Questions on Remainder & Integer Divisibility

In summary, when taking 102! modulo 103, we can use Wilson's theorem to find that the remainder is -1. This is because 103 is prime, so the integers taken modulo 103 form a field and each element has an inverse. When we multiply all the elements together, we get -1, which is the remainder. This can also be seen by comparing the exponents in the prime factorization of 102! to those in the prime factorization of 103, where we can see that -1 is the only number that is equal to its own inverse.
  • #1
clueles
15
0
I have 2 questions.

1)what is the remainder with 100! is divided by 103? explain your answer

2)a = 238000 = 2^4 x 5^3 x 7 x 17 and b=299880 = 2^3 x 3^2 x 5 7^2 17. Is there an integer so that a divides b^n? if so what is the smallest possibility for n?

the first one i have no idea how to even start it and the second one i know that the prime factorization helps to find out what n is.
 
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  • #2
I think I would start by asking what is the remainder when 102! is divided by 103. i.e. zero. Then let's see if that helps.

Since 103 is prime, then mod 103, the numbers from 1 to 102 forma multiplicative group, i.e. each has an inverse such thnat its product with that guy is 1. now there are two numbers equal to their own inverse 1 and 102 = -1. so no other number equals its own inverse. thus when we multiplky all these numbers to gether, we get -1. so the remainder of 102! is -1 mod 103. but 102! equals (102)(101)(100!),

so (102)(101)(100!) = -1 mod 103. does that help?
 
  • #3
can you explain it again

i'm still confused about it. how do you get the -1. I'm sorry would you be able to explain it a different way?
 
  • #4
1) have you seen Wilson's theorem? If so, mathwonk's post should look familiar

2) How do the exponents in the prime factorization tell you when one number is divisible by another? Can you answer this- does a divide b? Can you tell this just from comparing exponents?
 
  • #5
The integers taken modulo any prime p form a field. 103 is prime, so [tex]\mathbb{Z}_{103}[/tex] is a field. Thus, [tex] \forall x \in \mathbb{Z}_{103}, \exists x^{-1} \in \mathbb{Z}_{103} \ \mbox{s.t.} \ xx^{-1} = x^{-1}x = 1[/tex]. Now, if [tex]x=1 \ \mbox{or} \ x=102 = -1, \ \mbox{then} \ x^{-1} = x \ \mbox{in} \ \mathbb{Z}_{103}[/tex]. Thus when you multiply all the elements in [tex]\mathbb{Z}_{103}[/tex] together, ie. take [tex]102! = \prod_{\mathbb{Z}_{103} \ni i=1}^{102} i[/tex], you just get [tex](102)(1) = (-1)(1) = -1[/tex], and so [tex] 102! \equiv -1 \equiv 102 \ (\mbox{mod} 103)[/tex]
 
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1. What is the definition of a remainder?

A remainder is the amount left over after dividing a number by another number. For example, in the division problem 10 ÷ 3, the quotient is 3 and the remainder is 1.

2. How can I determine if a number is divisible by another number?

A number is divisible by another number if the remainder of the division is 0. This means that the number can be divided evenly by the other number without any leftover amount.

3. What is the difference between integer division and regular division?

Integer division only produces whole numbers as the quotient, while regular division can produce decimal numbers. In integer division, any decimal part of the quotient is dropped and not included in the answer.

4. How can I use remainder and divisibility rules to solve math problems?

Knowing the rules for divisibility by certain numbers can help you quickly determine if a number is divisible by another number. For example, if a number is divisible by both 2 and 3, it must also be divisible by 6.

5. Can remainders be negative?

No, remainders are always positive numbers. If the quotient is a negative number, the remainder would be the positive equivalent to make the division problem work. For example, in the division problem 10 ÷ -3, the quotient is -4 and the remainder is 2.

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