Simple math problem?

Question

when N is divided by 10 reminder is 1 and when N is divided by 3 remainder is 2. what is the remainder when N is divided by 30?

How to solve this?

Answer 1

check out with an example

say N is 41

clearly it fits the conditions

so when 41 is divided by 30 we get 11 as the remainder

N can also be 11

again the remainder is 11

one more example to resolve the ambiguities

71 fits N

once again remainder is 11

Answer 2

Let N = 11 + 30x Then N = 10 + 1 + 30x, or 10(3x + 1) + 1, and N = 9 + 2 + 30x, or 3(10x + 3) + 2, satisfying the 2 conditions. The remainder, upon division by 30, is 11.

Answer 3

Let's say N has the form of 30m + n. (0<= n < 30 for obvious reason)

1) When we divide N by 10
(30m + n) mod 10 = (30m mod 10) + (n mod 10) = 0 + (n mod 10)
n mod 10 = 1 <-> n =1, 11, or 21

2) When we divide N by 3
(30m + n) mod 3 = (30m mod 3) + (n mod 3) = 0 + n mod 3
From the choices of n = 1, 11 or 21, the only n that satisfies n mod 3 = 2 is n = 11.

Therefore the remainder when N is divided by 30 is 11.

Answer 4

the remainder is 11

Answer 5

20

Answer 6

First we state the congruences you were given as equations:

n=10x+1
n=3k+2

This allows us to solve by setting:

10x+1=3k+2

Restating this as a congruence relation:

10x+1≡2 mod 3

Which yields:

x≡1 mod 3

So substituting this back in:

n=10(1 + 3k) + 1
n=30k+11
n≡11 mod 30

And we are done.

Answer 7

the remainder will be 11
you solve using modulo arithmetic