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That is the smallest positive integer N  for which (12,500,000)n leaves a remainder of 111 when divided by 999,999,999?

 Jan 29, 2018

Best Answer 

 #3
avatar+26396 
+4

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)n111(mod999 999 999)|:(12 500 000)n111112 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)1a1(modm).

 

2.

greatest common divisor gcd(12 500 000,999 999 999)=1 

 

112 500 000(mod999 999 999)12 500 0001(mod999 999 999)12 500 000ϕ(999 999 999)1(mod999 999 999)12 500 000648 646 7041(mod999 999 999)12 500 000648 646 703(mod999 999 999)80(mod999 999 999)

 

3. n = ?

n111112 500 000(mod999 999 999)111(12 500 0001(mod999 999 999))11180(mod999 999 999)8880(mod999 999 999)

 

The smallest positive integer n is 8880.

 

12 500 0008880111(mod999 999 999)110 000 000 000111(mod999 999 999) 

 

laugh

 Jan 29, 2018
edited by heureka  Jan 29, 2018
 #1
avatar
+2

[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 !!!!

 Jan 29, 2018
edited by Guest  Jan 29, 2018
 #3
avatar+26396 
+4
Best Answer

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)n111(mod999 999 999)|:(12 500 000)n111112 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)1a1(modm).

 

2.

greatest common divisor gcd(12 500 000,999 999 999)=1 

 

112 500 000(mod999 999 999)12 500 0001(mod999 999 999)12 500 000ϕ(999 999 999)1(mod999 999 999)12 500 000648 646 7041(mod999 999 999)12 500 000648 646 703(mod999 999 999)80(mod999 999 999)

 

3. n = ?

n111112 500 000(mod999 999 999)111(12 500 0001(mod999 999 999))11180(mod999 999 999)8880(mod999 999 999)

 

The smallest positive integer n is 8880.

 

12 500 0008880111(mod999 999 999)110 000 000 000111(mod999 999 999) 

 

laugh

heureka Jan 29, 2018
edited by heureka  Jan 29, 2018
 #4
avatar
+2

Simple correction. heureka made a simple typo:

80 x 111 =8,880. The smallest integer.

 Jan 29, 2018
 #5
avatar+26396 
+2

Thank you Guest

laugh

heureka  Jan 29, 2018

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