We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+2
49
3
avatar+321 

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+520 
+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+22188 
+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

7 Online Users