A two digit number is obtained by either multiplying the sum of tue digits by 8 and adding 1 or by multiplying number. How much such numbers are these?
Answers
Answer:
the number =10y+x=10×4+1=41
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.
As, it is mentioned the number is a two digit number so,
let the number in tens place be x
and
let the number in digits place be y.
Number: 10x+y
Also given in the question in the first part; the two digit number is obtained by either multiply the sum of the digits by 8 and adding 1
So,
10x+y= 8(x+y)+1
10x+y= 8x+8y+1
10x-8x-8y-y=1
2x-7y=1
Let's mark 2x-7y=1 as equation. 1
Now,solving the second part, multiplying the difference of the digits by 13 and adding 2.
10x+y= 13(x+y)+2
10x+y=13x+13y+2
10x-13x+13y+y=2
-3x+14y=2
Let's mark -3x+14y=2 as equation 2
Now, multiply equation 1 by 2 and add in equation. 2
2x-7y =1=>1(multiply by 2)
4x-14y=2
Let's mark 2
4x-14y=2 as equation. 3
Now add equation 2 and 3,
-3x+14y=2
+ 4x-14y =2
____________
x =4
Now put x=4 in equation 1
2(4)-7y =1
8 -7y =1
8 -1 =7y
7y = 7
y =7/7
y =1
So, now look out for the values of x and y,
x=4
and
y=1
So, the number therefore=41