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How many ABC three digit numbers are there such that (A+B)^C has three digits and is a power of 2?

Which of the following is necessarily a divisor of the number $ntimes1000+2 times n$ where $nlt 500$?How many 7-digit ID numbers do not contain three consecutive sixes.How many three digit sequences using the numbers 0 … 9 with conditionsNumber of $9-$digit numbers, all digits are different and nonzero and having no consecutive digits in consecutive positions8 digit telephone number with 3 or 4 same consecutive digits, how many possible number?How many odd three-digits numbers are there whose all three digits are differentProblem dealing with the sum of digits of certain numbers.How many ways are there to choose strings of length $3$ from $S={1,2,3,4}$ for $3$ people if each each digit from $S$ has to appear at least once?Find all 3-digit numbers divisible by a sum of groups of its digitsHow many $4$-digit positive integers with distinct digits are there in which the sum of the first two digits equals the sum of the last two digits?

0

0

$begingroup$

I am about to give an exam soon for joining an algorithms course so they sent me a sample test just so I know what's it gonna like and I couldn't solve this problem. The question is the following:

How many $ABC$ three digit numbers are there such that $(A+B)^C$ has three digits and is a power of $2$ ?

I don't have any answer options and I am not sure if $A, B$ and $C$ can all be the same number or not. Seems like they cannot.
Would be thankful if you could help me out.

combinatorics elementary-number-theory logic square-numbers perfect-powers

share|cite|improve this question

edited Mar 15 at 19:09

DAVO

asked Mar 14 at 0:08

DAVODAVO

566

$endgroup$

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  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: how many powers of $2$ have three digits? Of those powers of two, how many have integral $C$th roots, for $C=0,1,dots,9$?
  $endgroup$
  – TomGrubb
  Mar 14 at 0:11
  

  
  
  
  


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  $begingroup$
  If it is an algorithms class then shouldn't you just write a program to check all 3 digit numbers?
  $endgroup$
  – DanielV
  Mar 14 at 10:41
  

  
  
  
  


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  $begingroup$
  @DanielV Well it was in the "Logic" section so probably not. Also I wanted like the mathematical version of this. Otherwise yea, I can just brute force :D
  $endgroup$
  – DAVO
  Mar 15 at 18:26
  

  
  
  
  

add a comment | 

0

0

$begingroup$

I am about to give an exam soon for joining an algorithms course so they sent me a sample test just so I know what's it gonna like and I couldn't solve this problem. The question is the following:

How many $ABC$ three digit numbers are there such that $(A+B)^C$ has three digits and is a power of $2$ ?

I don't have any answer options and I am not sure if $A, B$ and $C$ can all be the same number or not. Seems like they cannot.
Would be thankful if you could help me out.

combinatorics elementary-number-theory logic square-numbers perfect-powers

share|cite|improve this question

edited Mar 15 at 19:09

DAVO

asked Mar 14 at 0:08

DAVODAVO

566

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: how many powers of $2$ have three digits? Of those powers of two, how many have integral $C$th roots, for $C=0,1,dots,9$?
  $endgroup$
  – TomGrubb
  Mar 14 at 0:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If it is an algorithms class then shouldn't you just write a program to check all 3 digit numbers?
  $endgroup$
  – DanielV
  Mar 14 at 10:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielV Well it was in the "Logic" section so probably not. Also I wanted like the mathematical version of this. Otherwise yea, I can just brute force :D
  $endgroup$
  – DAVO
  Mar 15 at 18:26
  

  
  
  
  

add a comment | 

$begingroup$

I am about to give an exam soon for joining an algorithms course so they sent me a sample test just so I know what's it gonna like and I couldn't solve this problem. The question is the following:

How many $ABC$ three digit numbers are there such that $(A+B)^C$ has three digits and is a power of $2$ ?

I don't have any answer options and I am not sure if $A, B$ and $C$ can all be the same number or not. Seems like they cannot.
Would be thankful if you could help me out.

combinatorics elementary-number-theory logic square-numbers perfect-powers

share|cite|improve this question

edited Mar 15 at 19:09

DAVO

asked Mar 14 at 0:08

DAVODAVO

566

$endgroup$

share|cite|improve this question

edited Mar 15 at 19:09

DAVO

asked Mar 14 at 0:08

DAVODAVO

566

share|cite|improve this question

edited Mar 15 at 19:09

DAVO

asked Mar 14 at 0:08

DAVODAVO

566

asked Mar 14 at 0:08

DAVODAVO

566

asked Mar 14 at 0:08

DAVODAVO

566

2 Answers
2

active

oldest

votes

3

$begingroup$

Well. If $(A+B)^C$ is a power of two then $A+B$ is also a power of two, so $A+B = 1,2,4,8 $ or $16$.

