Question

The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is 109. If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order. Find the number.
a. 773
b. 683
c. 944
d. 863; The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is 109. If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order. Find the number.; a. 773; b. 683; c. 944; d. 863

Answer 1

Answer:863

Step-by-step explanation:

Let the number be abc.

Then a + b + c= 17 .....(1)

a^2+b^2+c^2=109 .....(2)

100a+10b+c -495 = 100c+10b+a ......(3)

From 3, we get a - c = 5

So the possibilities for (a, c, b) are (6,1,10), (7,2,8), (8,3,6), (9,4,4)

From the above, (8,3,6) satisfies the condition.

Number is 863

Answer 2

The number is 863 (Option d is the correct answer)

Given,

The sum of the digits of a three digit number is 17,

the sum of the squares of its digits is 109

subtract 495 from the number

To Find,

Find the number consisting of the same digits written in the reverse order

Solution,

Let's consider the three digit is : 100x+10y+z

According to the question,

a sum of the digits of a three-digit number is 17

so, x+y+z=17.................(i)

again, the sum of the squares of its digits is 109

so, x²+ y²+z²= 109 ................ (ii)

Now, if we subtract 495 from that number, we shall get a number consisting of the same digits written in the reverse order-

100x+10y+z−495=100z+10y+x

or, 99x−99z=495

or, x−z=5

or, x=z+5..............(iii)

From (i) and (iii) we can get:

(z+5)+y+z=17

y=12−2z...........(iv)

From (i),(iii), and (iv)

(z+5)² +(12−2z)² +z² =109

or, z²+25+10z+144+4z² −48z+z²=109

or, 3z² −19z+30=0 which is a quadratic equation.

from this we get, z= 10/3 , is not possible as digit of any number can't be a fraction

hence, z=3

from (iii) and (iv)

y=6 and x=8

Hence, the number is 863

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