+654 How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.
+981 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
+118608 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.
+118608 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.
