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Ivory writes down three two-digit numbers, each with units digit 7. The tens digit of the product of these three numbers is 5. What is the tens digit of the sum of the three numbers?

someone please explain

 Feb 17, 2019
 #1
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+2

By trial and error, I found 17, 27, and 67. 

 

17 x 27 x 67 = 30,753                      This satisfies the "tens digit of the product is 5" requirement

 

17 + 27 + 67 = 111                           The sum of the three numbers

 

The tens digit of the sum is 1  

 

Remember I said I found these by trial and error.  There might be other solutions. 

 Feb 17, 2019
 #2
avatar+32 
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Thank you so much these are a possible solution. I was also messing around with this problem and found that the number 37 also fits for all digits. Thanks for the help.

IDontKnowMath  Feb 18, 2019
 #3
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GOOD WORK.  I made a fundamental mistake.  I assumed that the three numbers had to be different.  Nowhere did the problem stipulate that.  Three 37's is a perfectly valid solution.

 

Your contribution got me to thinking.  I noted that in my answer 17, 27, and 67, the sum of the 10's digit is 9.  Likewise in your answer 37, 37, 37, the sum of the 10's digit is 9.  Does this suggest the same thing to you that it does to me?

 Feb 18, 2019
edited by Guest  Feb 18, 2019
 #4
avatar+100778 
-1

Thanks guest, nothing is wrongwith using trial and error  laugh

 

Ivory writes down three two-digit numbers, each with units digit 7. The tens digit of the product of these three numbers is 5. What is the tens digit of the sum of the three numbers?

 

Let the three numbers be

10a+7, 10b+7 and 10c +7 

Sum of the 3 numbers is   10a+7+ 10b+7 + 10c +7 = 10(a+b+c)+21

 

The product of these is  

(100ab+70a+70b+49)(10c+7)

I'm only interested in the 10s digit

so 100ab is too big to care about

 

(70a+70b+49)(10c+7)

700ac+700bc+490c+490a+490c+343

I don't care about the first 2

490c+490a+490c+343

90c+90a+90c+43

90(a+b+c)+43

The 10s digit will be the tens digit of      9(a+b+c)+4

9(a+b+c)+4=??5

9(a+b+c)=??1

So   9*(a+b+c)  ends in 1  

and     a+b+c <= 27

and    9(a+b+c) <= 243

what are the possibilities for 9(a+b+c)

81, 171, 261,   (but 261 is too big)

a+b+c = 9 or 19

 

The sum of the 3 numbers 

=10(a+b+c)+21

=10(9)+21     or     10(19)+21

=111      or     211

 

For both of these the tens digit is 1

 Feb 18, 2019

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