Latest TCS Aptitude Question SOLUTION: A two digit number is 18 less than the square of the sum of its digits. How many such numbers are there? (1)1 (2)2 (3)3 (4)4

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(#M40021891) TCS Logical Reasoning question Keep an EYE

A two digit number is 18 less than the square of the sum of its digits. How many such numbers are there?
(1)1
(2)2
(3)3
(4)4

Asked In TCS Ish Mishra (10 years ago)

Unsolved Read Solution (15)

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Read Solution (15) :

MANOJ KAR  (10 years ago)

2 is the answer. we have to see the numbers which become perfect squares..hence we have to check only 25,36,49,64,81,100 which are greater than 18.subtracting 18 from 81 we get 63 which fullfills the condition mentioned.........

Is this solution Helpfull? Yes (46) | No (17)  

ramanareddy  (10 years ago)

take 63 and 82.
xy+18=(x+y)^2.
63+18=(6+3)^2.
82+18=(8+2)^2.

Is this solution Helpfull? Yes (28) | No (19)  

puneeth kumar  (9 years ago)

10x+y=(x+y)^2-18
hence the above eq is a second degree polynomial it has two solutions
therefore the ans is 2

Is this solution Helpfull? Yes (22) | No   

rahul singh  (10 years ago)

xy+18=(x+y)^2
5^2=25,25-18=7
6^2=36,36-18=18 and 1+8=9
7^2=49
8^2=64,64-18=64
9^2=81,81-18=63 and 6+3=9,9^2=81 so xy=63 answer is 1.

Is this solution Helpfull? Yes (6) | No (24)  

Kumari guddu singh  (8 years ago)

For the above condition (a+b)^2=(10a+b)+18
where ab is the two digit number..

Possible values of (a+b)^2 are 16, 25, 36, 49, 64, 81, 100. from which 18 can be subtracted to get the possible values of the 2 digit. They are 18, 31, 56, 63, 82.

By trail and error, you get the the values for above conditions, which will be 63 & 82.

(6+3)^2=(63+18)=81
(8+2)^2=(82+18)=100

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Dhanraj  (8 years ago)

(a+b)^2=(10a+b)+18
Target- to find perfect square and substract 18 from that and after that satisfy the above equation
so values of perfect square for two digit - 16,25,36,49,81,100.
16-18=-8 can't take.25-18=7 can't take as it is not two digit no. 36-18=18 can take .49-18=31 can take. 81-18=63. and for 100-18=82.
Now corresponding values we get as 18,31,63,82.
Now satisfy the equation (a+b)^2=(10a+b)+18 with these values.
Only 63 and 82 is possible.
Thanx

Is this solution Helpfull? Yes (4) | No (2)  

Lalit Narayan Jadhav  (10 years ago)

xy+18=(x+y)^2.
by applying the values to x and y from 1 to 9 we get these two which satisfies the given condition.
63+18=(6+3)^2.
82+18=(8+2)^2.

you can check it by inserting this formulae in the basic c program

Is this solution Helpfull? Yes (3) | No (2)  

Danish Shaikh   (10 years ago)

36 and 81
cz
36–18=18
81–18 =63
8+1=6+3=9
both satisfies the condition

Is this solution Helpfull? Yes (1) | No (11)  

Pranai  (10 years ago)

answer is 4
two digit nos 36,49,64,81

Is this solution Helpfull? Yes (1) | No (21)  

Arjun  (10 years ago)

ans = 2 ;
63&81 and 82&100

Is this solution Helpfull? Yes (1) | No (3)  

kirthika  (9 years ago)

RAMANAREDDY why did u assume 63 and 82 ????

Is this solution Helpfull? Yes (1) | No (1)  

Danish Shaikh   (10 years ago)

63 satisfies the condition

63=(6+3)²–18

Is this solution Helpfull? Yes | No (6)  

pritam banerjee  (9 years ago)

@manokar what isthe logicbehind perfect square

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Dhanraj  (8 years ago)

(a+b)^2=(10a+b)+18
Target- to find perfect square and substract 18 from that and after that satisfy the above equation

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Dhananjay  (4 years ago)

Take N = 10a+b.
Given that, (10a+b)+18 = K2 = (a+b)2
Given number = K2

- 18 = (10a+b)

That means, when we add 18 to the given number it should be a perfect square. So K2
takes the

following values. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ....
1 to 16 are ruled out as if we subtract 18 from them, the resulting number is a single digit number.
Now 25 - 18 = 7
36 - 18 = 18
49 - 18 = 31
64 - 18 = 46
81 - 18 = 63
100 - 18 = 82
121 - 18 = 103
Now 63, 82 satisfies.
so (2) 2 is answer

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