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(#M40164660) ELITMUS Numerical Ability question Keep an EYE

the sum of a two digit number and the number obtained by reversing its digits is a square number.how many such numbers a re there??

Asked In Elitmus Sniper (7 years ago)

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Devendra Marghade  (7 years ago)

8 Numbers

Let the two digit number= (10x+y), then number obtained by reversing its digit=(10y+x)
Sum of these two numbers=(10x+y) + (10y+x) = 11(x+y)
The square number multiple of 11 is 121, where x,y lies between 1 and 9.
For 11(x+y)= 121, (x,y)=(2,9) (3,8) (4,7) (5,6) (6,5) (7,4) (8,3) (9,2) total 8 combinations.
So we can have 8 such numbers 29, 38, 47, 56, 65, 74, 83 and 92.

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Gaurishankar Kumar jaiswal  (7 years ago)

Let 2 digit no.=xy Where x,y are digits from 0 to 9
Reverse of no.=yx
According to question
Sum of no.+sum of reverse of no.=square of any no.
10x+y+10y+x=square
11*(x+y)=square
1

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