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 Post subject: Arithmetic II -- Question no 73Posted: Mon Jul 25, 2011 9:17 pm

Joined: Mon Jul 25, 2011 9:07 pm
Posts: 3
I think the solution provided for question 73 is wrong, although it somehow got the correct answer.

considering only 2 digits
67
* A8
.......
86
.......
Solution says 3 + A = 8 ; So, A = 5

rather is should be 3 + 7*A = 8; So, unit digit of 7*A = 5; So, A = 5

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 Post subject: Re: Arithmetic II -- Question no 73Posted: Wed Jul 27, 2011 5:57 pm

Joined: Thu May 06, 2010 3:17 pm
Posts: 28
Location: Everywhere
I think the solution used A to mean any unknown digit, rather than the tens digit of integer y specifically...I think that's why it looks a little off. I set it up like this:

67
*y8
----
536
zy0
------
v86

3+y=8
y=5

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 Post subject: Re: Arithmetic II -- Question no 73Posted: Sat Jul 28, 2012 7:23 am

Joined: Sat Jul 28, 2012 7:21 am
Posts: 1
So where did this "z" variable come from and how did you completely ignore "7" when doing the multiplication.

I'm with Josh. I think this should be 3 +7A = 8. Anyone else?

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 Post subject: Re: Arithmetic II -- Question no 73Posted: Tue Jul 31, 2012 5:00 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Here is a solution:

X = 67 (has tens digit of 6 and units digit of 7)
Y = A8 (has tens digit of “A” and units digit of 8)

Expanding the multiplication:

67
* A8
-----
536
B
------
86

B = units digit of A * 7

8 = B + 3 … because the tens digit of the product is 8

B = 5

If A = 5, A*7 = 35, units digit of A*7 = 5 as required, so A = 5

- Veritas Help

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