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?
Answers
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:
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.