All multiples of 10 are exceptions to this condition. Reason :

\(10^2 = 100\). On reversing 10 we get \(01\). Now square it we get \(01^2 = 1 = 01 = 001 = 0001 ..... and~so~on\). Now reversing it we get \(10,100,1000 .....and~so~on\). So, the multiples of \(10\) are exceptions. More clearly they will be in an undetermined form.

@Ram Mohith
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You can twist the question in the direction of palindromes. Say you give examples for palindromes and ask whether it works all the time ( For other numbers too). And you give three options. Yes, always , No, never and Yes , sometimes. But do not make it too obvious

There is a question on this (but I forgot and cannot search). The real pattern is 1 followed by some number of 2's and when you reverse it still makes sense

And if you take \( 14 \) in base \( 20 \) (so 24 in base 10) and square it, you get \( 18G_{20} \). If you square \( 42_{20} \) (81 in base 10) you get \( G81_{20} \).

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## Comments

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TopNewestYes. It is quite good.

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Try to find if there are any such numbers.

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Well I noticed that the 144 is 12 squared. And the prime factorization of 144 is 3^2 times 2^4 which is 4^2 and 3 and 4 are consecutive.

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AND THAT WAY WORKS

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Yes that is a good way finding such numbers. I too will try.

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P.S. I am 10 this my Brother's account

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You just take two consecutive numbers. Then multiply them. and then the rest works.

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My assumption WORKED. YAY

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I am thinking to frame a question based on these observations.

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May I assist

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Did you get still anymore numbers like these.

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I am trying to obtain a general form for these numbers or if there is some periodicity between the numbers .

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You cannot frame a question

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Why can't we ?

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It only works around 20

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All single digit integers will satisfy this condition the reason being when they are reversed the same number is obtained.

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True

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All multiples of 10 are exceptions to this condition. Reason :

\(10^2 = 100\). On reversing 10 we get \(01\). Now square it we get \(01^2 = 1 = 01 = 001 = 0001 ..... and~so~on\). Now reversing it we get \(10,100,1000 .....and~so~on\). So, the multiples of \(10\) are exceptions. More clearly they will be in an undetermined form.

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Make sense

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@Ram Mohith, There is a problem that goes something like this

\(12^2= 144 ; 21^2 = 441\)

\(122^2 = 14884; 221^2 = 48841\)

\(1222^2 = 1493284; 2221^2 = 4932841\)

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Did you know that both the endings of 31 and 19 end in 61.

31 times 31 = 961

and

19 times 19 = 361

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Square 20. We get 400.

Reverse 20. We get 02.

Square 02. We get 004

Reverse 400. We get 004

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Good one again

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It only works below 20

EDIT: The pattern is with 1 and a followed number of 2'sLog in to reply

Ok. Should see about this !!!

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WAIT 10 and 11 work

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Sorry. It is around 20

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20 means squaring and reversing

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The idea of making a problem out of this out of question

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Unless it is a proof stating question

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Yes. Your point is also correct.

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But my idea is not to write a proof based question.

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Can you help me in my other notes

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surely.

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Lets go to interesting prime powers relationship(the name)

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There is a question on this (but I forgot and cannot search). The real pattern is 1 followed by some number of 2's and when you reverse it still makes sense

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This also works with \( 13 \):

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My friend told me that on Friday but I forgot

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And if you take \( 14 \) in base \( 20 \) (so 24 in base 10) and square it, you get \( 18G_{20} \). If you square \( 42_{20} \) (81 in base 10) you get \( G81_{20} \).

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