Mixing
Sum of digits of n^n
$begingroup$
Can I receive some assistance please. I was going through some questions when is saw this one,
Let A be the sum of the digits of 16^16. Let B be the sum of the digits of A. What is the sum of the digits of B without finding A.
Well, initially I thought that the sum of the digits of 16^n was (6n + 1) but that fails for n>10. Help?
elementary-number-theory
edited Jan 28 at 0:41
Omnomnomnom
129k792185
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
$endgroup$
add a comment |
$begingroup$
Can I receive some assistance please. I was going through some questions when is saw this one,
Let A be the sum of the digits of 16^16. Let B be the sum of the digits of A. What is the sum of the digits of B without finding A.
Well, initially I thought that the sum of the digits of 16^n was (6n + 1) but that fails for n>10. Help?
elementary-number-theory
edited Jan 28 at 0:41
Omnomnomnom
129k792185
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
$endgroup$
add a comment |
$begingroup$
Can I receive some assistance please. I was going through some questions when is saw this one,
Let A be the sum of the digits of 16^16. Let B be the sum of the digits of A. What is the sum of the digits of B without finding A.
Well, initially I thought that the sum of the digits of 16^n was (6n + 1) but that fails for n>10. Help?
elementary-number-theory
edited Jan 28 at 0:41
Omnomnomnom
129k792185
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
$endgroup$
Can I receive some assistance please. I was going through some questions when is saw this one,
Let A be the sum of the digits of 16^16. Let B be the sum of the digits of A. What is the sum of the digits of B without finding A.
Well, initially I thought that the sum of the digits of 16^n was (6n + 1) but that fails for n>10. Help?
elementary-number-theory
elementary-number-theory
edited Jan 28 at 0:41
Omnomnomnom
129k792185
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
edited Jan 28 at 0:41
Omnomnomnom
129k792185
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
edited Jan 28 at 0:41
Omnomnomnom
129k792185
edited Jan 28 at 0:41
Omnomnomnom
129k792185
edited Jan 28 at 0:41
Omnomnomnom
129k792185
129k792185
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
asked Jan 28 at 0:24
Varaun RamgoolieVaraun Ramgoolie
11
11
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
A little estimating shows that $16^{16}$ has $20$ digits. Therefore $A≤180$. Therefore $B≤27$.
Now, if you kept on iterating the digit sum, you'd get to $7$ since $16^{16}equiv 7 pmod 9$. Thus $Bequiv 7 pmod 9$ So $Bin {7,16,25}$.
It follows that the sum of the digits of $B$ is $7$.
Worth noting: a simple search through the possible values for $A$ shows that $25$ is not possible, but both $7$ and $16$ are. I don't see an easy way to eliminate either of these. In truth, $B=16$ but I don't see how to get there without heavier computation.
answered Jan 28 at 0:34
lulululu
43.2k25080
$endgroup$
$begingroup$
You don't even need to check that 25 is not possible as a value for $B$, since in that case the sum of digits of $B$ is still 7. All you need is $B < 79$
$endgroup$
– Rolf Hoyer
Jan 28 at 0:54
$begingroup$
@RolfHoyer You mean $A$? But $A=88$ as it happens. I agree that you don;'t need to eliminate $25$ to see that the digit sum of $B$ is $7$ (and have edited accordingly).
$endgroup$
– lulu
Jan 28 at 0:55
$begingroup$
Well $79$ is the smallest value possible for $B$ that is equal to 7 modulo 9 but with a digital sum greater than 7. The corresponding bound on $A$ would be $A < 799999999$, ie the smallest number equal to 7 modulo 9 with a digital sum at least 79.
$endgroup$
– Rolf Hoyer
Jan 28 at 1:00
$begingroup$
Wouldn’t B < or equal to 17 and not 27?
$endgroup$
– Varaun Ramgoolie
Jan 28 at 12:05
$begingroup$
@VaraunRamgoolie Oh, you can definitely get a better estimate on $B$, no question. As I say (for different reasons) it's not hard to show that $B$ is either $7$ or $16$. I just don't see an elementary way to rule out $7$. Just tightening the bound on $B$ isn't going to do that, so far as I can see.
$endgroup$
– lulu
Jan 28 at 12:46
add a comment |
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$begingroup$
A little estimating shows that $16^{16}$ has $20$ digits. Therefore $A≤180$. Therefore $B≤27$.
Now, if you kept on iterating the digit sum, you'd get to $7$ since $16^{16}equiv 7 pmod 9$. Thus $Bequiv 7 pmod 9$ So $Bin {7,16,25}$.
It follows that the sum of the digits of $B$ is $7$.
Worth noting: a simple search through the possible values for $A$ shows that $25$ is not possible, but both $7$ and $16$ are. I don't see an easy way to eliminate either of these. In truth, $B=16$ but I don't see how to get there without heavier computation.
answered Jan 28 at 0:34
lulululu
43.2k25080
$endgroup$
$begingroup$
You don't even need to check that 25 is not possible as a value for $B$, since in that case the sum of digits of $B$ is still 7. All you need is $B < 79$
$endgroup$
– Rolf Hoyer
Jan 28 at 0:54
$begingroup$
@RolfHoyer You mean $A$? But $A=88$ as it happens. I agree that you don;'t need to eliminate $25$ to see that the digit sum of $B$ is $7$ (and have edited accordingly).
