Question

if x,y,z are digits such that (100x+10y+z)(x+y+z)=2005,then find x,y,z. Ace only can find​

Answer 1

Answer:

we know that from 2 to 9 2005 is divisble by only 5,

so that we have x+y+z = 5

so that 2005/5 = 401

we have to set 100x+10y+z this equation that this value is = 401

x+y+z = 5

100x + 10y + z = 401

put x=4, y=0,z=1 that satisfy both equation.

(100*(4) + 10*(0) + 1) * (4+0+1) = 401*5 =2005

Answer 2

Answer:

$x=4,y=0$ and $z=1$

Step-by-step explanation:

Let $N_1 =100x+10y+z$ . clearly, N is a three digit number(as any three digit no. can be expressed is this fashion.

say $N_{2}= x+y+z.$ This is sum of the digits of the same number.

Now intutively look that 2005  can be decomposed into product of two primes 5 and 401.  If we take 1 and 2005 OR 2005 and 1 then we are not getting three digit number. so possible combination will be $N_1 = 401$ and $N_2 = 5$. This gives values of x,y and z as 4,0 and 1 respectively.