This problem from China is almost 2000 years old: Find a number that when divided by 3 gives a remainder of 2, when divided by 5 gives a remainder of 3, and when divided by 7 gives a remainder of 4.

Answer 1

look up "Chinese Remainder Theorem"
and you will find many pages for this kind of problem

You will have to know how to do arithmetic in "modular arithmetic" , and this forum is not suitable to teach you that.
I just worked out the above problem using that method and got

263

If you have enough patience, here is another way to do that question which is very easy to understand:

divided by 3 gives a remainder of 1 :
4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 ..... 263

when divided by 5 gives a remainder of 4 :
4 9 14 19 24 29 34 39 43 49 53 59 ..... 263

when divided by 7 gives a remainder of 2 :
2 9 16 23 30 37 44 51 58 ...... 263

Answer 2

ignore the 2nd part of my answer to the above, the numbers don't even match. I cut and pasted from a previous solution but the numbers were not even the same

Also note that even though 263 is a solution, it is one of an infinite number. But it said to find "a number", so the answer of 263 is a correct answer.
If you want others, especially the smallest such number, proceed as follows
looking the the divisiors of 3, 5, and 7, we would have a LCM of 105
so 263 - 105 = 158 is another solution
158 - 105 = 53 is another, and the smallest.

the long and tedious way would be"

divided by 3 gives a remainder of 2 :
2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 ..... 263

when divided by 5 gives a remainder of 3 :
3 8 13 18 23 28 33 38 43 48 53 58 ..... 263

when divided by 7 gives a remainder of 4 :
4 11 18 25 32 39 46 53 60 ...... 263

notice that 53 is the first such number.

Answer 3

Thank you so much for the help!