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Each of the 6 circles contains a different counting number. The sum of all 6 numbers is 21.The sum of the 3 numbers along each side of the triangle is shown in the diagram. so What is the sum of the numbers in the shaded circles?

Note that obviously the answer is easy. But what kind of ways you suggest for this problem?

enter image description here

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    $\begingroup$ 'i think this question by simple face': can anybody decipher this? $\endgroup$
    – TonyK
    Jul 7, 2014 at 17:25

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We have $$8+8+14=({\rm sum \space of \space white \space balls})+2({\rm sum \space of \space shaded \space balls}).$$ You can take it from here.

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  • $\begingroup$ @bluebird7: this answer looks good to me. $\endgroup$
    – TonyK
    Jul 7, 2014 at 17:06
  • $\begingroup$ see the latest answer ! $\endgroup$
    – Mazdak
    Jul 7, 2014 at 17:11
  • $\begingroup$ @bluebird7: 'the latest answer': Do we have to work out which one that is, based on the time of you comment? I'd rather you just told us which answer you mean. $\endgroup$
    – TonyK
    Jul 7, 2014 at 17:24
  • $\begingroup$ @bluebird7 It's not false but it's not an answer, really, more like a hint. He's telling you that summing the sides overcounts the shaded balls (doublecounts them, to be more specific). $\endgroup$
    – Lisa
    Jul 7, 2014 at 21:22
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Let $X_1,X_2,X_3$ the shaded corners. $Y_1,Y_2,Y_3$ the white ones.

$$X_1+Y_1+X_2=8$$ $$X_2+Y_2+X_3=8$$ $$X_1+Y_3+X_3=14$$ $$X_1+X_2+X_3+Y_1+Y_2+Y_3=21$$

Adding the three first equations:

$$2X_1+2X_2+2X_3+Y_1+Y_2+Y_3=30$$

And substract this equation with the fourth one:

$$X_1+X_2+X_3=9$$

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9.

The numbers in the triangle are:

5

2 6

1 4 3

Two of the sides sum to 8 and one side sums to 14, and the whole thing sums to 21 as specified.

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    $\begingroup$ From the question: "Each of the 6 circles contains a different counting number." $\endgroup$
    – marvinthemartian
    Jul 7, 2014 at 17:14
  • $\begingroup$ Yes, you're right, Marvin. $\endgroup$
    – user153918
    Jul 7, 2014 at 17:51
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$$2\text{ (Sum of shaded balls)}+\text{sum of unshaded balls}=8+8+14=30 \tag1$$ A/Q $$\text{(sum of shaded balls)}+\text{sum of unshaded balls}=21 \tag2$$

Subtracting (1) & (2), Sum of shaded balls=9.

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  • $\begingroup$ Welcome to the site. Your answer could be formatted better. See math notation guide. $\endgroup$
    – user147263
    Aug 8, 2014 at 21:33

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