Let a two digit integer be of the form 9 x for some integer x .
If we reverse the digits of the two digit number, it will be of the form 9 x .
What is the value of x ?
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Correct use of what we call commonsense!!!
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Thanks sir .
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Nice solution. Btw, make ab and ba as a b and b a to be more clear. Otherwise ab and ba sound the same.
It is said that "ab" is a two-digit number. So,in 9^x, the only exponent that will make it a two-digit number is 2.
Wow. Quite simple.
Since 9^x is a two digit integer, so :
x can't be 1 or 3, 4, ......
So x = 2 , which is valid also with the two digit integer 9 x
x = 2.
Ora, 9^2 = 81;
9 x 2 = 18.
A = 8;
B = 1.
it is said that its a two digit number so 9^x must not 3 digit number so the solution is just simple putting x=2 we can find the answer.. ab=81 ,, ba=18
listing possible powers of 9 that is element of Z: 9 ^ 0 = 1 9 ^ 1 = 9 9 ^ 2 = 81 9 ^ 3 = 729 ... x would never be negative because 9^x cannot produce a negative value which can satisfy 9x notice that 9^2 is the only one that has 2 digits hence, x = 2, to check : 9^2 = 81 ; 9*2 = 18 where a = 8 and b = 1 in the form 10a+b
i'm good at prediction
That's awessome . ( ⌣ ¨ )
Using trial and error method,first we go with the smallest no.
9 2 = 8 1
And,
9 × 2 = 1 8
And also, you might get this instinctively.! ;)
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Since x is a natural number , only one value of x satisfies the equation , that is 2 . Values more than 2 will result in the RHS of 3,4,5..... digits . 9 2 = 8 1 9 × 2 = 1 8