When a two digit number is added to its reverse, the result is 143. The number is 3 less than the sum of the squares of its digits. Calculate this number.
bhalaraneha:
number is 94.
Answers
Answered by
5
let the number be N = a b where b is the units digit. a = the tens digit.
the value of the number N = 10 * a + b
number after reversing : b a
the value of the number after reversing : 10 * b + a
sum of the two numbers = 10 a + b + 10 b + a = 11 (a + b)
given, 11 ( a + b) = 143 => a + b = 13 --- (1)
sum of squares of digits = a² + b² = (a+b)² - 2 a b
given, N = 10 a + b = (a² + b²) - 3
=> 10 a + b = (13² - 2 a b) - 3 = 166 - 2 a b
Hence, 10 a + b + 2 a b = 166 --- (2)
do (2) - 10 * (1) : - 9 b + 2 a b = 36
so b (2a -9) = 36 --- (4) .
substitute value of b from (1) in (4) :
(13 - a) ( 2a - 9) = 36
=> -2 a² + 35 a - 117 = 36
=> 2 a² - 35 a + 153 = 0
discr: 35² - 8 * 153 = 1
a = (35 +- 1)/ 4 = 9 or 8.5
we take the integer value only as a and b are integers less than 10.
so a = 9 and b = 4
so the number is 94.
the value of the number N = 10 * a + b
number after reversing : b a
the value of the number after reversing : 10 * b + a
sum of the two numbers = 10 a + b + 10 b + a = 11 (a + b)
given, 11 ( a + b) = 143 => a + b = 13 --- (1)
sum of squares of digits = a² + b² = (a+b)² - 2 a b
given, N = 10 a + b = (a² + b²) - 3
=> 10 a + b = (13² - 2 a b) - 3 = 166 - 2 a b
Hence, 10 a + b + 2 a b = 166 --- (2)
do (2) - 10 * (1) : - 9 b + 2 a b = 36
so b (2a -9) = 36 --- (4) .
substitute value of b from (1) in (4) :
(13 - a) ( 2a - 9) = 36
=> -2 a² + 35 a - 117 = 36
=> 2 a² - 35 a + 153 = 0
discr: 35² - 8 * 153 = 1
a = (35 +- 1)/ 4 = 9 or 8.5
we take the integer value only as a and b are integers less than 10.
so a = 9 and b = 4
so the number is 94.
Answered by
2
let the tens digit of the no. be 'x'. and ones digit be 'y'
the no. will be 10x+y
according to the question,,10x+y + 10y+x=143
11x+11y=143
by solving we get x+y=13
y=x-13 ..............eq 1
according to the question,,10x+y+3=x²+y² ............................eq 2
substituting eq 1 in eq 2
10x+13-x+3=x²+(13-x)²
by solving we get 2x²-35x+153=0
Solving the quadratic equation
a=2 b=-35 c=153
determinant=b²-4ac
=1225-1224
=1
√D =1
x=-b+√D÷2a or x=-b-√D÷2a
x=9 or x=34/4(ruled out since digits cannot
be fractions)
hence x=9
y=13-9
=4
the number is 94.
hope this helps u...
the no. will be 10x+y
according to the question,,10x+y + 10y+x=143
11x+11y=143
by solving we get x+y=13
y=x-13 ..............eq 1
according to the question,,10x+y+3=x²+y² ............................eq 2
substituting eq 1 in eq 2
10x+13-x+3=x²+(13-x)²
by solving we get 2x²-35x+153=0
Solving the quadratic equation
a=2 b=-35 c=153
determinant=b²-4ac
=1225-1224
=1
√D =1
x=-b+√D÷2a or x=-b-√D÷2a
x=9 or x=34/4(ruled out since digits cannot
be fractions)
hence x=9
y=13-9
=4
the number is 94.
hope this helps u...
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