4, 7 and 9 are mutually coprime, so you can use the Chinese remainder theorem.
Start with
![x=7\cdot9+4\cdot2\cdot9+4\cdot7\cdot5](https://tex.z-dn.net/?f=x%3D7%5Ccdot9%2B4%5Ccdot2%5Ccdot9%2B4%5Ccdot7%5Ccdot5)
Taken mod 4, the last two terms vanish and we're left with
![x\equiv63\equiv64-1\equiv-1\equiv3\pmod4](https://tex.z-dn.net/?f=x%5Cequiv63%5Cequiv64-1%5Cequiv-1%5Cequiv3%5Cpmod4)
We have
, so we can multiply the first term by 3 to guarantee that we end up with 1 mod 4.
![x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5](https://tex.z-dn.net/?f=x%3D7%5Ccdot9%5Ccdot3%2B4%5Ccdot2%5Ccdot9%2B4%5Ccdot7%5Ccdot5)
Taken mod 7, the first and last terms vanish and we're left with
![x\equiv72\equiv2\pmod7](https://tex.z-dn.net/?f=x%5Cequiv72%5Cequiv2%5Cpmod7)
which is what we want, so no adjustments needed here.
![x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5](https://tex.z-dn.net/?f=x%3D7%5Ccdot9%5Ccdot3%2B4%5Ccdot2%5Ccdot9%2B4%5Ccdot7%5Ccdot5)
Taken mod 9, the first two terms vanish and we're left with
![x\equiv140\equiv5\pmod9](https://tex.z-dn.net/?f=x%5Cequiv140%5Cequiv5%5Cpmod9)
so we don't need to make any adjustments here, and we end up with
.
By the Chinese remainder theorem, we find that any
such that
![x\equiv401\pmod{4\cdot7\cdot9}\implies x\equiv149\pmod{252}](https://tex.z-dn.net/?f=x%5Cequiv401%5Cpmod%7B4%5Ccdot7%5Ccdot9%7D%5Cimplies%20x%5Cequiv149%5Cpmod%7B252%7D)
is a solution to this system, i.e.
for any integer
, the smallest and positive of which is 149.
21, it says it right in the name.
aline it like that then do 6×0 then put that answer ,you know what I'm jut going to show you. 60×
16
360
600
↑then add 360 and 600 and that's all. Sorry that the numbers look all funny.