what 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
i 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
It 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.
what 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.
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To find a number that satisfies these conditions, we can use the concept of solving simultaneous congruences. In this case, we need to find a number that satisfies three different congruences:
1. Number ≡ 2 (mod 3)
2. Number ≡ 3 (mod 4)
3. Number ≡ 4 (mod 5)
To solve this, we can start by listing the multiples of 3, adding 2 to each, and looking for a multiple of 4 and 5:
2, 5, 8, 11, 14, 17, 20, ...
Going through this list, we can see that 14 satisfies the first and second congruences. Let's check if it also satisfies the third congruence:
14 ≡ 4 (mod 5)
Since 14 satisfies all three congruences, the number we're looking for is 14.