Find the smallest positive integer that leaves a remainder of 5 when divided by 7, a remainder of 6 when divided by 11, and a remainder of 4 when divided by 13.
Why did the number go to therapy? Because it had too many unresolved remainders! But fear not, my friend, I've got just the solution for this remainder riddle. Let's find the smallest positive integer that satisfies all these conditions.
Since the integer leaves a remainder of 5 when divided by 7, we can start our journey with 5 and add multiples of 7 until we hit the other two remainders.
Checking 5, 12, 19, 26... we find that none of them give a remainder of 6 when divided by 11.
But don't despair, let's add multiples of (7 x 11) = 77 to our search. The next number on our list, 82, satisfies the remainder of 6 when divided by 11.
Now, let's see if it also leaves a remainder of 4 when divided by 13. Adding multiples of (7 x 11 x 13) = 1001 to our number, we have 82, 1083, 2084, and so on.
Finally, we find that 82 + 1001 = 1083 works!
So the smallest positive integer that leaves a remainder of 5 when divided by 7, a remainder of 6 when divided by 11, and a remainder of 4 when divided by 13 is 1083. Ta-da!
To find the smallest positive integer that satisfies these conditions, we can start by considering the remainders. Let's call the unknown integer "x".
Given:
- x leaves a remainder of 5 when divided by 7.
- x leaves a remainder of 6 when divided by 11.
- x leaves a remainder of 4 when divided by 13.
To solve this, we can set up a system of linear congruences using modular arithmetic.
First, let's write down the equations:
x ≡ 5 (mod 7)
x ≡ 6 (mod 11)
x ≡ 4 (mod 13)
We can solve this system using the Chinese Remainder Theorem, which states that there is a unique solution modulo the product of the moduli (7 * 11 * 13). In this case, the modulus is 1001 (7 * 11 * 13).
Now, let's solve the system step-by-step:
Step 1: Using the Chinese Remainder Theorem, find the inverse of each modulus.
To find the inverse of 7 modulo 11, we look for a number "a" such that (7 * a) ≡ 1 (mod 11). In this case, a=8 satisfies the equation.
To find the inverse of 11 modulo 13, we look for a number "b" such that (11 * b) ≡ 1 (mod 13). In this case, b=4 satisfies the equation.
Step 2: Multiply the remainders by the product of the other moduli.
Multiply 5 by (11 * 13) = 715.
Multiply 6 by (7 * 13) = 546.
Multiply 4 by (7 * 11) = 308.
Step 3: Combine the results from Step 2.
(5 * 715 * 8) + (6 * 546 * 4) + (4 * 308 * 4) = 21340 + 13104 + 4928 = 39372.
Step 4: Take the result from Step 3 modulo the product of the moduli to get the smallest solution.
The smallest solution to the system of congruences is 39372 ≡ 117 (mod 1001).
Therefore, the smallest positive integer that satisfies the given conditions is 117.
To find the smallest positive integer that satisfies these conditions, we can use the Chinese Remainder Theorem.
The Chinese Remainder Theorem states that if we have a set of congruences:
x ≡ a₁ (mod n₁)
x ≡ a₂ (mod n₂)
...
x ≡ aₖ (mod nₖ)
where n₁, n₂, ..., nₖ are pairwise coprime, then there exists a unique solution modulo (n₁ * n₂ * ... * nₖ).
In this case, we have the following congruences:
x ≡ 5 (mod 7)
x ≡ 6 (mod 11)
x ≡ 4 (mod 13)
The moduli, 7, 11, and 13, are pairwise coprime, so we can apply the Chinese Remainder Theorem.
To find the solution, we can use the Extended Euclidean Algorithm. Let's solve for the first pair of congruences:
x ≡ 5 (mod 7)
x ≡ 6 (mod 11)
By using the Extended Euclidean Algorithm, we can find coefficients s and t such that:
7s + 11t = 1
We can solve this equation to find s and t:
7s + 11t = 1
Solving this equation, we find that s = 3 and t = -2.
Now, we can express the solution as:
x = (5 * 11 * -2 + 6 * 7 * 3) mod (7 * 11)
x = (-110 + 126) mod 77
x = 16 mod 77
So, the solution for the first pair of congruences is x ≡ 16 (mod 77).
Now, we can solve the second pair of congruences:
x ≡ 16 (mod 77)
x ≡ 4 (mod 13)
Using the Extended Euclidean Algorithm, we can find coefficients s and t such that:
77s + 13t = 1
Solving this equation, we find that s = 5 and t = -29.
Now, we can express the solution as:
x = (16 * 13 * -29 + 4 * 77 * 5) mod (77 * 13)
x = (-6032 + 1540) mod 1001
x = -4492 mod 1001
x = 509 mod 1001
So, the smallest positive integer that satisfies all three congruences is x = 509.
leaving remainder of 5 when divided by 7 ...
must be multiples of 7 plus 5 :
12 19 26 33 40 47 54 61 68 75 82 89 96 103 110 ..
remainder of 6 when divided by 11
17 28 39 50 61 72 83 94 105 116 127 ...
remainder of 4 when divided by 13
17 30 43 56 69 82 95 108 121 ...
noting so far, I will let you go on for a bit more.
Looking for a common number in all 3 lists,