A three digit number is such that the sum of the digit in the hundred’s place and the ten’s place is 1 more than the digit in the unit’s place. It is also given that the digit in the ten’s place exceeds the square of the digit in the hundred’s place by 1, and that the square of the digit in the units place diminished by 7 is the same as the sum of the squares of the other two digits. What is the number?
(1) 346 (2) 256 (3) 458 (4) 526 (5) None of These.
option 2 256
ReplyDeleteWorked through the options :P
Sorry:)
the problem is pretty easy let the number be xyz, x can't be zero
ReplyDeletex+y=z+1
y=x^2+1
z^2-7=x^2+y^2
subtract first two
we get z=x(x+1) as x and z are digits, and x not 0
then obviously z=2,6 as these two are the only solutions possible
but with z=2 z^2-7 is negative , hence rejected
so x=2 z=6 and y=5
our number is 256
the problem is pretty easy let the number be xyz, x can't be zero
ReplyDeletex+y=z+1
y=x^2+1
z^2-7=x^2+y^2
subtract first two
we get z=x(x+1) as x and z are digits, and x not 0
then obviously z=2,6 as these two are the only solutions possible
but with z=2 z^2-7 is negative , hence rejected
so x=2 z=6 and y=5
our number is 256