Saturday

October 10, 2015
Posted by **Jacob** on Sunday, November 4, 2007 at 5:45pm.

- Math -
**Jacob**, Sunday, November 4, 2007 at 6:12pmi ment what is a number that divied by 3 has a remainder of 2 also that same number divided by 4 has a remainder 3 and that same number divided by 5 has a remainder of 4

- Math -
**drwls**, Sunday, November 4, 2007 at 6:20pmIt must be 2 more than a number evenly divisible by 3, such as

5,8,11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59 ...

It must also be 3 more than a number evenly divisible by 4, such as

7,11,15,19,23,27,31,35,39, 43,47, 51, 55, 59...

It must also be 4 more than a number evenly divisble by 5, such as

9,15,19,24, 29, 34, 39, 44, 49, 54, 59, 64, 69..

The smallest number that satisfies all requirements is 59. There will probably be larger numbers also.

- Math -
**tchrwill**, Sunday, November 4, 2007 at 6:25pmwhat is a number that divied by 3 has a remainder of 2 also that smae number divided by 4 has a remainder 3 and that same number divided by 5 has a remainder of 4

Here is another version of the famous Chinese Remainder puzzles which should give you an idea as to how to go about solving it. There are other methods.

Lets now attack the problem where three divisors and remainders are involved.

What is the smallest number that will leave a remainder of 1 when divided by 5, a remainder of 2 when divided by 7, and a remainder of 3 when divided by 9?

Let

1--N/5 = A + 1/5 or N = 5A + 1

2--N/7 = B + 2/7 or N = 7B + 2

3--N/9 = C + 3/9 or N = 9C + 3

4--Combining (1) and (2), 5A + 1 = 7B + 2 or 5A - 7B = 1 or B = (5A - 1)/7.

4--Derive the first values of A and B by trial and error. Subsequent values of A differ by the coefficient of B or 7 and subsequent values of B differ by the coefficient of A, 5.

5--A.....3.....10.....17.....24.....31.....38.....45

....B.....2......7.....12.....17.....22.....27.....32

6--Combining (2) and (3), 7B + 2 = 9C + 3 or 7B - 9C = 1 or C = (7B - 1)/9.

7--Derive the first values of B and C by trial and error. Subsequent values of B differ by the coefficient of C or 9 and subsequent values of C differ by the coefficient of B, or 7.

8--B.....4.....13.....22.....31.....40.....49.....58

....C....3......10.....17.....24.....31.....38.....45

9--Combining (3) and (1), 9C + 3 = 5A + 1 or 5A - 9C = 2 or A = (9C + 2)/5

10--Derive the first values of C and A by trial and error. Subsequent values of C differ by the coefficient of A or 5 and subsequent values of A differ by the coefficient of C, or 9.

11--C.....2.....7.....12.....17.....22.....27.....32

.....A.....4....13....22......31....40.....49......58

12--We must now search the tabular data to find a consistent set of values acrossall three tables.

13--The lowest consistent set is A = 31, B = 22 and C = 17.

14--This then leads to the minimum N = 5(31) + 1 = 7(22) + 2 = 9(17) + 3 = 156.

- Math -
**Macey Davis**, Friday, October 10, 2008 at 7:58pmhbjbjgingginkgofgmogf