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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
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nileshk wrote:
Hello,

I understood the rule, but i didn't get how you conclude 53N is divisible by 3.

Can you please elaborate?


Since N is 3 digit number, where all digit are of the same value

eg. 111 , 222, 333, .... 999 : these all are divisible by 3 in short N has to be divisible by 3

Hence, 53N should be a multiple of 3

Therefore, 53*N = consist of 1, 2, 4, 6, 7, 9 . if we add all the digit it should be divisible by 3 ( Divisibility property of 3)

1 + 2 + 4 + 6 + 7 + 9 = 29 - This is not divisible by 3 but if we remove digit 2 then it is divisible 3


***
Multiply 53*333 = 17,649 = you get a 5 digit no. except 2. Remaining digits are 1,4,6,7 and 9, as mentioned in the ques.

You can try out with all the nos. by taking N =111 .... , 999 and compare out the result
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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
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Thanks for elaborating.
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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
this took me so long to get my head around...
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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
I just started multiplying all three-digit same character numbers by 53 on my calculator. 333 was the one and I got the exact number.
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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
170896 wrote:
I just started multiplying all three-digit same character numbers by 53 on my calculator. 333 was the one and I got the exact number.


Great idea!

Cheers,
Brent
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N is a positive 3-digit integer in which all three digits ha [#permalink]
GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
N is a positive 3-digit integer in which all three digits have the same value. The product 53N is a 5-digit integer comprised of five of the following six digits: 1, 2, 4, 6, 7, 9

Which of the following digits is NOT in the product 53N?
A) 1
B) 2
C) 4
D) 6
E) 7


KEY PROPERTY #1: There's a nice divisor property that says: If n is divisible by d, then kn is divisible by d (assuming n, d and k are INTEGERS)
For example, since 33 is divisible by 11, we also know that (33)(7) is divisible by 11

Since 111 is divisible by 3, we know that (2)(111) is divisible by 3, and (3)(111) is divisible by 3, and (4)(111) is divisible by 3, and (5)(111) is divisible by 3, etc.
In other words, we know that N is divisible by 3.
So, from property #1, we also know that 53N is divisible by 3.

KEY PROPERTY #2: If a number is divisible by 3, the sum of its numbers is also be divisible by 3
So, we must choose 5 numbers (from 1, 2, 4, 6, 7, 9) so that the sum of the digits is divisible by 3.
1 + 2 + 4 + 6 + 7 + 9 = 29
So, if we DON'T select the digit 2, the remaining digits add to 27, which IS divisible by 3.

Answer: B

Cheers,
Brent


If we multiply 53 with 444, we get 23,532 which contains 2. Can you explain how is this wrong?
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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
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shubhammathur98 wrote:

If we multiply 53 with 444, we get 23,532 which contains 2. Can you explain how is this wrong?


We are told that the product 53N is a 5-digit integer comprised of five of the following six digits: 1, 2, 4, 6, 7, 9

23,532 is a 5 digit-integer, BUT it is not comprised of five of the following six digits: 1, 2, 4, 6, 7, 9
So, it does not meet the conditions given in the question.
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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
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Re: N is a positive 3-digit integer in which all three digits ha [#permalink]
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