How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.

supermanaccz
Apr 29, 2018

#2**+1 **

Hey supermanaccz!

If \(127 \equiv 7 \pmod{n}\), then n is a divisor of 127 - 7 = 120.

The prime factorization of 120 is \(2^3 \cdot 3 \cdot 5\)

which has \((3 + 1)(1 + 1)(1 + 1) = 16\) positive divisors.

Therefore, there are 16 possible values of n.

I hope this helps,

Gavin.

There was the same question 1 day ago, https://web2.0calc.com/questions/how-many-positive-integers-n-satisfy-127-equiv-7

GYanggg
Apr 30, 2018

#3**+1 **

How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.

How many positive integers \(n\) satisfy \(127 \equiv 7 \pmod{n}\)? n=1 is allowed.

I am just thinking on paper - not sure where it will go.

factor(127) = 127 so 127 is a prime number and so is 7

127 = 7(mod 120)

120=0(mod n)

so n is a factor of 120 i think

factor(120) = (2^3*3)*5

What factors are bigger than 7

8, 10, 12, 15, 20, 24, 30, 40, 60, 120

So the answer is 10 That is if n is a positive.

Melody
Apr 30, 2018

#4**+1 **

I have been asked why 1,2,3,4,5 and 6 are not also answers.

The mod of a number is the remainder that you get when you divide by the modulus.

If the mod is less then 7 then the remainder cannot be 7 or more.

127=0(mod1)

127=1(mod2)

127=1 (mod3)

127=3(mod4)

127=2(mod5)

127=1(mod6)

127=7(mod8) Bingo, this is the smallest.

Melody
Apr 30, 2018