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 Find the area (Posted on 2011-04-08)
If RE*ER= AREA evaluate the AREA .

Please evaluate using pencil and paper only.

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 Mod Squad (spoiler) | Comment 3 of 5 |
Well, there are really only 45 different numbers to try:
11*11, 12*21, 13*31, etc.   Not too much to do by pencil and paper, but it's more fun to get analytical

Consider the equation AREA = (RE)*(ER), using mod 9.
Mod 9: A + (R + E) + A = (R + E)*(R + E)
Let x = (R + E) (mod 9)
Solving, A = x(x-1)/2

x can take one of only 9 values, and each determines a unique A:

x A = x(x+1)/2
- -----------------
0 9 (A cannot be 0)
1 9 (A cannot be 0)
2 1
3 3
4 6
5 1
6 6
7 3
8 1

Since A cannot be 0, it must be 1, 3, 6, or 9

Consider the four possibilities separately.

1) If A equals 1, then R+E = 2 or 5 or 8 (mod 9)
The only one digit pairs whose product ends in 1 are (1,1), (3,7) and (9,9)
Of these, only (1,1) sums to 2, 5, or 8 mod 9.
But 11*11 = 121, so A cannot be 1.

3) If A equals 3, then R+E = 3 or 7 (mod 9)
The only one digit pairs whose product ends in 3 are (1,3) and (7,9)
Of these, only (7,9) sums to 3 or 7 mod 9.
But 79*97 = 7663, so A cannot be 3.

6) If A equals 3, then R+E = 4 or 6 (mod 9)
The only one digit pairs whose product ends in 6 are (1,6),(2,3),(2,8),(4,4),(4,9),(6.6), and (7,8)
Of these, only (4,9) and (7,8) sum to 4 or 6 mod 9.
49*94 = 4606 <-- Not a solution
78*87 = 6786 <-- A SOLUTION!

6) If A equals 9, then R+E = 0 or 1 (mod 9)
The only one digit pairs whose product ends in 9 are (1,9), (3,3), and (7,7)
Of these, only (1,9) sums to 0 or 1 mod 9.
But 19*91 =  1729, so A cannot be 9

So, the only solution is 78*87 = 6786, and we only had to do 5 multiplications out of 45 to find it!

 Posted by Steve Herman on 2011-04-09 01:18:26

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