You are here:

Advanced Math/Division algorithm

Advertisement


Question
Hey Paul

(page 523)
When 15 is divided by the positive integer k, the remainder is 3. For how many different values of k is this true?
a) One b) Two c)Three d) Four e) Five

I don't get why does 'the remainder is 3' mean.
Does it mean like this? : 15/k = x+ 3? If this is correct, how do i solve x to find how many different values for k? (x is the number 15/k equals plus remainder 3)

Thanks in advance

Answer
Questioner:   Emily
Category:  Advanced Math
Private:  No
 
Subject:  How many different values for k?
Question:  Hey Paul

(page 523)  << what is that?
When 15 is divided by the positive integer k, the remainder is 3. For how many different values of k is this true?
a) One b) Two c)Three d) Four e) Five

I don't get why does 'the remainder is 3' mean.
Does it mean like this? : 15/k = x+ 3? If this is correct, how do i solve x to find how many different values for k? (x is the number 15/k equals plus remainder 3)

Thanks in advance
.......................................
Hi, Emily,

The 'division algorithm' for integers says:

A = k q + r, where   r is between  0 and k-1


A is called the dividend,
k is called the divisor,
q is called the quotient
r is called the remainder.

Here you have:

15 = kq + 3.

So the divisor cannot be smaller than 4.

So try some values

15 divided by  4 gives r = 3, because  15 = 4(3) + 3  [good]
15 divided by  5 gives r = 0  [no good]
15 divided by  6 gives r = 3  [good]
15 divided by  7 gives r = 1  [no good]
15 divided by  8 gives r = 7  [no good]
15 divided by  9 gives r = 6  [no good]
15 divided by 10 gives r = 5  [no good]
15 divided by 11 gives r = 4  [no good]
15 divided by 12 gives r = 3  [good]
15 divided by 13 gives r = 2  [no good]
15 divided by 14 gives r = 1  [no good]
15 divided by 15 gives r = 0  [no good]

15 divided by something larger than 15 gives r = 15  [no good]

Looks like the answer is 3 possible divisors.

Advanced Math

All Answers


Answers by Expert:


Ask Experts

Volunteer


Paul Klarreich

Expertise

I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience

I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

Education/Credentials
-----------

©2012 About.com, a part of The New York Times Company. All rights reserved.