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

supermanaccz  Apr 29, 2018
 #1
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Can somebody translate this into proper LaTex?

Guest Apr 29, 2018
 #2
avatar+942 
+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
edited by GYanggg  Apr 30, 2018
edited by GYanggg  Apr 30, 2018
edited by GYanggg  Apr 30, 2018
 #3
avatar+93352 
+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
avatar+93352 
+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

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