Q: How many 3-digit numbers (in base 10, with no leading zero) when add with 497, the digit of the sum is exactly reverse with the digit of the original number.
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Any hints are appreciated.
Thanks in advance.
Q: How many 3-digit numbers (in base 10, with no leading zero) when add with 497, the digit of the sum is exactly reverse with the digit of the original number.
====
Any hints are appreciated.
Thanks in advance.
Hello, yutiwu!
As written, the problem has no solution.
I must assume that there is a typo . . .
How many 3-digit numbers (in base 10, with no leading zero) when added to 495,
the digits of the sum is exactly reverse with the digit of the original number.
The original number is: .$\displaystyle N \:=\:100a + 10b + c$
Its reversal is: .$\displaystyle N^* \:=\:100c + 10b + a$
We are told: .$\displaystyle N + 495 \:=\:N^*$
We have: .$\displaystyle 100a + 10b + c + 495 \:=\:100c + 10b + a \quad\Rightarrow\quad 495 \:=\:99c - 99a $
. . . . . . . . $\displaystyle 99(c-a) \:=\:495 \quad\Rightarrow\quad c-a \:=\:5$
There are 4 choices: .$\displaystyle \begin{array}{cc} a & c \\ \hline 1 & 6 \\ 2 & 7 \\ 3 & 8 \\ 4 & 9 \end{array}$
And $\displaystyle b$ can be any of the 10 available digits.
Therefore, there are 40 such numbers.
Thanks man, Soroban! That actually cleared my doubts.
Hmm, a typo, eh.
I kindna figure it out somewhere along the line when I'm coming out with a solution.
But, well, since it is question for my assignment, I guess I'll put a statement to prove it has no real answers till I can reconfirm the question from my lecturer.