Question 168491
Note: I'm going to let a=first digit, b=second digit, c=third digit etc, until I get to j=tenth digit



So the telephone number looks like this:


(a b c) d e f - g h i j



Now let's do some translations:


"The Product of the sixth and seventh numbers is the third number. " 


translates to


f*g=c


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"The fourth, eigth, ninth, and tenth numbers are multiples of 3"


translates to 


d, h, i, j are multiples of 3 (giving us the possible values 0, 3, 6, 9)



Note: 0 is a multiple since 3*0=0


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"The sum of the fourth and sixth numbers is the same as the sum of the fifth and eigth numbers. "


translates to


d+f = e+h



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"The second, third, sixth, and seventh numbers are powers of 2."


translates to


b, c, f, g are powers of 2 which gives the possible values: 1, 2, 4, 8



Note: 1 is a power of 2 since {{{2^0=1}}}


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"The first, fifth, seventh, and tenth numbers are prime."


translates to


a, e, g, j are prime 


List of (single digit) prime numbers: 2, 3, 5, 7



You will need to refer back to this list so it might help to write the list on a separate sheet of paper and cross out any digits that we've used


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Note: remember, each digit is different which means that NO repeats are allowed.



Since digit j is both a multiple of 3 AND is prime, this means that the only possible digit for j is 3 (since 3 is both a multiple of 3 AND prime)



So <font color="red">j=3</font>


The same can be done with g. Since g is both a power of 2 AND is prime, the only candidate is 2 (since 2 is both a power of 2 AND is prime)



So <font color="red">g=2</font>



Looking at f*g=c, we know that all three are powers of 2. So this means that f*2=some power of 2



So let's try some values of f


f=1 (CANNOT do since 1*2=2; a repeat would occur where g=c)
f=2 (CANNOT do since 2 is already taken)
f=4 (CAN do since 4*2=8; all three are powers of 2)
f=8 (CANNOT do since 8*2=16; 16 is comprised of 2 digits)


So <font color="red">f=4</font> and <font color="red">c=8</font>



Since we know that <font color="red">g=2</font>, <font color="red">f=4</font>, and <font color="red">c=8</font> (and there are ONLY 4 digits that are powers of 2), this means that <font color="red">b=1</font> (by elimination)


Just to recap, we've taken the digits 3, 2, 4, 8, and 1 so far (5 digits taken already).



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Now let's move onto the equation 


d+f = e+h


We know that f=4, so let's plug that in:


d+4 = e+h



Here's where we have to try some combinations


The possible values for d are 0, 6, and 9 (remember d is a multiple of 3 and the digit 3 is already taken)


The possible values for e are 5 and 7 (e is prime and the digits 2 and 3 are already taken)


The possible values for h are 0, 6, and 9 (remember h is a multiple of 3 and the digit 3 is already taken)



So we could have the following:


d=0 OR d=6 OR d=9


e=5 OR e=7


h=0 OR h=6 OR h=9



So let's add some combinations:



Combination 1:

If d=0, e=5, and h=6 (0 already taken), then


0+4=5+6
4=11 ... which is false (cross this combination out)



Combination 2:

If d=0, e=5, and h=9, then


0+4=5+9
4=14 ... which is false (cross this combination out)



Combination 3:

If d=0, e=7, and h=6 (0 already taken), then


0+4=7+6
4=13 ... which is false (cross this combination out)



Combination 4:

If d=0, e=7, and h=9, then


0+4=7+9
4=16 ... which is false (cross this combination out)



Combination 4:

If d=6, e=5, and h=0 (6 already taken), then


6+4=5+0
10=5 ... which is false (cross this combination out)



Combination 5:

If d=6, e=5, and h=9 (6 already taken), then


6+4=5+6
10=11 ... which is false (cross this combination out)



Combination 6:

If d=6, e=7, and h=0 (6 already taken), then


6+4=7+0
10=7 ... which is false (cross this combination out)



Combination 7:

If d=6, e=7, and h=9 (6 already taken), then


6+4=7+9
10=16 ... which is false (cross this combination out)



Combination 8:

If d=9, e=7, and h=0 (9 already taken), then


9+4=7+0
13=7 ... which is false (cross this combination out)



Combination 9:

If d=9, e=7, and h=6 (9 already taken), then


9+4=7+6
13=13 ... which works




So after trying 8 combinations of numbers, we get

<font color="red">d=9</font>, <font color="red">e=7</font> and <font color="red">h=6</font>


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So to recap, we've taken the digits 3, 2, 4, 1, and 8 (from the first part) and the digits 9, 7, and 6 (from the second part)


Altogether, we know that 


<font color="red">j=3</font>, <font color="red">g=2</font>, <font color="red">f=4</font>, <font color="red">b=1</font>, and <font color="red">c=8</font> (from part 1)


<font color="red">d=9</font>, <font color="red">e=7</font> and <font color="red">h=6</font> (from part 2)



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Since <font color="red">d=9</font>, <font color="red">h=6</font>, and <font color="red">j=3</font>, using elimination we get <font color="red">i=0</font> (note: I'm looking at the "multiples of 3" group)




Finally, since <font color="red">e=7</font>, <font color="red">g=2</font>, and <font color="red">j=3</font>, this means that <font color="red">a=5</font> (note: I'm looking at the "prime numbers" group)



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Answer:


So we have the following numbers:


<font color="red">a=5</font>
<font color="red">b=1</font>
<font color="red">c=8</font>


<font color="red">d=9</font>
<font color="red">e=7</font>
<font color="red">f=4</font>


<font color="red">g=2</font>
<font color="red">h=6</font>
<font color="red">i=0</font>
<font color="red">j=3</font>



which means that the telephone number is 


(518) 974-2603