2x0x1 is a five-digit number, what is x such that the number is divisible by 7? Please help.?

You can do this by trial and error....

and you'll get x = 6 (so, 26061).

Or realise that the last digit of the multiplier must be 3

so 2x0x / 7 must have a remainder of 2, and keep working back from there, which still gets you to x = 6

20001 + x * 1010 = 7k

20001 =>

19600 + 401 =>

19600 + 350 + 51 =>

19600 + 350 + 49 + 2 =>

7m + 2

7m + 2 + 1010x = 7k

2 + 1010x = 7 * (k - m)

x = 0, 7 * (k - m) = 2, so false

x = 1 , 7 * (k - m) = 1012, false

x = 2 , 7 * (k - m) = 2022 , false

x = 3 , 7 * (k - m) = 3032

x = 4 , 7 * (k - m) = 4042

x = 5 , 7 * (k - m) = 5052

x = 6 , 7 * (k - m) = 6062

x = 7 , 7 * (k - m) = 7072

x = 8 , 7 * (k - m) = 8082

x = 9 , 7 * (k - m) = 9092

Test each possible number and see if it's divisible by 7. For instance, 2, 1012 , 2022 , and 7072 are not divisible by 7, so x = 0 , 1 , 2 , or 7 doesn't work.

2x0x1=20000+1000x+10x+1 in value=>

1000x+10x=-20001(mod 7)

10x(100+1)=5(mod 7)=>

1010x=5(mod 7)=>

x=6 (mod 7)=>

x=6+7k, k=0, +/-(1, 2, 3, ...)

In this case, k=0 only

=>

x=6

=>

the 5-digit number is 26061.