Using the digits 1 through 9 without repetition, fill out the number grid above. What is the product of the numbers in the 4 corners?
Note: The order in which this grid calculates is left-to-right/top-to-bottom unlike the usual order of operations. E.g., 1 + 2 × 3 = ( 1 + 2 ) × 3 = 9 .
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Why 8 - 4 x 2 = 8 ?! Must be ZERO (0)
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It is given in the problem.
Note: The order in which this grid calculates is left-to-right/top-to-bottom unlike the usual order of operations.
8-4=4 *2=8
Yes, using logic the blanks can be filled up one by one, as shown above, in the following sequence; 1,3,9,2,6,4,8,7,5.
Label the rows 1 , 2 , 3 from top to bottom and the columns a , b , c from left to right. So ( 1 , a ) is the top left corner and ( 3 , c ) is the bottom right. Now a step-by-step approach:
( 3 , b ) must be 1 or 3 since it must be a factor of 3 , but if it were 3 then ( 3 , a ) and 3 , c ) would have to be the same. Thus ( 3 , b ) is 1 .
Then ( 3 , a ) , ( 3 , c ) would have to be either 6 , 2 or 9 , 3 . But if ( 3 , c ) were 2 then ( 1 , c ) , ( 2 , c ) would have to be some ordering of 2 , 4 or 8 , 1 , which can't be since both 1 and 2 would have already been used. Thus ( 3 , a ) , ( 3 , c ) must be 9 , 3 .
With ( 3 , c ) being 3 we must have ( 1 , c ) , ( 2 , c ) being some ordering of 2 , 6 . But since ( 2 , c ) must be a factor of 8 , it must be 2 , and thus ( 1 , c ) must be 6 .
Since ( 2 , c ) is 2 , we must have ( 2 , a ) − ( 2 , b ) = 4 , and since 1 , 2 , 3 , 6 and 9 have already been used, we are left with ( 2 , a ) = 8 and ( 2 , b ) = 4 .
Plugging in ( 1 , a ) = 5 and ( 1 , b ) = 7 then satisfies all the remaining equations.
Thus the product of the corner numbers is
( 1 , a ) ∗ ( 1 , c ) ∗ ( 3 , a ) ∗ ( 3 , c ) = 5 ∗ 6 ∗ 9 ∗ 3 = 8 1 0 .
Hello Sir Can you ilibrate it in a simple way...
I have one question like this Can you solve that for me?
For this solution, call the digits a,b,c,d,e,f,g,h,i.
Start in the middle column, and note that b-e is a whole number and therefore h is an odd number, either 1 or 3.
Move to the bottom row. The only combinations of numbers g.h/i = 3 and satisfying h = 1 or 3 is 9, 1, 3. Therefore g is 9, h is 1 and i is 3.
We now know that c.f = 12 and that therefore they are both 2 or 6. No combination of the remaining digits can make (d-e).6 = 8 so c is 6 and f is 2.
Continuing, a+b = 12 and therefore one is 7 and the other is 5, but because we have already used the two, b must be 7, e must be 4 and a must be 5.
Finally, d = 8.
Therefore a.c.g.i = 5.6.9.3 = 810.
Good explanation. Thanks for your solution
Thnx for xplanation
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But 2nd row is confuseable
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It is not after you have established that f is 2 and I will explain that at the end how we reached that. To satisfy the second row d-e should be 4 and there is only one combination that is d being 8 and e being 4. Now going back to why f is 2. In order to satisfy third column cf must be 12. There are only 4 combinations possible 3x4, 4x3, 2x6 and 6x2. First two are not possible because 3 has already been used. Now f can be either 2 or 6 but in order to satisfy the second row it should be a factor of 8 so that only leaves 2 and the rest fits in.
Thanks for above solutions!!
I think there is a mistake in the question. The result in the bottom right corner is not 1. Its 4.
Whoops. Fixed.
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I think its better to write answer example like a * b * c * d=answer or something like that i just answer 5693, 5 (i think input each number) and i just dont know its just mutiply all answer.
6 + 2 - 2 = 6
+ - *
2 - 1 * 8 = 8
- * /
4 * 3 / 4 = 3
= = =
4 3 4
Gives 192
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In your solution you have only used 1,2,3,4,6 and 8 - you have to use ALL the digits from 1 to 9 and use each digit only once.
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Using the same logic as Brian Charlesworth . The solution is as follows: