That is the smallest positive integer N for which (12,500,000)⋅n leaves a remainder of 111 when divided by 999,999,999?
What is the smallest positive integer N for which
(12,500,000)⋅n
leaves a remainder of 111 when divided by 999,999,999?
1.
(12 500 000)⋅n≡111(mod999 999 999)|:(12 500 000)n≡111⋅112 500 000(mod999 999 999)
According to Euler's theorem,if a is coprime to m, that is, gcd(a,m)=1, then aϕ(m)≡1(modm),where ϕ is Euler's totient function.Therefore, a modular multiplicative inverse can be found directly:aϕ(m)−1≡a−1(modm).
2.
greatest common divisor gcd(12 500 000,999 999 999)=1
112 500 000(mod999 999 999)≡12 500 000−1(mod999 999 999)≡12 500 000ϕ(999 999 999)−1(mod999 999 999)≡12 500 000648 646 704−1(mod999 999 999)≡12 500 000648 646 703(mod999 999 999)≡80(mod999 999 999)
3. n = ?
n≡111⋅112 500 000(mod999 999 999)≡111⋅(12 500 000−1(mod999 999 999))≡111⋅80(mod999 999 999)≡8880(mod999 999 999)
The smallest positive integer n is 8880.
12 500 000⋅8880≡111(mod999 999 999)110 000 000 000≡111(mod999 999 999) ✓
[12500000*n] mod 999999999 =111, solve for n
One obvious N =999,999,999 + 111 =1,000,000,110 because:
1,000,000,110 mod 999,999,999 =111 !!!, But:
n = 1,000,000,110 / 12,500,00
n =80.0000088, which is non-integer, but when multiplied by 12,500,00 you get:
N =1,000,000,110 which is an integer !!!! I don't think your teacher will buy this !!!!.
Now, onward to another rediculous answer !!!
If we wish to have both N and n whole integers, will try this:
n =111 x [999,999,999 / 111 + 1,000,000,000 / 12,500,000]
n = 111 x [9,009,009 + 80]
n =1,000,008,879, which is a whole integer!!!!. And...........
N =1,000,008,879 x 12,500,000
N =12,500,110,987,500,000, which is a whole integer !!!!! And.......
12,500,110,987,500,000 mod 999,999,999 =111 !!!!
Melody and CPhill : Take a whack at this one !!!!
What is the smallest positive integer N for which
(12,500,000)⋅n
leaves a remainder of 111 when divided by 999,999,999?
1.
(12 500 000)⋅n≡111(mod999 999 999)|:(12 500 000)n≡111⋅112 500 000(mod999 999 999)
According to Euler's theorem,if a is coprime to m, that is, gcd(a,m)=1, then aϕ(m)≡1(modm),where ϕ is Euler's totient function.Therefore, a modular multiplicative inverse can be found directly:aϕ(m)−1≡a−1(modm).
2.
greatest common divisor gcd(12 500 000,999 999 999)=1
112 500 000(mod999 999 999)≡12 500 000−1(mod999 999 999)≡12 500 000ϕ(999 999 999)−1(mod999 999 999)≡12 500 000648 646 704−1(mod999 999 999)≡12 500 000648 646 703(mod999 999 999)≡80(mod999 999 999)
3. n = ?
n≡111⋅112 500 000(mod999 999 999)≡111⋅(12 500 000−1(mod999 999 999))≡111⋅80(mod999 999 999)≡8880(mod999 999 999)
The smallest positive integer n is 8880.
12 500 000⋅8880≡111(mod999 999 999)110 000 000 000≡111(mod999 999 999) ✓