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A positive integer is divisible by 3 if and only if the sum
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Updated on: 03 Jun 2014, 12:38
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A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have? A 2 B. 3 C. 4 D. 5 E 10
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Originally posted by NoHalfMeasures on 13 May 2014, 10:07.
Last edited by Bunuel on 03 Jun 2014, 12:38, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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Re: divisibility by 3
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14 May 2014, 22:14
MensaNumber wrote: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?
1. 2 2. 3 3. 4 4. 5 5 10 Divisibility by 3 can be easily checked without adding all the digits of a large number: Say the number is 36275981720160 Now rather than adding all the digits, simply ignore digits if they are multiples of 3. Also ignore some digits if their sum is a multiple of 3. What I mean is this: Go one digit at a time 3  Ignore 6  Ignore 2 + 7 = 9  Ignore 5  Hold 9  Ignore 8 + 1  Ignore 7 + 2  Ignore 1 + 5 (which you were holding) = 6  Ignore 6  Ignore 0  Ignore You are not holding anything so the number is divisible by 3. Similarly, when you have a number such as 1k2k24, 1+2 = 3  Ignore 2+4 = 6  Ignore So if k is a multiple of 3, then the number will be divisible by 3, else it will not be. There are 4 single digit multiples of 3: 0, 3, 6, 9 Hence k can take 4 values.
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Re: divisibility by 3
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14 May 2014, 03:58
n will be divisible by 3 if (1+k+2+k+2+4=2k+9) is divisible by 3. Since 9 is divisible by 3, we have to find the values of k (k is single digit) for which 2k is divisible by 3. k can be 3,6 or 9, hence possible values of n can be 1 32 324 or 1 62 624 or 1 92 924. Hence number of possible values of n is 3. Answer is 2
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Re: divisibility by 3
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14 May 2014, 14:01
kundan232003 wrote: n will be divisible by 3 if (1+k+2+k+2+4=2k+9) is divisible by 3. Since 9 is divisible by 3, we have to find the values of k (k is single digit) for which 2k is divisible by 3. k can be 3,6 or 9, hence possible values of n can be 132324 or 162624 or 192924. Hence number of possible values of n is 3. Answer is 2 I think it is actually 4. You did not include 0 as a value of k. Edit: 4 as in 4 values of k, which is really answer choice 3.



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Re: divisibility by 3
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21 May 2014, 01:49
Thank you Karishma for sharing this great little tip! Here are a few more questions to practice the just learnt tip: 1. If the fivedigit integer n is divisible by 3 and 10, and n is of the form 4x13y, where x and y represent digits, how many values could n have? (A)1 (B) 2 (C) 3 (D) 4 (E) 5 2. If the fivedigit integer n is divisible by 3 and 5, and n is of the form 4x13y, where x and y represent digits and x>1, how many values could n have?(A) 2 (B) 3 (C) 4 (D) 5 (E) 6 3. If the fivedigit integer n is divisible by 3 and 2, and n is of the form 4x13y, where x and y represent digits and y>2, how many values could n have?(A) 8 (B) 9 (C) 10 (D) 15 (E) 36
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Re: divisibility by 3
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21 May 2014, 02:10
ANSWERS 1. C 2. D 3. C SOLUTIONS 1. For n to be divisible by 10, y has to be zero. So, n = 4x130 As suggested by Karishma above, when doing the divisibility test for 3, we don't need to add all digits: 4+x+1+3+0. We can simply add the digits that are nonmultiples of 3 or whose sum is not a multiple of 3. So, in this case, we need to only add 4+x+1 = 5+x For what values of x will 5+x be divisible by 3? For x=1, 4, 7. So, in all there are only 3 possible numbers that will be divisible by both 3 and 10: 41130, 44130, 47130 (C) is the answer. 2. For n to be divisible by 5, y could either be 0 or 5. Case 1: y= 0So, n= 4x130 Using the divisibility rule for 3, we find that possible values of x are 1, 4, 7. However, it is given that x>1. So, in total, only 2 values of x are possible. Case 2: y=5So, n= 4x135 Here sum of 1+3+5 = 9, a multiple of 3. So, we need to only consider the digits 4 and x for the divisibility test. Possible values of x are 2, 5, 8. So, 3 values are possible. Thus, total possible values of n are 5: 44130, 47130 42135, 45135, 48135 3. For n to be divisible by 2, y= 0, 2,4,6,8. However, it is given that y>2. So, allowed values of y: 4, 6, 8 Case 1: y= 4So, n= 4x134 Possible values of x = 0,3,6,9. Therefore, 4 values. Case 2: y=6So, n= 4x136 We need to only consider the digits 4, x, 1 for the divisibility rule (since 3 and 6 are already multiples of 3). Possible values of x= 1,4,7. Therefore, 3 values Case 3: y=8So, n= 4x138 We need to consider only the digits 4 and x for the divisibility rule (since 3 and 8+1 are already multiples of 3). Possible values of x = 2, 5, 8. Therefore, 3 values Total values of n possible = 4+3+3 = 10
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Re: divisibility by 3
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21 May 2014, 03:13
Please find below the complete list of divisibility rules: Divisible by: 2 IF: The last digit is even (0,2,4,6,8) 3 IF: The sum of the digits is divisible by 3 4 IF: The last 2 digits are divisible by 4 5 IF: The last digit is 0 or 5 6 IF: The number is divisible by both 2 and 3 7 IF: If you double the last digit and subtract it from the rest of the number and the answer is: 0, or divisible by 7 (Note: you can apply this rule to that answer again if you want) Eg: 672 (Double 2 is 4, 674=63, and 63÷7=9) Yes 8 IF: The last three digits are divisible by 8 9 IF: The sum of the digits is divisible by 9 10 IF: The number ends in 0 11 IF: If you sum every second digit and then subtract all other digits and the answer is: 0, or divisible by 11 Eg: 3729 ((7+9)  (3+2) = 11) Yes; 25176 ((5+7)  (2+1+6) = 3) No 12 IF: The number is divisible by both 3 and 4 Cheers, Rajat
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Re: divisibility by 3
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24 May 2014, 22:55
rajatjain14 wrote: Please find below the complete list of divisibility rules:
7 IF: If you double the last digit and subtract it from the rest of the number and the answer is: 0, or divisible by 7
Cheers, Rajat To further clarify on this: It is not a one time operation but one needs to keep subtracting double of last digit till it results in a 2 digit number. Example: 1603 > 1602(3)=154 > 152(4)=7, so 1603 is divisible by 7. Look up here for more divisibility checks: http://www.savory.de/maths1.htm



