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laxieqv
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 11 Nov 2005, 23:40
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5109094x171709440000 = 21!, find x.

A. 1
B. 2
C.3
D.4
E.5



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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 00:01
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B 2

21! has to be divided by 3

add all digits gives 61 + x
to divide by 3 it has to be 63
x = 2
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laxieqv
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 00:32
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duttsit
B 2

21! has to be divided by 3

add all digits gives 61 + x
to divide by 3 it has to be 63
x = 2
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how about x= 5?! 66 is divisible by 3 also :roll:
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 01:06
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how do u know that 21! has to be divisible by 3

duttsit
B 2

21! has to be divided by 3

add all digits gives 61 + x
to divide by 3 it has to be 63
x = 2
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 07:19
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i did it the same as duttsit did it. the number is a multiple of the primes 1...21. i tried 11 and got 2 for x.
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 07:56
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christoph
i did it the same as duttsit did it. the number is a multiple of the primes 1...21. i tried 11 and got 2 for x.
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How can you determine whether this number is divisible by 11 Christoff?

I understand Duttsit`s logic, but why not use 5 to get a numerical sum of 66?
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 08:57
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GMATT73
christoph
i did it the same as duttsit did it. the number is a multiple of the primes 1...21. i tried 11 and got 2 for x.

How can you determine whether this number is divisible by 11 Christoff?

I understand Duttsit`s logic, but why not use 5 to get a numerical sum of 66?
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How can you determine whether this number is divisible by 11 Christoff?
--> 21!= 1*2*3*4*....*11*12*......*20*21
that's why 21! is divisible by 11

but why not use 5 to get a numerical sum of 66 ----> you're right! :wink:
It would be better to try 9 instead of 3 . The divisibility indicator of 9 is that the sum of all digits forming the number is divisible by 9. Only 63 satisfies this. Thus, x=2.
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 09:09
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laxieqv
GMATT73
christoph
i did it the same as duttsit did it. the number is a multiple of the primes 1...21. i tried 11 and got 2 for x.

How can you determine whether this number is divisible by 11 Christoff?

I understand Duttsit`s logic, but why not use 5 to get a numerical sum of 66?

How can you determine whether this number is divisible by 11 Christoff?
--> 21!= 1*2*3*4*....*11*12*......*20*21
that's why 21! is divisible by 11

but why not use 5 to get a numerical sum of 66 ----> you're right! :wink:
It would be better to try 9 instead of 3 . The divisibility indicator of 9 is that the sum of all digits forming the number is divisible by 9. Only 63 satisfies this. Thus, x=2.
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So essentially the divisibility rule for 3 and 9 are the same---->sum of the digits/3 or 9

Got it!! 8-)
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5109094x171709440000 = 21!, find x. A. 1 B. 2 C.3 D.4 12 Nov 2005, 09:15
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GMATT73
laxieqv
GMATT73
[quote="christoph"]i did it the same as duttsit did it. the number is a multiple of the primes 1...21. i tried 11 and got 2 for x.

How can you determine whether this number is divisible by 11 Christoff?

I understand Duttsit`s logic, but why not use 5 to get a numerical sum of 66?

How can you determine whether this number is divisible by 11 Christoff?
--> 21!= 1*2*3*4*....*11*12*......*20*21
that's why 21! is divisible by 11

but why not use 5 to get a numerical sum of 66 ----> you're right! :wink:
It would be better to try 9 instead of 3 . The divisibility indicator of 9 is that the sum of all digits forming the number is divisible by 9. Only 63 satisfies this. Thus, x=2.

So essentially the divisibility rule for 3 and 9 are the same---->sum of the digits/3 or 9

Got it!! 8-)[/quote]

question:

since it's a factorial: 1*2*3*4*5...*20*21

could you (in theory) use any number as divisor to find the missing digit?

of course below a certain number, because you cannot add more than a certain number, if it will change the next digit...

right?



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