$endgroup$
– lulu
Jan 28 at 0:55
$begingroup$
Well $79$ is the smallest value possible for $B$ that is equal to 7 modulo 9 but with a digital sum greater than 7. The corresponding bound on $A$ would be $A < 799999999$, ie the smallest number equal to 7 modulo 9 with a digital sum at least 79.
$endgroup$
– Rolf Hoyer
Jan 28 at 1:00
$begingroup$
Wouldn’t B < or equal to 17 and not 27?
$endgroup$
– Varaun Ramgoolie
Jan 28 at 12:05
$begingroup$
@VaraunRamgoolie Oh, you can definitely get a better estimate on $B$, no question. As I say (for different reasons) it's not hard to show that $B$ is either $7$ or $16$. I just don't see an elementary way to rule out $7$. Just tightening the bound on $B$ isn't going to do that, so far as I can see.
$endgroup$
– lulu
Jan 28 at 12:46
add a comment |
$begingroup$
A little estimating shows that $16^{16}$ has $20$ digits. Therefore $A≤180$. Therefore $B≤27$.
Now, if you kept on iterating the digit sum, you'd get to $7$ since $16^{16}equiv 7 pmod 9$. Thus $Bequiv 7 pmod 9$ So $Bin {7,16,25}$.
It follows that the sum of the digits of $B$ is $7$.
Worth noting: a simple search through the possible values for $A$ shows that $25$ is not possible, but both $7$ and $16$ are. I don't see an easy way to eliminate either of these. In truth, $B=16$ but I don't see how to get there without heavier computation.
answered Jan 28 at 0:34
lulululu
43.2k25080
$endgroup$
$begingroup$
You don't even need to check that 25 is not possible as a value for $B$, since in that case the sum of digits of $B$ is still 7. All you need is $B < 79$
$endgroup$
– Rolf Hoyer
Jan 28 at 0:54
$begingroup$
@RolfHoyer You mean $A$? But $A=88$ as it happens. I agree that you don;'t need to eliminate $25$ to see that the digit sum of $B$ is $7$ (and have edited accordingly).
$endgroup$
– lulu
Jan 28 at 0:55
$begingroup$
Well $79$ is the smallest value possible for $B$ that is equal to 7 modulo 9 but with a digital sum greater than 7. The corresponding bound on $A$ would be $A < 799999999$, ie the smallest number equal to 7 modulo 9 with a digital sum at least 79.
$endgroup$
– Rolf Hoyer
Jan 28 at 1:00
$begingroup$
Wouldn’t B < or equal to 17 and not 27?
$endgroup$
– Varaun Ramgoolie
Jan 28 at 12:05
$begingroup$
@VaraunRamgoolie Oh, you can definitely get a better estimate on $B$, no question. As I say (for different reasons) it's not hard to show that $B$ is either $7$ or $16$. I just don't see an elementary way to rule out $7$. Just tightening the bound on $B$ isn't going to do that, so far as I can see.
$endgroup$
– lulu
Jan 28 at 12:46
add a comment |
$begingroup$
A little estimating shows that $16^{16}$ has $20$ digits. Therefore $A≤180$. Therefore $B≤27$.
Now, if you kept on iterating the digit sum, you'd get to $7$ since $16^{16}equiv 7 pmod 9$. Thus $Bequiv 7 pmod 9$ So $Bin {7,16,25}$.
It follows that the sum of the digits of $B$ is $7$.
Worth noting: a simple search through the possible values for $A$ shows that $25$ is not possible, but both $7$ and $16$ are. I don't see an easy way to eliminate either of these. In truth, $B=16$ but I don't see how to get there without heavier computation.
answered Jan 28 at 0:34
lulululu
43.2k25080
$endgroup$
A little estimating shows that $16^{16}$ has $20$ digits. Therefore $A≤180$. Therefore $B≤27$.
Now, if you kept on iterating the digit sum, you'd get to $7$ since $16^{16}equiv 7 pmod 9$. Thus $Bequiv 7 pmod 9$ So $Bin {7,16,25}$.
It follows that the sum of the digits of $B$ is $7$.
Worth noting: a simple search through the possible values for $A$ shows that $25$ is not possible, but both $7$ and $16$ are. I don't see an easy way to eliminate either of these. In truth, $B=16$ but I don't see how to get there without heavier computation.
answered Jan 28 at 0:34
lulululu
43.2k25080
edited Jan 28 at 0:54
answered Jan 28 at 0:34
lulululu
43.2k25080
answered Jan 28 at 0:34
lulululu
43.2k25080
answered Jan 28 at 0:34
lulululu
43.2k25080
43.2k25080
$begingroup$
You don't even need to check that 25 is not possible as a value for $B$, since in that case the sum of digits of $B$ is still 7. All you need is $B < 79$
$endgroup$
– Rolf Hoyer
Jan 28 at 0:54
$begingroup$
@RolfHoyer You mean $A$? But $A=88$ as it happens. I agree that you don;'t need to eliminate $25$ to see that the digit sum of $B$ is $7$ (and have edited accordingly).
