Number magic – Mathematical puzzle
If you multiply me by 2, subtract 1, and read the reverse the result you’ll find me.
Which numbers can I be?
37, 397, 3997, 39997, 399…..997 etc.
37 * 2 – 1 = 73
397 * 2 – 1 = 793
399….997 * 2 – 1 = 799….993
Method to find the number,
We first try to find the possible number which consist of two digits.
Let the number be x , such that x = AB (where A and B are two digits)
In other words, x = 10 * A + B,
It holds that 2 * x – 1 = 20*A + 2*B – 1
If we reverse this number, we’ll get the original number AB back.
But then it of course also hold that if you reverse the original number (that becomes 10 * B + A), you will get 20*A + 2*B – 1.
20 * A + 2 * B – 1 = 10 * B + A
We can equate this as-
19 * A = 8 * B + 1.
The right hand side of this equation is always odd and hence the left part should be odd as well.
This implies that A should be odd.
The right part is always smaller than 73.
So A cannot be larger than 3.
If we try A = 1,
we do not find a solution.
But A = 3 has a solution with B = 7.
So the number we search for is 37. This is the only number with two digits that satisfies our requirements.
Nice answer, but you forgot to include the special case of “1”: 2*1 – 1 = 1. 🙂
Thanks @dougbell, Yup 1 would also be in answer
Why you deleted your’s answer
I deleted my answer because it was incomplete. Your answer is much better.
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