Question

Determine all natural numbers x, y and z such that x < y < z, and xyz + xy + xz + yz + x + y +2+1 = 2020​

Answer 1

Answer:

Step-by-step explanation:

$Given,\\xyz+xy+xz+yz+x+y+z+1=2020\\=>(xyz+xy)+(xz+x)+(yz+y)+(z+1)=2020\\=>xy(z+1)+x(z+1)+y(z+1)+(z+1)=2020\\=>(z+1)[(xy+x)+(y+1)]=2020\\=>(x+1)(y+1)(z+1)=2020$

Knowing that x, y and z are natural numbers, we look for the possible factors of 2020:

they are : 1,2,2,5,101

so possible combination of (x,y,z) are : (2,10,101),(4,5,101),(2,5,202),(2,2,505)

Answer 2

Given :  xyz + xy + xz + yz + x + y +z+1 = 2020​

natural numbers x, y and z such that x < y < z

To Find : Determine all natural numbers satisfying the given condition

Solution:

xyz + xy + xz + yz + x + y +z+1 = 2020​

=> xyz + xy + xz + x  + yz  + y +z+1 = 2020​

=> xy(z + 1) + x(z + 1) + y(z + 1) + 1(z + 1) = 2020

=> (z + 1)(xy + x + y + 1) = 2020

=> (z + 1)(x(y + 1) + 1(y + 1)) = 2020

=> (x + 1)(y + 1)(z +1) = 2020

2020 = 2 * 2 * 5 * 101

x + 1          y + 1           z + 1

2                5                202

2                10               101

4                 5                101

Hence x , y  , z   are

x      y     z

1      4     201

1      9     100

3     4      100

(1 , 4 , 201) , (1 , 9 , 100) and ( 3 , 4 , 100)

Verification :

xyz + xy + xz + yz + x + y +z+1 = 2020​

(1 , 4 , 201)

LHS

1 * 4 * 201  + 1 * 4 + 1 * 201 + 4 * 201 +  1 + 4 + 201 + 1

= 804 + 4 + 201 + 804  + 207

= 2020

Similarly for others

(x , y , z )  =   (1 , 4 , 201) , (1 , 9 , 100) and ( 3 , 4 , 100)

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