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A number is said to be a “digifac” if each of its digits is a factor
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09 Sep 2015, 23:32
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A number is said to be a “digifac” if each of its digits is a factor
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10 Sep 2015, 00:23
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. A number is said to be a “digifac” if each of its digits is a factor of the number itself. What is the sum of the missing digits of the following fivedigit digifac: 9, 5, 3 _ _ ? (A) 5 (B) 7 (C) 9 (D) 10 (E) 14 Let the missing digits be A, B. Then we should find fivedigit digifac 9, 5, 3, A, B. First of all, it should be a multiple of 5. Since a multiple of 5 should have unit digit 0 or 5, B should be 5(since 0 cannot be a factor of an integer). Furthermore the sum of all digits of a multiple of 3 or 9 is also a multiple of 3 or 9. So 9+5+3+A+5 should be a multiple of 9(it’s enough to consider only 9, since a multiple of 9 is also a multiple of 3). Therefore 13+A should be multiple of 9. So A is 5. The answer is 5+5=10 > D.
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Re: A number is said to be a “digifac” if each of its digits is a factor
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10 Sep 2015, 04:07
Bunuel wrote: A number is said to be a “digifac” if each of its digits is a factor of the number itself. What is the sum of the missing digits of the following fivedigit digifac: 9, 5, 3 _ _ ?
(A) 5 (B) 7 (C) 9 (D) 10 (E) 14
Kudos for a correct solution. Solution: The number should be multiple of 5. So, the last digit is 5 or 0. But 0 cant be a factor. So, Last digit is 5. The sum of all digits should be a multiple of 9. Now the sum is 22. So, 4th digit should be 5. Therefore, required sum = 10 Option D



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Re: A number is said to be a “digifac” if each of its digits is a factor
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10 Sep 2015, 05:37
Bunuel wrote: A number is said to be a “digifac” if each of its digits is a factor of the number itself. What is the sum of the missing digits of the following fivedigit digifac: 9, 5, 3 _ _ ?
(A) 5 (B) 7 (C) 9 (D) 10 (E) 14
Kudos for a correct solution. For the number to be divisible by 9, its sum should be a multiple of 9. This number will also be divisible by 3. The sum of digits we have now is 9+5+3 = 17 For number to be divisible by 5, it's unit digit should be 0 or 5. But 0 cannot be a factor of any number as 0 multiplied with any number will be 0. So, the unit's digit will be 5. The sum now becomes, 17+5 = 22 We need a multiple of 9 greater than 22 i.e 27. so, the ten's digit will be 2722 = 5. and the sum of missing digits will be 5+5=10 Answer: D



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Re: A number is said to be a “digifac” if each of its digits is a factor
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10 Sep 2015, 06:06
Bunuel wrote: A number is said to be a “digifac” if each of its digits is a factor of the number itself. What is the sum of the missing digits of the following fivedigit digifac: 9, 5, 3 _ _ ?
(A) 5 (B) 7 (C) 9 (D) 10 (E) 14
Kudos for a correct solution. For the number to be divisible by 3 and 9, the sum of the digits should be divisible by 3 and 9. Only D=10 satisfies. The sum is 27 Answer D



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Re: A number is said to be a “digifac” if each of its digits is a factor
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13 Sep 2015, 07:31
Bunuel wrote: A number is said to be a “digifac” if each of its digits is a factor of the number itself. What is the sum of the missing digits of the following fivedigit digifac: 9, 5, 3 _ _ ?
(A) 5 (B) 7 (C) 9 (D) 10 (E) 14
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Here, the term “digifac” should look intimidating. You probably haven’t studied digifacs before, so how should you approach this problem? Well, keep in mind that digifacs aren’t being tested; in fact, the author of this question just made that term up, and then defined it for you. What makes this question hard is that the nonchallengeseeker (I think I just made that term up, too…) will see the unfamiliar term “digifac” and lose faith immediately. “I don’t know what that is!” She who finds the challenge in the GMAT fun, however, will read the definition and think “got it – I need to find the two digits that ensure that 9, 5, and 3 are both factors of the overall number, and that the remaining two digits are also factors”. And work from there. The number must be divisible by 5, so the only units digits that work are 0 or 5. And the number must be divisible by 9 (and also 3), so we need the sum of all digits to be a multiple of 9. 9 + 5 + 3 = 17, so our only options are to get the sum to 18 (by adding 1) or to 27 (by adding 10). A quick glance at the answer choices shows that 0 1 isn’t an option. Why not? That would require 0 to be one of the digits…and 0 isn’t a factor of anything. So the units digit must be 5, making the tens digit 5, and we have 95,355. That number is a multiple of 5, 3, and 9, so it works: the correct answer is D, and more importantly this fun challenge required no “trivial” information about digifacs…that term only existed to obscure the link between the given information and the path to the answer.
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Re: A number is said to be a “digifac” if each of its digits is a factor
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13 Sep 2015, 07:35
Bunuel wrote: Bunuel wrote: A number is said to be a “digifac” if each of its digits is a factor of the number itself. What is the sum of the missing digits of the following fivedigit digifac: 9, 5, 3 _ _ ?
(A) 5 (B) 7 (C) 9 (D) 10 (E) 14
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Here, the term “digifac” should look intimidating. You probably haven’t studied digifacs before, so how should you approach this problem? Well, keep in mind that digifacs aren’t being tested; in fact, the author of this question just made that term up, and then defined it for you. What makes this question hard is that the nonchallengeseeker (I think I just made that term up, too…) will see the unfamiliar term “digifac” and lose faith immediately. “I don’t know what that is!” She who finds the challenge in the GMAT fun, however, will read the definition and think “got it – I need to find the two digits that ensure that 9, 5, and 3 are both factors of the overall number, and that the remaining two digits are also factors”. And work from there. The number must be divisible by 5, so the only units digits that work are 0 or 5. And the number must be divisible by 9 (and also 3), so we need the sum of all digits to be a multiple of 9. 9 + 5 + 3 = 17, so our only options are to get the sum to 18 (by adding 1) or to 27 (by adding 10). A quick glance at the answer choices shows that 0 1 isn’t an option. Why not? That would require 0 to be one of the digits…and 0 isn’t a factor of anything. So the units digit must be 5, making the tens digit 5, and we have 95,355. That number is a multiple of 5, 3, and 9, so it works: the correct answer is D, and more importantly this fun challenge required no “trivial” information about digifacs…that term only existed to obscure the link between the given information and the path to the answer. Check other Special Numbers and Sequences questions in our Special Questions Directory.
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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: A number is said to be a “digifac” if each of its digits is a factor
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27 Jul 2018, 12:41
Hmm good question, got me thinking a bit. It's just recognizing divisibility rules. If each digit is a factor, then the number must be divisible by 3, 5, 9 (all provided). Everything divisible by 9 is divisible by 3 so ignore 3. If we take 9 + 5 + 3, we get 17. The only way to get a number divisible by 9 and 5 is to have it be a multiple of 9 and end in 5 or 0. Only adding 10 (D) gives a choice that is divisible by 9 and ends in 0.
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Re: A number is said to be a “digifac” if each of its digits is a factor &nbs
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