$1^C$ is never $3$ digits.

$2^C$ is $3$ digits if $7le C le 9$ so $(A,B)$ may be $(2,0)$ or $(1,1)$ and $C$ may be $7,8,$ or $9$ so there are $6$ ways.

$4^C$ is three digits if $C=4$ so $(A,B)$ may be $(4,0); (3,1); (2,2); (1,3)$ so there are $4$ ways to do that.

$8^C$ is three digits if $C=3$ and from $(8,0)$ to $(1,7)$ there are $8$ things $A,B$ may be.

$16^C$ is three digits if $C=2$. And from $(A,B)$ may be $(9,7)$ or $(8,8)$ or $(7,9)$. So there are three such.

So there are $6 + 4 + 8 + 3 = 21$ such numbers. They are:

$207,208,209,117,118,119, 404,314,224,134, 803,713, 623,533,443,353,263,173, 972,882,792.$

share|cite|improve this answer

edited Mar 15 at 19:15

DAVO

566

answered Mar 14 at 0:22

fleabloodfleablood

73k22789

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  WoW, this problem really had to be checked by few conditions. :D Thanks a lot for the answer.
  $endgroup$
  – DAVO
  Mar 15 at 19:01
  

  
  
  
  

add a comment | 

1

$begingroup$

Note that $A > 0$. Since $(A+B)^C$ has 3 digits and is a power of $2$, $A+B in {2, 4, 8, 16}$ (since $A+B=1$ yields only 1 digit integers).

Case 1: $A+B=2$

In this case, we must have $7 leq C leq 9$, and $(A,B) in {(2,0), (1,1)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3 times 2 = 6$ possible integers that can be made here.

Case 2: $A+B=4$

In this case, we must have $C = 4$, and $(A,B) in {(4,0), (3,1), (2,2), (1,3)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $4$ possible integers that can be made here.

Case 3: $A+B=8$

In this case, we must have $C = 3$, and $(A,B) in {(8,0), (7,1), (6,2), (5,3), (4,4), (3,5), (2,6), (1,7)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $8$ possible integers that can be made here.

Case 4: $A+B=16$

In this case, we must have $C = 2$, and $(A, B) in {(9,7), (8,8), (7, 9)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3$ possible integers that can be made here.

Therefore, we have a total of $6 + 4 + 8 + 3 = boxed{21}$ satisfying 3 digit integers, namely $117, 118, 119, 134, 173, 207, 208, 209, 224, 263, 314, 353, 404, 443, 533, 623, 713, 792, 803, 882$ and $972$.

share|cite|improve this answer

edited Mar 14 at 0:30

answered Mar 14 at 0:17

Sharky KesaSharky Kesa

803413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Case 3: has $8$, not $7$ options. $A$ may be anything from $1$ to $8$ and $B = 8-A$.
  $endgroup$
  – fleablood
  Mar 14 at 0:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood My bad, thanks for that
  $endgroup$
  – Sharky Kesa
  Mar 14 at 0:28
  

  
  
  
  


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  $begingroup$
  As long as I'm critiquing... $7 le C ge 9$ doesn't really make sense.
  $endgroup$
  – fleablood
  Mar 14 at 0:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood was fixing it as you wrote that, thanks anyways
  $endgroup$
  – Sharky Kesa
  Mar 14 at 0:31
  

  
  
  
  

add a comment | 

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3

$begingroup$

Well. If $(A+B)^C$ is a power of two then $A+B$ is also a power of two, so $A+B = 1,2,4,8 $ or $16$.

$1^C$ is never $3$ digits.

$2^C$ is $3$ digits if $7le C le 9$ so $(A,B)$ may be $(2,0)$ or $(1,1)$ and $C$ may be $7,8,$ or $9$ so there are $6$ ways.

$4^C$ is three digits if $C=4$ so $(A,B)$ may be $(4,0); (3,1); (2,2); (1,3)$ so there are $4$ ways to do that.

$8^C$ is three digits if $C=3$ and from $(8,0)$ to $(1,7)$ there are $8$ things $A,B$ may be.

$16^C$ is three digits if $C=2$. And from $(A,B)$ may be $(9,7)$ or $(8,8)$ or $(7,9)$. So there are three such.

