in a 3 digit number sum of digits is 9and the units digit is twice the tens digit on adding 99 to the number the digits are reversed find the no.
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In a three digit number, the sum of the digits is 9 and the unit digit is twice the tens digit adding 99 to the number when the digits are reversed. What is the number?
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Harry Mouat
Answered Jun 21
The best way to answer this is by putting the number in truncated form, where xx is your number:
x=100m+10n+px=100m+10n+p
We can then make the equation:
m+n+p=9m+n+p=9
To do the rest of the equations is more difficult, let’s first call your reversed number yy:
y=100p+10n+my=100p+10n+m
When we add 99 to it:
y=100p+10(n+9)+(m+9)y=100p+10(n+9)+(m+9)
As we can see by this, if n≥1n≥1 then the hundreds digit will increase by 1 and if m≥1m≥1 then the tens digit increases by 1. For this reason we must consider the situations of n=0n=0 and m=0m=0 separately.
First, if m=0,n=0m=0,n=0. The units digit cannot be twice the tens digit since they are both 9.
If m=0,n≥1m=0,n≥1. The units digit is 9, so it cannot be twice the tens digit since then the tens digit would not be an integer.
If n=0,m≥1n=0,m≥1. The tens digit is 0, since (n+9)(n+9) is 9, but then we have to add 1 to it, because there is a carry from the units column due to (m+9)(m+9) being greater than 9. This means the units digit must be 0 as well to be half of it. Therefore m−1=0m−1=0and m=1m=1. Subsitute into the first equation to find 1+0+p=91+0+p=9, and p=8p=8. So our first solution is x=100∗1+10∗0+8=108x=100∗1+10∗0+8=108. We can check this solution quite trivially. The digits obviously sum to 9, 1+0+8=91+0+8=9, reversing we get 801, adding 99 to get 900. The tens digit is then 0 and the units is 0, which is twice 0.
Finally, to check if there are any more solutions, we shall consider m≥1,n≥1m≥1,n≥1. This means there is a carry from the units to the tens, so y=100p+10(n+10)+(m−1)y=100p+10(n+10)+(m−1), and there is another carry from the tens to the hundreds, so y=100(p+1)+10n+(m−1)y=100(p+1)+10n+(m−1) using this we can find that for the units to be twice the tens, m−1=2nm−1=2n. To solve this, we can just substitute nn as all the integers from 1 to 4, since any greater than 4 and m is a 2 digit number, which is impossible. We find solutions of m=3,n=1m=5,n=2m=7,n=3m=9,n=4m=3,n=1m=5,n=2m=7,n=3m=9,n=4. Using our first equation, the last two sets of solutions will not work, since they require ppto be negative, which it cannot be. With our first two we find solutions of p=5p=5 and p=2p=2 respectively. Again, we can check trivially. Our two numbers are 315 and 522. The digits obviously sum to 9 again. 315 reversed is 513 and adding 99 we get 612. 2 is twice 1, so this works as a solution. 522 reversed is 225, adding 99 we get 624, 4 is twice 2, so again this works as a solution.
So, all of the solutions are: 108, 315, and 522
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1 ANSWER

Harry Mouat
Answered Jun 21
The best way to answer this is by putting the number in truncated form, where xx is your number:
x=100m+10n+px=100m+10n+p
We can then make the equation:
m+n+p=9m+n+p=9
To do the rest of the equations is more difficult, let’s first call your reversed number yy:
y=100p+10n+my=100p+10n+m
When we add 99 to it:
y=100p+10(n+9)+(m+9)y=100p+10(n+9)+(m+9)
As we can see by this, if n≥1n≥1 then the hundreds digit will increase by 1 and if m≥1m≥1 then the tens digit increases by 1. For this reason we must consider the situations of n=0n=0 and m=0m=0 separately.
First, if m=0,n=0m=0,n=0. The units digit cannot be twice the tens digit since they are both 9.
If m=0,n≥1m=0,n≥1. The units digit is 9, so it cannot be twice the tens digit since then the tens digit would not be an integer.
If n=0,m≥1n=0,m≥1. The tens digit is 0, since (n+9)(n+9) is 9, but then we have to add 1 to it, because there is a carry from the units column due to (m+9)(m+9) being greater than 9. This means the units digit must be 0 as well to be half of it. Therefore m−1=0m−1=0and m=1m=1. Subsitute into the first equation to find 1+0+p=91+0+p=9, and p=8p=8. So our first solution is x=100∗1+10∗0+8=108x=100∗1+10∗0+8=108. We can check this solution quite trivially. The digits obviously sum to 9, 1+0+8=91+0+8=9, reversing we get 801, adding 99 to get 900. The tens digit is then 0 and the units is 0, which is twice 0.
Finally, to check if there are any more solutions, we shall consider m≥1,n≥1m≥1,n≥1. This means there is a carry from the units to the tens, so y=100p+10(n+10)+(m−1)y=100p+10(n+10)+(m−1), and there is another carry from the tens to the hundreds, so y=100(p+1)+10n+(m−1)y=100(p+1)+10n+(m−1) using this we can find that for the units to be twice the tens, m−1=2nm−1=2n. To solve this, we can just substitute nn as all the integers from 1 to 4, since any greater than 4 and m is a 2 digit number, which is impossible. We find solutions of m=3,n=1m=5,n=2m=7,n=3m=9,n=4m=3,n=1m=5,n=2m=7,n=3m=9,n=4. Using our first equation, the last two sets of solutions will not work, since they require ppto be negative, which it cannot be. With our first two we find solutions of p=5p=5 and p=2p=2 respectively. Again, we can check trivially. Our two numbers are 315 and 522. The digits obviously sum to 9 again. 315 reversed is 513 and adding 99 we get 612. 2 is twice 1, so this works as a solution. 522 reversed is 225, adding 99 we get 624, 4 is twice 2, so again this works as a solution.
So, all of the solutions are: 108, 315, and 522
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