$endgroup$
– lulu
Jan 28 at 0:55
$begingroup$
Well $79$ is the smallest value possible for $B$ that is equal to 7 modulo 9 but with a digital sum greater than 7. The corresponding bound on $A$ would be $A < 799999999$, ie the smallest number equal to 7 modulo 9 with a digital sum at least 79.
$endgroup$
– Rolf Hoyer
Jan 28 at 1:00
$begingroup$
Wouldn’t B < or equal to 17 and not 27?
$endgroup$
– Varaun Ramgoolie
Jan 28 at 12:05
$begingroup$
@VaraunRamgoolie Oh, you can definitely get a better estimate on $B$, no question. As I say (for different reasons) it's not hard to show that $B$ is either $7$ or $16$. I just don't see an elementary way to rule out $7$. Just tightening the bound on $B$ isn't going to do that, so far as I can see.
$endgroup$
– lulu
Jan 28 at 12:46
add a comment |
$begingroup$
You don't even need to check that 25 is not possible as a value for $B$, since in that case the sum of digits of $B$ is still 7. All you need is $B < 79$
$endgroup$
– Rolf Hoyer
Jan 28 at 0:54
$begingroup$
@RolfHoyer You mean $A$? But $A=88$ as it happens. I agree that you don;'t need to eliminate $25$ to see that the digit sum of $B$ is $7$ (and have edited accordingly).
$endgroup$
– lulu
Jan 28 at 0:55
$begingroup$
Well $79$ is the smallest value possible for $B$ that is equal to 7 modulo 9 but with a digital sum greater than 7. The corresponding bound on $A$ would be $A < 799999999$, ie the smallest number equal to 7 modulo 9 with a digital sum at least 79.
$endgroup$
– Rolf Hoyer
Jan 28 at 1:00
$begingroup$
Wouldn’t B < or equal to 17 and not 27?
$endgroup$
– Varaun Ramgoolie
Jan 28 at 12:05
$begingroup$
@VaraunRamgoolie Oh, you can definitely get a better estimate on $B$, no question. As I say (for different reasons) it's not hard to show that $B$ is either $7$ or $16$. I just don't see an elementary way to rule out $7$. Just tightening the bound on $B$ isn't going to do that, so far as I can see.
$endgroup$
– lulu
Jan 28 at 12:46
$begingroup$
You don't even need to check that 25 is not possible as a value for $B$, since in that case the sum of digits of $B$ is still 7. All you need is $B < 79$
$endgroup$
– Rolf Hoyer
Jan 28 at 0:54
$begingroup$
You don't even need to check that 25 is not possible as a value for $B$, since in that case the sum of digits of $B$ is still 7. All you need is $B < 79$
$endgroup$
– Rolf Hoyer
Jan 28 at 0:54
$begingroup$
@RolfHoyer You mean $A$? But $A=88$ as it happens. I agree that you don;'t need to eliminate $25$ to see that the digit sum of $B$ is $7$ (and have edited accordingly).
$endgroup$
– lulu
Jan 28 at 0:55
$begingroup$
@RolfHoyer You mean $A$? But $A=88$ as it happens. I agree that you don;'t need to eliminate $25$ to see that the digit sum of $B$ is $7$ (and have edited accordingly).
$endgroup$
– lulu
Jan 28 at 0:55
$begingroup$
Well $79$ is the smallest value possible for $B$ that is equal to 7 modulo 9 but with a digital sum greater than 7. The corresponding bound on $A$ would be $A < 799999999$, ie the smallest number equal to 7 modulo 9 with a digital sum at least 79.
$endgroup$
– Rolf Hoyer
Jan 28 at 1:00
$begingroup$
Well $79$ is the smallest value possible for $B$ that is equal to 7 modulo 9 but with a digital sum greater than 7. The corresponding bound on $A$ would be $A < 799999999$, ie the smallest number equal to 7 modulo 9 with a digital sum at least 79.
$endgroup$
– Rolf Hoyer
Jan 28 at 1:00
$begingroup$
Wouldn’t B < or equal to 17 and not 27?
$endgroup$
– Varaun Ramgoolie
Jan 28 at 12:05
$begingroup$
Wouldn’t B < or equal to 17 and not 27?
$endgroup$
– Varaun Ramgoolie
Jan 28 at 12:05
$begingroup$
@VaraunRamgoolie Oh, you can definitely get a better estimate on $B$, no question. As I say (for different reasons) it's not hard to show that $B$ is either $7$ or $16$. I just don't see an elementary way to rule out $7$. Just tightening the bound on $B$ isn't going to do that, so far as I can see.
$endgroup$
– lulu
Jan 28 at 12:46
$begingroup$
@VaraunRamgoolie Oh, you can definitely get a better estimate on $B$, no question. As I say (for different reasons) it's not hard to show that $B$ is either $7$ or $16$. I just don't see an elementary way to rule out $7$. Just tightening the bound on $B$ isn't going to do that, so far as I can see.
$endgroup$
– lulu
Jan 28 at 12:46
add a comment |
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