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# 10^25тАУ 7 is divisible by a) 2 b) 3

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CEO
Joined: 15 Aug 2003
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10^25Ñ‚ÐÐ£ 7 is divisible by a) 2 b) 3 [#permalink]

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11 Oct 2003, 05:57
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10^25тАУ 7 is divisible by

a) 2
b) 3
c) 9
d) Both (2) and (3)

Last edited by Praetorian on 13 Oct 2003, 01:15, edited 1 time in total.
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11 Oct 2003, 13:38
praetorian123 wrote:
10^25тАУ 7 is divisible by

1) 2
2) 3
3) 9
4) Both (2) and (3)

10^25 = 10 & 25 0's (10 00000 00000 00000 00000 00000)
subtract 7 and u get 24 nines and 1 three (9999 99999 99999 99999 99999 3)

1. div by 2 is out
2. div by 3 if sum is div by 3 - sum = 9*24 = 216 + 3 = 219
219/3 = 73 - so yes it is div by 3
3. div by 9 if the sum is div by 9 --> 219 is not div by 9

Manager
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12 Oct 2003, 08:57
I'd say (2) also.

I quick method... 10^25 - 7 will give all 9's except the last digit. This value is divisble by 3 leaving 1 in the answer. On basis of that I figured since all 9's will be divided evenly by 9, last three won't be. So you end up with 1/3.
SVP
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13 Oct 2003, 01:18
agree with (2)

my approach:

1) 2 is out, for even (10^N)- odd (7)=odd
2) 3 is OK. A number is divisible by 3 if the sum of its digits is divisible by 3. Our number leaver many 9s and 3, the sum of which is divisible by 3
3) 9 is out. A number is divisible by 9 if the sum of its digits is divisible by 9. 10^25 is a 26-digit number. Minus 7 gives a 25-digit number: 24*9+3. Clearly not divisible by 9.
CEO
Joined: 15 Aug 2003
Posts: 3454
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Kudos [?]: 874 [0], given: 781

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13 Oct 2003, 01:19
wonder_gmat wrote:
I'd say (2) also.

I quick method... 10^25 - 7 will give all 9's except the last digit. This value is divisble by 3 leaving 1 in the answer. On basis of that I figured since all 9's will be divided evenly by 9, last three won't be. So you end up with 1/3.

correct.

its not divisible by 2, last digit is 3

its not divisible by 9 , as the sum of digits is 9*24 +3 = 219 ...which is not divisible by 9

Sum of digits is divisible by 3 ,so yes, 3 is the answer.

thanks
praetorian
13 Oct 2003, 01:19
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