+32 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
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.
+32 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.
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?
+118608 Thanks guest, nothing is wrongwith using trial and error 
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
