What number when divided by 3 leave a remainder of 1 divided by 4 leaves a remainder of 2 divided by 5 leaves a remainder of 3 and divided by 6 leaves a remainder of 4?

Answer 1

This is actually a little bit tricky to work, without just trying numbers. The first answer is 58, however there are an infinite amount of correct answers. For example 118, 178, and 238 are the next correct answers in sequence. In fact, the general solution to this answer is:n = 58 + 60k, where k is just any positive integer (0, 1, 2, 3, 4, 5, ...)

In the equation n = 58 + 60k, the 60 comes from the fact that it is the least common multiple of 3, 4, 5, and 6. The 58 is a little harder to explain. Here is how to algebraic solve this. It involves the modulus operation, so you should be familiar with that before reading the steps on how to systematically solve this problem.

n = dk + r, where n is the answer, d is the number we are dividing by, k is some integer, and r is the remainder.

k will vary from each condition, so let a,b,c,d = some integer

1. n = 3a + 1
2. n = 4b + 2
3. n = 5c + 3
4. n = 6d + 4

If we rewrite these as a modulus we have (1 below corresponds to 1 above):

1. n%3 = 1 (% = modulus)
2. n%4 = 2
3. n%5 = 3
4. n%6 = 4

Now take one of the n terms and place it into a different modulus rewrite:

We'll use the first term (#1) and the second term (#2).

n = 3a + 1 AND n%4 = 2

( 3a + 1 ) % 4 = 2

For simplicity, let's make the remainder zero, instead of two.

( 3a + 1 - 2 ) % 4 = 2 - 2

3a - 1 % 4 = 0

We need the left hand side to be some multiple of 4 for this equation to hold true. If a = -1 (find -1 by trial and error)

3(-1) - 1 = -3 - 1 = -4, and -4%4=0 is true, so this value of a works

Let j = some integer

a = -1 + 4j (we can add any multiple of 4 to our solution)

Now we need to do the exact same thing with the second two conditions (terms #3 and terms #4):

n = 5c + 3 AND n%6 = 4

5c + 3 % 6 = 4

5c - 1 % 6 = 0

Once again, if we try c = -1, we end up with:

5(-1) - 1 % 6 = 0

-5 - 1 % 6 = 0

-6%6=0 which holds true

Let k be some integer

Therefore, c = -1 + 6k

Plug a back into it's orginial equation:

n = 3a + 1, where a = -1 + 4j

n = 3(-1 + 4j) + 1 = 12j - 2

And Plug c back into it's originial equation:

n = 5c + 3, where c = -1 + 6k

n = 5(-1 + 6k) + 3 = 30k - 2

Now we have:

n = 12j - 2

n = 30k - 2

We have to do the same process as above with the new terms:

n = 12j - 2 is the same as saying n%12=-2

However, that doesn't make since, so add a multiple of 12 to it

n%12=-2(+12)=10

We only have to convert on term to the modulus form, and plug the other form into it, just like before:

n%12=10 AND n = 30k - 2

30k-2%12=10

30k-12%12=0

because 12 is just a multiple of itself, 12, we can drop the -12 (if you don't drop it you will end up with the same answer)

30k%12=0

Simplify your answer (divide both terms by 6):

5k%2=0

We need to find a value of k, where 5k%2=0 holds true:

if k=2: 10%2=0 (which is true)

Let x = some integer

So k=2x

Plug k back into an equation:

n = 30k - 2, where k=2x

n = 30(2x) - 2 = 60x - 2

Because we have now included every condition, and reduced our equation to only one independent variable, we have our answer:

n = 60x - 2, where x is some integer

if x=0: n=-2 (which all conditions hold true on)

if x=1: n=58 (which is the answer I provided up top)

if x=2: n=118

if x=3: n=178

if x=4: n=238

and so on and so forth

n = 60x - 2 is the same as the equation I provided above, it's just offset by 60.