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16. The three digit number XYZ when divided by 8, gives as quotient the two
YZ when divided by 8, gives as quotient the two digit number Zx
remainder Y. The number XYZ is
um nf its digits.
Answers
Answer:these conditions won't be forming a no
The proof is given below
Step-by-step explanation:
Since the 3 digit number is given as XYZ
We can express it as 100x+10y+z
Now as given xyz/8= yz
And also, xyz/8= zx+2 (since it's given as 2 is remainder)
So we can deduce that
yz=zx+2
Now if we consider the three equations and express them in the form of 100x+10y+z
Then, 100x+ 10y+ z= 80y+8z
Or, 100x - 70y-7z=0
Again 100x + 10y + z = 80z + 8x + 2
Or, 92x + 10y - 79z = 16
Lastly 10y+ z=10z+x+2
Or, -x+10y - 9z = 2
If we go on solving for x y and z we will get decimal values as
x= 0.15, y= 0.21 and z= 0
Hence as these Values are unlikely for being digits of a no, hence these conditions won't be forming a no
Answer:
XYZ = 435
435 = 54 * 8 + 3
XYZ = ZX * 8 + Y
Step-by-step explanation:
Correct Question is : The three digit number XYZ when divided by 8. gives as quotient the two digt number zx , remainder Y. The number XYZ is
XYZ = 100X + 10Y + Z
when divided by 8. gives as quotient the two digt mmber ZX & Remainder = Y
=> 8 * (10Z + X) + Y = 100X + 10Y + Z
=> 80Z + 8X + Y = 100X + 10Y + Z
=> 92X + 9Y = 79Z
=> 9Y = 79Z - 92X
=> 9Y = 81Z - 2Z - 90X - 2X
=> 9Y = 9(9Z - 10X) - (2X + 2Z)
=> Y = (9Z - 10X) - (2/9)(X + Z)
to have integral value :
=> X + Z = 9
9Z - 10X - 2 < 10 to have Y as single digit & 9Z - 10X - 2 > 0
=> 9Z - 10X < 12
=> 9Z -10(9-Z) < 12
=> 9Z - 90 + 10Z < 12
=> 19Z < 102
=> Z ≤ 5
9Z - 10X - 2 > 0
=> 9Z - 10(9-Z) > 2
=> 9Z - 90 + 10Z > 2
=> 19Z > 92
=> Z ≥ 5
from Both only possible Solution Z = 5
=> X = 4
& Y = (9Z - 10X) - (2/9)(X + Z)
=> Y = (9*5 - 10*4) - (2/9)(4 + 5)
=> Y = 5 - 2
=> Y = 3
Hence XYZ = 435
435 = 54 * 8 + 3
XYZ = ZX * 8 + Y