BrainDen.com - Brain Teasers

# Sum It Up

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The host posts a random three digit number and 6 other number (four random from 1 to 10 and another two from 25, 50, 75 & 100)

Then everyone else would have to use the six number to reach the three digit one using +, -, / and *. The first one to get the number wins, or the one with the closes total, and become the next host.

For example 485 with 3, 4, 6, 9, 25, 100.

4*100 = 400

3*25 = 75

400 + 75 = 475

475 + 9 = 484.

I will start:

Total to reach: 524

Number to use: 2, 3, 7, 9, 50 & 100

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523 = (100 * (7 - 2)) + 50 - (3 * 9)

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Question. Does the answer have to be an integer value?

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523.777.... = ((50-3)*100+(2*7))/9

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No. Lets say rounded to the nearest 1000th.

And I forgot to put in my first post that everybody is allowed only one try per round. (So its a question of timing. Do I put my almost answer there, or do I try to get closer and hope no one in the mean time post the almost answer.)

sorry...

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my fault that I didn't say it earlier. So I will accept the 523.778 answer.

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50*100+3*9-7+2

Can you use the same number twice?

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No can't use the same number twice.

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Can you use squares and powers and such.

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525 = 50 x 9 + 100 x (7/2(3))

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no squares and powers. only +, /, - and *.

(maybe we can added others if it get boring?)

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Well curr3nt you had the closes answer you up next.

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1, 4, 7, 10, 50, 75 ?= 612

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612.5 = 75(7+1) + 50/4

10 - 7 = 3

3 * 4 = 12

50 + 1 = 51

12*51 = 612

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Well, mike is automatically up next.

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2, 4, 5, 8, 50, 75 to reach 591.

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591 = 75 x 8 - 5 - 4

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Bit too easy that one. I guess with random number it sometimes is.

TheCube is next.

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I used random numbers too. Also, we don't have to use all the numbers?

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No don't need to use all numbers.

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alright

983

75, 25, 4, 9, 1, 7

GO!

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988= (75+25)(9)-7-4-1

Edit- CRAP!! I misread the number.

Edited by Brainiac100
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(75 * (4 + 9)) + 7 + 1

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