the sum of the digits of a two-digit number is 12. when the digits are reversed, the new number is 18 less than the original number. check your answer. HELP~
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If the number has the form \(a_0a_1\), then we can make the system:
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I'm lost :c
...this exhibits my love/hate relationship with Diophantine equations. Nice formulation Mr. Math :)
Let me explain. You have a number of two digits, call it \(a_0a_1\), where \(a_0\) is the 10's digit and \(a_1\) is the 1's digit. Now we're given that the sum of the two digits is 12. We can write that as \(a_0+a_1=12\). Following so far?
ohh yeah, ohkay keep going..
Good. The second information we are given is that when reversing the digits [that's writing it as \(a_1a_0\)], the new number is 18 less than the original number. First you need to notice that if the original number is x, we can write \(x=10a_0+a_1\). Call the new number y, then \(y=10a_1+a_0\). Hence the second equation of the system is:
All you're left to do here is to solve the system for \(a_0\) and \(a_1\).
Can you do the steps and explain to me please? I seem to learn better by looking at the steps done.
\[a_0+a_1=12\to a_0=12-a_1\]\[10a_0+a_1=10a_1+a_0+18\to9a_0-9a_1=18\]multiply first eqn by 9\[9a_0+9a_1=108\]\[9a_0-9a_1=18\]add these together...\[18a_0=126\to a_0=7\]now we can find the other number with the formula for a_1 on the first line\[a_0+a_1=12\to a_1=12-a_0=12-7=5\]hence \[a_0=7\]\[a_1=5\]so the numbers are\[57,75\]as UnkleRhaukus said
the numbers are 57, and 75 but which is the original?