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How many 2 digits number are there which when subtracted from the number formed by reversing its digits as well as when added to the number formed by reversing its digit, result is a perfect square.
Read Solution (Total 7)
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- 65 is one such number.
If xy is number, then 11(x+y) and 9*(x-y) should be perfect square.
which is possible when
x+y=11 and x-y = 1,4,9
value of x,y is valid only when x-y=1, then x=6,y=5
so number is 65. - 11 years agoHelpfull: Yes(34) No(7)
- ans 56
yx-xy=9(y-X)
yx+xy=11(x+y)
y-x=1,4,9
y+x=11
possible xy=56
- 11 years agoHelpfull: Yes(12) No(3)
- let 10x+y be two digit no
the 10x+y+10y+x=11x+11y
10x+y-10y-x=9x-9y
both should be perfect suare
ans is 65 - 11 years agoHelpfull: Yes(8) No(4)
- I think the answer of @Siddharth is correct. Please notice the language of the question. The number is subtracted FROM the number formed by reversing its digits.
- 11 years agoHelpfull: Yes(8) No(0)
- 65 is one such number.
If xy is number, then 11(x+y) and 9*(x-y) should be perfect square.
which is possible when
x+y=11 and x-y = 1,4,9
value of x,y is valid only when x-y=1, then x=6,y=5
so number is 65. - 11 years agoHelpfull: Yes(6) No(3)
- let the number be 10x+y, therefore as per problem 10x+y-10y-x=9(x-y)
and 10x+y+10y+x=11(x+y) this implies that x+y should be 11 and x-y can be 1 or 9...65 satisfies both the condition.... - 11 years agoHelpfull: Yes(2) No(2)
- 65 is right
- 11 years agoHelpfull: Yes(1) No(2)
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