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A four-digit hexadecimal integer is written on a napkin such that the units digit is illegible. The first three digits are 4, A, and 7. If the integer is a multiple of 17 base ten, what is the units digit?

 May 1, 2019
 #1
avatar+2551 
+3

I do not know how to solve this problem, so I will partially solve it and let someone else who knows how to solve it finish my solution.

 

 

Converting hexadecimal to base-ten

 

4 * 16^0

 

a * 16^1

 

7 * 16^2

 

x (ones digit)

 

x * 16^3

 

So now we have 4 + 16a + 1792 + 4096x

 

So we have 4096x+16a+1796, and we have to find the value of x and a in which it is divisible by 17.

 

Notice how x and a can only be 1 through 9.

 

So trial and error?

 May 1, 2019
edited by CalculatorUser  May 1, 2019
edited by CalculatorUser  May 1, 2019
 #2
avatar
+2

Well, you have already done most of the work!
Just go up the numbers from 0 to 9 and see which one is a multiple of 17.
Base 16                     Base 10
4A70                          19,056
4A71                           19,057, Bingo!, since it is multiple of 17. So, the Hex number ends in "1", or 4A71.

 May 1, 2019
 #3
avatar+23788 
+3

A four-digit hexadecimal integer is written on a napkin such that the units digit is illegible.

The first three digits are 4, A, and 7.

If the integer is a multiple of 17 base ten, 

what is the units digit?

 

\(\begin{array}{|rcll|} \hline \mathbf{4A7x_{16}} &=& \mathbf{ ( ( 4\cdot 16+A )\cdot 16+7)\cdot16+x } \quad | \quad A = 10 \\ &=& ((4\cdot 16+10)\cdot 16+7)\cdot16+x \\ &=& (74\cdot 16+7)\cdot16+x \\ &=& 1191\cdot16+x \\ &=& \mathbf{19056 +x},\qquad x=0,1,\ldots, 15 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{19056 +x} &=& \mathbf{n\cdot 17},\ n \in \mathbf{Z} \\\\ 19056 +x &\equiv& 0 \pmod {17} \quad | \quad 19056 \pmod {17} = 16 \\ 16 +x &\equiv& 0 \pmod {17} \\ 16 +x &=& n\cdot 17 \\ x &=& n\cdot 17 -16 \quad | \quad n = 1 \\ x &=& 1\cdot 17 -16 \\ \mathbf{x} &=& \mathbf{1} \\ \hline \end{array}\)

 

laugh

 May 2, 2019
edited by heureka  May 2, 2019

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