again she was left with 4, 5 and 6 toffees. What is the minimum number of tofees she had?
N is the number of toffes
N/5=R3 (remainder 3)
N/6=R4
N/7=R5
N/8=R6
http://www.youtube.com/watch?v=3PkxN_r9up8 is a good example of how to use the Chinese Remainder Theorem to solve this modulus problem. Repost if you can't work it out.
To find the minimum number of toffees Meenu had, we need to find the common remainder when the number of toffees is divided by the numbers of 5, 6, 7, and 8.
Let's start by finding the common remainder between 5 and 6:
- We know that Meenu had a remainder of 3 toffees when dividing by 5, meaning the number of toffees is congruent to 3 (mod 5).
- Similarly, Meenu had a remainder of 4 toffees when dividing by 6, meaning the number of toffees is congruent to 4 (mod 6).
Now we can find the common remainder between 5 and 6 by checking the numbers congruent to both 3 (mod 5) and 4 (mod 6). The common remainder is 9.
Next, we find the common remainder between 9 and 7:
- Meenu had a remainder of 9 toffees when dividing by 5 and 6, meaning the number of toffees is congruent to 9 (mod 5) and 9 (mod 6).
- We can check the numbers congruent to both 9 (mod 5) and 9 (mod 6). The common remainder is 39.
Finally, we find the common remainder between 39 and 8:
- Meenu had a remainder of 39 toffees when dividing by 5, 6, and 7, meaning the number of toffees is congruent to 39 (mod 5), 39 (mod 6), and 39 (mod 7).
- We can check the numbers congruent to 39 (mod 5), 39 (mod 6), and 39 (mod 7). The common remainder is 39.
Therefore, the minimum number of toffees Meenu had is 39.
To find the minimum number of toffees Meenu had, we need to find the least common multiple (LCM) of the given differences (3, 4, 5, 6).
The LCM of two or more numbers is the smallest number that is divisible by each of the given numbers without any remainder.
First, let's determine the LCM of 3 and 4:
Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, ...
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, ...
From the list, we can see that the smallest number divisible by both 3 and 4 is 12.
Now, let's determine the LCM of 12 and 5:
Multiples of 12: 12, 24, 36, 48, 60, ...
Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, ...
From the list, we can see that the smallest number divisible by both 12 and 5 is 60.
Finally, let's determine the LCM of 60 and 6:
Multiples of 60: 60, 120, 180, 240, ...
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, ...
From the list, we can see that the smallest number divisible by both 60 and 6 is 60.
Therefore, the minimum number of toffees Meenu had is 60