Question 721839
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If the product of the three numbers is 120, since 120 ends in 0, one of the three numbers must be either 5 or 10.  If one of the numbers is 10, then the product of the other 2 is 12 and they must add up to 10.  No such numbers.  So one of the numbers has to be 5.   Dividing 120 by 5 is 24.  The two single digit factors of 24 are either 4 and 6 or 3 and 8.  No combination of 4, 5, and 6 exists where the sum of two of the digits is the third.  But 3, 5, and 8 work because 3 plus 5 is 8.  Recap:  3 + 5 + 8 = 16.  3 * 5 * 8 = 120.  and 3 + 5 = 8.  All conditions satisfied.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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