Simon was given a number and asked to divide it by 120. He divided the number by 5,6 and 7 and got 3,2 and 2 as remainders respectively. The remainder when the number is divided by 120 is

Let the number be $$N$$.

When the number is successively divided by $$5,6,$$ and $$7$$, the remainders are $$3,2,$$ and $$2$$ respectively.

Therefore,

$$N=5q_1+3$$

The quotient $$q_1$$ is then divided by $$6$$:

$$q_1=6q_2+2$$

The quotient $$q_2$$ is then divided by $$7$$:

$$q_2=7q_3+2$$

Now substitute backwards.

From

$$q_2=7q_3+2,$$

we get

$$q_1=6(7q_3+2)+2$$

$$=42q_3+12+2$$

$$=42q_3+14.$$

Therefore,

$$N=5(42q_3+14)+3$$

$$=210q_3+70+3$$

$$=210q_3+73.$$

So the number can be written as

$$N=210q_3+73.$$