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Re: A positive integer is divisible by 3 if and only if the sum
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03 Jun 2014, 20:52
MensaNumber wrote: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?
A 2 B. 3 C. 4 D. 5 E 10 1k2k24 Taking sum of the numericals = 1 + 2 + 2 + 4 = 9 We require the values of K such that they are also divisible by 3 1 02 024 1 32 324 1 62 624 1 92 924 Answer = 4 = C
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Re: divisibility by 3
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03 Jun 2014, 21:07
MavenQ wrote: Thank you Karishma for sharing this great little tip!
Here are a few more questions to practice the just learnt tip:
1. If the fivedigit integer n is divisible by 3 and 10, and n is of the form 4x13y, where x and y represent digits, how many values could n have?
(A)1 (B) 2 (C) 3 (D) 4 (E) 5
LCM of 3 & 10 = 30 To make 4x13y divisible by 30, y has to be 0 Now try to find values for x 4 + 1 + 3 + 0 = 8 We require the addition of digits to be divisible by 3, so x can be 1 or 4 or 7 No. created would be: 4 1130 4 4130 4 7130 Answer = 3 = C
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Re: divisibility by 3
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03 Jun 2014, 21:18
MavenQ wrote: Thank you Karishma for sharing this great little tip!
2. If the fivedigit integer n is divisible by 3 and 5, and n is of the form 4x13y, where x and y represent digits and x>1, how many values could n have?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6 To make no. divisible by 3, sum of all digits should be divisible by 3 To make no. divisible by 5, y has to be either 0 OR 5 When y = 0, x can be 4 or 7 (1 has to be ignored as x>1) 4 413 04 713 0When y = 5, x can be 2, 5 or 8 4 213 54 513 54 813 5Total 5 combinations possible Answer = 5 = D
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Re: divisibility by 3
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03 Jun 2014, 21:58
MavenQ wrote: 3. If the fivedigit integer n is divisible by 3 and 2, and n is of the form 4x13y, where x and y represent digits and y>2, how many values could n have?
(A) 8 (B) 9 (C) 10 (D) 15 (E) 36 When y>2, it may be 4 , 6 or 8 When y = 4, x may be 0, 3, 6, 9 When y = 6, x may be 1, 4, 7 When y = 8, x may be 2, 5, 8 Total combinations = 4 + 3+ 3 = 10 Answer = C
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Re: divisibility by 3
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04 Jun 2014, 09:44
Bunuel wrote: MensaNumber wrote: MensaNumber wrote: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?
1. 2 2. 3 3. 4 4. 5 5 10 HI Bunuel, How can I add gmatclub time tracker and other stat tool of percentage right/wrong etc to this question? This one is from gmatprep question pack. Also pls could update the permalink? In its present form it is not searchable. Thanks Done. Can you please check whether you copied the question exactly as it is in the source? Thank you. P.S. Posting rules: rulesforpostingpleasereadthisbeforeposting133935.htmlThanks Bunuel! I think I did. But I think I didnt make full use of the subject line. Thanks for sharing the positing rules also. Will follow them exactly now.
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Re: A positive integer is divisible by 3 if and only if the sum
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30 Nov 2014, 14:08
VeritasPrepKarishma wrote: MensaNumber wrote: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?
1. 2 2. 3 3. 4 4. 5 5 10 Divisibility by 3 can be easily checked without adding all the digits of a large number: Say the number is 36275981720160 Now rather than adding all the digits, simply ignore digits if they are multiples of 3. Also ignore some digits if their sum is a multiple of 3. What I mean is this: Go one digit at a time 3  Ignore 6  Ignore 2 + 7 = 9  Ignore 5  Hold 9  Ignore 8 + 1  Ignore 7 + 2  Ignore 1 + 5 (which you were holding) = 6  Ignore 6  Ignore 0  Ignore You are not holding anything so the number is divisible by 3. Similarly, when you have a number such as 1k2k24, 1+2 = 3  Ignore 2+4 = 6  Ignore So if k is a multiple of 3, then the number will be divisible by 3, else it will not be. There are 4 single digit multiples of 3: 0, 3, 6, 9 Hence k can take 4 values. Hi Karishma, So question for you here  if the above example ended at 3627598172  we would be holding 5. Does that mean that the number is NOT divisible by 3?