So there are $6 + 4 + 8 + 3 = 21$ such numbers. They are:

$207,208,209,117,118,119, 404,314,224,134, 803,713, 623,533,443,353,263,173, 972,882,792.$

share|cite|improve this answer

edited Mar 15 at 19:15

DAVO

566

answered Mar 14 at 0:22

fleabloodfleablood

73k22789

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  WoW, this problem really had to be checked by few conditions. :D Thanks a lot for the answer.
  $endgroup$
  – DAVO
  Mar 15 at 19:01
  

  
  
  
  

add a comment | 

3

$begingroup$

Well. If $(A+B)^C$ is a power of two then $A+B$ is also a power of two, so $A+B = 1,2,4,8 $ or $16$.

$1^C$ is never $3$ digits.

$2^C$ is $3$ digits if $7le C le 9$ so $(A,B)$ may be $(2,0)$ or $(1,1)$ and $C$ may be $7,8,$ or $9$ so there are $6$ ways.

$4^C$ is three digits if $C=4$ so $(A,B)$ may be $(4,0); (3,1); (2,2); (1,3)$ so there are $4$ ways to do that.

$8^C$ is three digits if $C=3$ and from $(8,0)$ to $(1,7)$ there are $8$ things $A,B$ may be.

$16^C$ is three digits if $C=2$. And from $(A,B)$ may be $(9,7)$ or $(8,8)$ or $(7,9)$. So there are three such.

So there are $6 + 4 + 8 + 3 = 21$ such numbers. They are:

$207,208,209,117,118,119, 404,314,224,134, 803,713, 623,533,443,353,263,173, 972,882,792.$

share|cite|improve this answer

edited Mar 15 at 19:15

DAVO

566

answered Mar 14 at 0:22

fleabloodfleablood

73k22789

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  WoW, this problem really had to be checked by few conditions. :D Thanks a lot for the answer.
  $endgroup$
  – DAVO
  Mar 15 at 19:01
  

  
  
  
  

add a comment | 

$begingroup$

Well. If $(A+B)^C$ is a power of two then $A+B$ is also a power of two, so $A+B = 1,2,4,8 $ or $16$.

$1^C$ is never $3$ digits.

$2^C$ is $3$ digits if $7le C le 9$ so $(A,B)$ may be $(2,0)$ or $(1,1)$ and $C$ may be $7,8,$ or $9$ so there are $6$ ways.

$4^C$ is three digits if $C=4$ so $(A,B)$ may be $(4,0); (3,1); (2,2); (1,3)$ so there are $4$ ways to do that.

$8^C$ is three digits if $C=3$ and from $(8,0)$ to $(1,7)$ there are $8$ things $A,B$ may be.

$16^C$ is three digits if $C=2$. And from $(A,B)$ may be $(9,7)$ or $(8,8)$ or $(7,9)$. So there are three such.

So there are $6 + 4 + 8 + 3 = 21$ such numbers. They are:

$207,208,209,117,118,119, 404,314,224,134, 803,713, 623,533,443,353,263,173, 972,882,792.$

share|cite|improve this answer

edited Mar 15 at 19:15

DAVO

566

answered Mar 14 at 0:22

fleabloodfleablood

73k22789

$endgroup$

share|cite|improve this answer

edited Mar 15 at 19:15

DAVO

566

answered Mar 14 at 0:22

fleabloodfleablood

73k22789

edited Mar 15 at 19:15

DAVO

566

edited Mar 15 at 19:15

DAVO

566

answered Mar 14 at 0:22

fleabloodfleablood

73k22789

answered Mar 14 at 0:22

fleabloodfleablood

73k22789

1

$begingroup$

Note that $A > 0$. Since $(A+B)^C$ has 3 digits and is a power of $2$, $A+B in {2, 4, 8, 16}$ (since $A+B=1$ yields only 1 digit integers).

Case 1: $A+B=2$

In this case, we must have $7 leq C leq 9$, and $(A,B) in {(2,0), (1,1)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3 times 2 = 6$ possible integers that can be made here.

Case 2: $A+B=4$

In this case, we must have $C = 4$, and $(A,B) in {(4,0), (3,1), (2,2), (1,3)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $4$ possible integers that can be made here.

Case 3: $A+B=8$

In this case, we must have $C = 3$, and $(A,B) in {(8,0), (7,1), (6,2), (5,3), (4,4), (3,5), (2,6), (1,7)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $8$ possible integers that can be made here.

Case 4: $A+B=16$

In this case, we must have $C = 2$, and $(A, B) in {(9,7), (8,8), (7, 9)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3$ possible integers that can be made here.

