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NYC5648
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When the product of 3,070,956 and n is divided by 720 there 07 Jun 2012, 12:40
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E
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Difficulty:

75% (hard)

Question Stats:

56% (02:38) correct 44% (02:15) wrong based on 327 sessions

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When the product of 3,070,956 and n is divided by 720 there will be no remainder. If n > 1, what is the smallest value of n?

A. 10
B. 20
C. 30
D. 40
E. 60
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E
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When the product of 3,070,956 and n is divided by 720 there 07 Jun 2012, 14:08
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NYC5648
When the product of 3,070,956 and n is divided by 720 there will be no remainder. If n > 1, what is the smallest value of n?

A. 10
B. 20
C. 30
D. 40
E. 60
Show more

Given: \(\frac{3,070,956*n}{720}=integer\).

Factorize the divisor: \(720=2^4*3^2*5\).

Now, since 56 (the last two digits of 3,070,956) is divisible by 4, then 3,070,956 is divisible by 4=2^2, and since 956 (the last three digits of 3,070,956) is NOT divisible by 8, then 3,070,956 is NOT divisible by 8=2^3. That means that n must have 2^2=4 as its factor (3,070,956 is divisible only by 2^2 so in order 3,070,956*n to be divisible by 2^4 n must have 2^2 as its factor);

Similarly since the sum of the digits of 3,070,956 is 3+0+7+0+9+5+6=30 then 3,070,956 divisible by 3 but not by 3^2, so n must have 3 as its factor;

And finally since the units digit of 3,070,956 is 6 then 3,070,956 is not divisible by 5, so n must have 5 as its factor.

Therefore the least value of \(n\) is \(2^2*3*5=60\).

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).
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When the product of 3,070,956 and n is divided by 720 there Updated on: 07 Jun 2012, 14:23
Originally posted by geometric on 07 Jun 2012, 14:19.
Last edited by geometric on 07 Jun 2012, 14:23, edited 1 time in total.
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Bunuel
NYC5648
When the product of 3,070,956 and n is divided by 720 there will be no remainder. If n > 1, what is the smallest value of n?

A. 10
B. 20
C. 30
D. 40
E. 60

Given: \(\frac{3,070,956*n}{720}=integer\).

Factorize the divisor: \(720=2^4*3^2*5\).

Now, since 56 (the last two digits of 3,070,956) is divisible by 4, then 3,070,956 is divisible by 4=2^2, and since 956 (the last three digits of 3,070,956) is NOT divisible by 8, then 3,070,956 is NOT divisible by 8=2^3. That means that n must have 2^2=4 as its factor (3,070,956 is divisible only by 2^2 so in order 3,070,956*n to be divisible by 2^4 n must have 2^2 as its factor);

Similarly since the sum of the digits of 3,070,956 is 3+0+7+0+9+5+6=30 then 3,070,956 divisible by 3 but not by 3^2, so n must have 3 as its factor;

And finally since the units digit of 3,070,956 is 6 then 3,070,956 is not divisible by 5, so n must have 5 as its factor.

Therefore the least value of \(n\) is \(2^2*3*5=60\).

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).
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Your explanations blow my mind, Bunuel. In other words, they are outstanding. All I knew was that this is a prime factorization problem, but wasn't sure how to break down the problem. Would you say this is about a 650+ difficulty?
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When the product of 3,070,956 and n is divided by 720 there 07 Jun 2012, 14:20
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Bunuel
NYC5648
When the product of 3,070,956 and n is divided by 720 there will be no remainder. If n > 1, what is the smallest value of n?

A. 10
B. 20
C. 30
D. 40
E. 60

Given: \(\frac{3,070,956*n}{720}=integer\).

Factorize the divisor: \(720=2^4*3^2*5\).

Now, since 56 (the last two digits of 3,070,956) is divisible by 4, then 3,070,956 is divisible by 4=2^2, and since 956 (the last three digits of 3,070,956) is NOT divisible by 8, then 3,070,956 is NOT divisible by 8=2^3. That means that n must have 2^2=4 as its factor (3,070,956 is divisible only by 2^2 so in order 3,070,956*n to be divisible by 2^4 n must have 2^2 as its factor);

Similarly since the sum of the digits of 3,070,956 is 3+0+7+0+9+5+6=30 then 3,070,956 divisible by 3 but not by 3^2, so n must have 3 as its factor;

And finally since the units digit of 3,070,956 is 6 then 3,070,956 is not divisible by 5, so n must have 5 as its factor.

Therefore the least value of \(n\) is \(2^2*3*5=60\).

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).

Your explanations blow my mind, Bunuel. In other words, they are outstanding. All I knew is that this was a prime factorization problem, but wasn't sure how to break down the problem. Would you say this is about a 650+ difficulty?
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This one is more like 700+.
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When the product of 3,070,956 and n is divided by 720 there 26 Jun 2013, 02:25
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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When the product of 3,070,956 and n is divided by 720 there 26 Jun 2013, 22:09
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NYC5648
When the product of 3,070,956 and n is divided by 720 there will be no remainder. If n > 1, what is the smallest value of n?

A. 10
B. 20
C. 30
D. 40
E. 60
Show more

3070956 * n/720

lets make it

3070956 * n/72 *10
since we can see that 10 is a common factor with the answers

now divide 3070956/72 = 1535478/36 = 767739/18 = 255913/6

which cannot be divided any further

Thus, 6*10 = 60 is the smallest value of n , Ans: E
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When the product of 3,070,956 and n is divided by 720 there 27 Jun 2013, 00:10
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NYC5648
When the product of 3,070,956 and n is divided by 720 there will be no remainder. If n > 1, what is the smallest value of n?

A. 10
B. 20
C. 30
D. 40
E. 60
Show more

3070956 is divisible by 4 and 3 and thus by 12. Also, it is not divisible by 8 and 9. Also, it is given that \(\frac {3070956*n} {12*6*10}\) is an integer. Note that once 3070956 gets factored by 12 in the denominator, the remaining integer will neither be divisible by 2 nor by 3.Thus, for factoring the 60 in the denominator, n has to contain at-least a 60,and hence the answer is E.
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When the product of 3,070,956 and n is divided by 720 there 08 Jun 2024, 04:46
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