Question

$\huge\orange{\boxed{\sf QuEStiON}}$

A two digit number is obtained by either multiplying the sum of the digits by 8 & adding 1, or by multiplying the difference of the digits by 13 & adding 2. Find the number. How many such numbers are there?​

Answer 1

Answer:

Let the digit at units place be x and the digit at ten's place be y. Then,

Number =10y+x

According to the given conditions, we have

10y+x=8(x+y)+1⇒7x−2y+1=0

and, 10y+x=13(y−x)+2⇒14x−3y−2=0

By using cross-multiplication, we have

−2×−2−(−3)×1x

= 7×−2−14×1−y

= 7×−3−14×−21

⇒ 4+3x

= −14−14−y

= −21+281

⇒ 7x

= 28y

= 71

⇒x= 77

=1 and y= 728

=4

Hence, the number =10y+x=10×4+1=41.

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Answer 2

Answer:

Step-by-step explanation:

Let the digit at units place be x and the digit at ten's place be y. Then,

Number =10y+x

According to the given conditions, we have

10y+x=8(x+y)+1⇒7x−2y+1=0

and, 10y+x=13(y−x)+2⇒14x−3y−2=0

By using cross-multiplication, we have

−2×−2−(−3)×1

x

​

=

7×−2−14×1

−y

​

=

7×−3−14×−2

1

​

⇒

4+3

x

​

=

−14−14

−y

​

=

−21+28

1

​

⇒

7

x

​

=

28

y

​

=

7

1

​

⇒x=

7

7

​

=1 and y=

7

28

​

=4

Hence, the number =10y+x=10×4+1=41.