Therefore, we have a total of $6 + 4 + 8 + 3 = boxed{21}$ satisfying 3 digit integers, namely $117, 118, 119, 134, 173, 207, 208, 209, 224, 263, 314, 353, 404, 443, 533, 623, 713, 792, 803, 882$ and $972$.

share|cite|improve this answer

edited Mar 14 at 0:30

answered Mar 14 at 0:17

Sharky KesaSharky Kesa

803413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Case 3: has $8$, not $7$ options. $A$ may be anything from $1$ to $8$ and $B = 8-A$.
  $endgroup$
  – fleablood
  Mar 14 at 0:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood My bad, thanks for that
  $endgroup$
  – Sharky Kesa
  Mar 14 at 0:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As long as I'm critiquing... $7 le C ge 9$ doesn't really make sense.
  $endgroup$
  – fleablood
  Mar 14 at 0:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood was fixing it as you wrote that, thanks anyways
  $endgroup$
  – Sharky Kesa
  Mar 14 at 0:31
  

  
  
  
  

add a comment | 

1

$begingroup$

Note that $A > 0$. Since $(A+B)^C$ has 3 digits and is a power of $2$, $A+B in {2, 4, 8, 16}$ (since $A+B=1$ yields only 1 digit integers).

Case 1: $A+B=2$

In this case, we must have $7 leq C leq 9$, and $(A,B) in {(2,0), (1,1)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3 times 2 = 6$ possible integers that can be made here.

Case 2: $A+B=4$

In this case, we must have $C = 4$, and $(A,B) in {(4,0), (3,1), (2,2), (1,3)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $4$ possible integers that can be made here.

Case 3: $A+B=8$

In this case, we must have $C = 3$, and $(A,B) in {(8,0), (7,1), (6,2), (5,3), (4,4), (3,5), (2,6), (1,7)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $8$ possible integers that can be made here.

Case 4: $A+B=16$

In this case, we must have $C = 2$, and $(A, B) in {(9,7), (8,8), (7, 9)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3$ possible integers that can be made here.

Therefore, we have a total of $6 + 4 + 8 + 3 = boxed{21}$ satisfying 3 digit integers, namely $117, 118, 119, 134, 173, 207, 208, 209, 224, 263, 314, 353, 404, 443, 533, 623, 713, 792, 803, 882$ and $972$.

share|cite|improve this answer

edited Mar 14 at 0:30

answered Mar 14 at 0:17

Sharky KesaSharky Kesa

803413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Case 3: has $8$, not $7$ options. $A$ may be anything from $1$ to $8$ and $B = 8-A$.
  $endgroup$
  – fleablood
  Mar 14 at 0:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood My bad, thanks for that
  $endgroup$
  – Sharky Kesa
  Mar 14 at 0:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As long as I'm critiquing... $7 le C ge 9$ doesn't really make sense.
  $endgroup$
  – fleablood
  Mar 14 at 0:29
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood was fixing it as you wrote that, thanks anyways
  $endgroup$
  – Sharky Kesa
  Mar 14 at 0:31
  

  
  
  
  

add a comment | 

$begingroup$

Note that $A > 0$. Since $(A+B)^C$ has 3 digits and is a power of $2$, $A+B in {2, 4, 8, 16}$ (since $A+B=1$ yields only 1 digit integers).

Case 1: $A+B=2$

In this case, we must have $7 leq C leq 9$, and $(A,B) in {(2,0), (1,1)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3 times 2 = 6$ possible integers that can be made here.

Case 2: $A+B=4$

In this case, we must have $C = 4$, and $(A,B) in {(4,0), (3,1), (2,2), (1,3)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $4$ possible integers that can be made here.

Case 3: $A+B=8$

In this case, we must have $C = 3$, and $(A,B) in {(8,0), (7,1), (6,2), (5,3), (4,4), (3,5), (2,6), (1,7)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $8$ possible integers that can be made here.

Case 4: $A+B=16$

In this case, we must have $C = 2$, and $(A, B) in {(9,7), (8,8), (7, 9)}$ in order for $(A+B)^C$ to be a 3 digit power of 2. Thus, there are $3$ possible integers that can be made here.

Therefore, we have a total of $6 + 4 + 8 + 3 = boxed{21}$ satisfying 3 digit integers, namely $117, 118, 119, 134, 173, 207, 208, 209, 224, 263, 314, 353, 404, 443, 533, 623, 713, 792, 803, 882$ and $972$.

share|cite|improve this answer

edited Mar 14 at 0:30

answered Mar 14 at 0:17

Sharky KesaSharky Kesa

803413

$endgroup$

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edited Mar 14 at 0:30

answered Mar 14 at 0:17

Sharky KesaSharky Kesa

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answered Mar 14 at 0:17

Sharky KesaSharky Kesa

803413

answered Mar 14 at 0:17

Sharky KesaSharky Kesa

803413