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by sanjoy18 » Thu Sep 12, 2013 1:10 pm
A natural number consists of only 0’s and 1’s. If the number is divisible by 375, then what is the least possible number of 0’s and 1’s in the number respectively?
(a) 6 and 3 (b) 3 and 6 (c) 6 and 9 (d) 3 and 4 e) None of these
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by batwaraanirudh » Thu Sep 12, 2013 4:26 pm
A natural number consists of only 0’s and 1’s. If the number is divisible by 375, then what is the least possible number of 0’s and 1’s in the number respectively?
Prime factors of 375 = 5*5*5*3

So the natural number that we are talking about should be divisible by 3

For any number to be divisible by the sum of the digits should be divisible by 3. Since the digits can be only 0s or 1s, minimum number of 1s require is 3.

the number should also be divisible by three 5s. So the number should have atleast 3 0s at the end. Since we cannot have any 5s.

Therefore the least possible number is 111000. This number is divisible by 375


Therefore choose E
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by ganeshrkamath » Thu Sep 12, 2013 7:21 pm
sanjoy18 wrote:A natural number consists of only 0’s and 1’s. If the number is divisible by 375, then what is the least possible number of 0’s and 1’s in the number respectively?
(a) 6 and 3 (b) 3 and 6 (c) 6 and 9 (d) 3 and 4 e) None of these
375 = 125 * 3
So the number has to end with 3 zeros and the sum of the digits of the number should be a multiple of 3.
The least such number = 111000

Choose E

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by [email protected] » Thu Sep 12, 2013 9:51 pm
Hi All,

I'm sure this goes without saying, but since it's been happening a lot as of late, this is NOT a GMAT question.

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Rich
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by ani781 » Fri Sep 13, 2013 3:43 am
Hi Rich,
Just curious, how do you earmark the questions as being GMAT or Non-GMAT ?
And based on the same chores, do you find this question above or below GMAT level ?

Regards.
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by Brent@GMATPrepNow » Fri Sep 13, 2013 6:24 am
I initially thought this question was out of scope until I saw ganeshrkamath's and batwaraanirudh's great solutions. I just want to elaborate on their solutions a bit so that others can see how they reached their conclusions.

Since 375 = (3)(125), we know that the answer must be divisible by both 125 and 3.

Let's take a closer look at numbers that are divisible by 125.
Notice that:
(1)(125) = 125
(2)(125) = 250
(3)(125) = 375
(4)(125) = 500
(5)(125) = 625
(6)(125) = 750
(7)(125) = 875
(8)(125) = 000
(9)(125) = 1125
.
.
.
So, as you can see, the last 3 digits of numbers divisible by 125 must be 125, 250, 375, 500, 625, 750, 875, or 000

Notice that ONLY ONE of these cases is such that the number consists of only 0’s and 1’s
So, the number in question must end in 000

Since the number in question is ALSO divisible by 3, the sum of its digits must be divisible by 3.
So, the least possible number must be 111000

Answer: E

Cheers,
Brent
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by rahul.sehgal@btgchampion » Sat Sep 14, 2013 7:57 am
Brent@GMATPrepNow wrote:I initially thought this question was out of scope until I saw ganeshrkamath's and batwaraanirudh's great solutions. I just want to elaborate on their solutions a bit so that others can see how they reached their conclusions.

Since 375 = (3)(125), we know that the answer must be divisible by both 125 and 3.

Let's take a closer look at numbers that are divisible by 125.
Notice that:
(1)(125) = 125
(2)(125) = 250
(3)(125) = 375
(4)(125) = 500
(5)(125) = 625
(6)(125) = 750
(7)(125) = 875
(8)(125) = 000
(9)(125) = 1125
.
.
.
So, as you can see, the last 3 digits of numbers divisible by 125 must be 125, 250, 375, 500, 625, 750, 875, or 000

Notice that ONLY ONE of these cases is such that the number consists of only 0’s and 1’s
So, the number in question must end in 000

Since the number in question is ALSO divisible by 3, the sum of its digits must be divisible by 3.
So, the least possible number must be 111000

Answer: E

Cheers,
Brent

As always, excellent explaination Brent. I have been learning a lot from you.

As I was reading through your explaination, I was just thinking - would that help if we can add a 1 before the zeroes while answering - (8) (125) to make it go with your statement - 'Notice that ONLY ONE of these cases is such that the number consists of only 0’s and 1’s'.

Just my 2 cents. Please do let me know in case I am missing anything here.
Best Regards,
Rahul Sehgal
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