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Re: A positive integer is divisible by 3 if and only if the sum
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30 Nov 2014, 22:04
russ9 wrote: VeritasPrepKarishma wrote: MensaNumber wrote: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?
1. 2 2. 3 3. 4 4. 5 5 10 Divisibility by 3 can be easily checked without adding all the digits of a large number: Say the number is 36275981720160 Now rather than adding all the digits, simply ignore digits if they are multiples of 3. Also ignore some digits if their sum is a multiple of 3. What I mean is this: Go one digit at a time 3  Ignore 6  Ignore 2 + 7 = 9  Ignore 5  Hold 9  Ignore 8 + 1  Ignore 7 + 2  Ignore 1 + 5 (which you were holding) = 6  Ignore 6  Ignore 0  Ignore You are not holding anything so the number is divisible by 3. Similarly, when you have a number such as 1k2k24, 1+2 = 3  Ignore 2+4 = 6  Ignore So if k is a multiple of 3, then the number will be divisible by 3, else it will not be. There are 4 single digit multiples of 3: 0, 3, 6, 9 Hence k can take 4 values. Hi Karishma, So question for you here  if the above example ended at 3627598172  we would be holding 5. Does that mean that the number is NOT divisible by 3? Yes, then the number would not be divisible by 3 and on dividing by 3 the remainder would be 2 (the remainder left on dividing 5 by 3).
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Re: A positive integer is divisible by 3 if and only if the sum
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01 May 2017, 14:46
n = 1k2,k24 1+2+2+4=9 So, for n to be divis by 3, k must either be 0 or a multiple of 3 between 0 and 9. Therefore, k can be 0, 3, 6, or 9. Answer C.



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Re: A positive integer is divisible by 3 if and only if the sum
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01 Jul 2017, 04:23
I think the trick would be to consider '0' as potential candidate.
Hence four.



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Re: A positive integer is divisible by 3 if and only if the sum
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14 May 2018, 16:54
NoHalfMeasures wrote: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?
A 2 B. 3 C. 4 D. 5 E 10 We know the following: (1 + 2 + 2 + 4 + 2k)/3 = integer (9 + 2k)/3 = integer Thus, k must be a multiple of 3, so it can be 0, 3, 6, or 9. Answer: C
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Re: A positive integer is divisible by 3 if and only if the sum
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22 May 2018, 05:42
NoHalfMeasures wrote: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the sixdigit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?
A 2 B. 3 C. 4 D. 5 E 10 Digits : 0,1,2,3,4,5,6,7,8,9
We can try every digit and see which fits perfect.
Only 0,3,6, and 9 works.
Hence n can have 4 values.
(C)
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Re: A positive integer is divisible by 3 if and only if the sum
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28 Aug 2018, 15:13
Am I the only one who found the question confusing? It asked "how many values could N have"  not K. I understand K can take 4 values, but doesn't any combination of 0, 3, 6, 9 mean a different value for N?




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