asdfasdfasdf
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- Jan 13, 2017
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The sum of the three digits is 9. The tens digit is 1 more than the hundreds digit. When the digits are reversed, the new number is 99 less than the original number. Find the original number.
I did the number as abc
so a+b+c=9
and a+(a+1)+c=9
so 2a+c=8
also I have cba+99=abc
That's all I have.
Thanks Harry the Cat, helped a lot.
My final work :
Let the number be abc ... this statement is still incorrect
a+b+c=9
a+(a+1)+c=9
2a+c=8
100a+10b+c-99=100c+10b+a
99a-99=99c
Substitute (bolded)
99a-99=-198a+792
-891=-297a
a=3
Going back to top (underlined)
3+3+1+c=9
7+c=9
c=2
b=4
So the number is 342
The sum of the three digits is 9. The tens digit is 1 more than the hundreds digit. When the digits are reversed, the new number is 99 less than the original number. Find the original number.
Then maybe try this lesson, which reflects the methodology you were given here.I don't think that is the answer to my question... Thanks Anyways.
This is the right answer because the tens digit is 2x the hundredths digit and the units digit is 1 more than the hundredths digit and they add up to 9Finished!
Thanks Harry the Cat, helped a lot.
My final work :
Let the number be abc
a+b+c=9
a+(a+1)+c=9
2a+c=8
100a+10b+c-99=100c+10b+a
99a-99=99c
Substitute (bolded)
99a-99=-198a+792
-891=-297a
a=3
Going back to top (underlined)
3+3+1+c=9
7+c=9
c=2
b=4
So the number is 342
This is the right answer because the tens digit is 2x the hundredths digit and the units digit is 1 more than the hundredths digit and they add up to 9