Day-1 MOCK TEST-1 Conceptual Quiz on Divisibility and Prime numbers. Number of Questions-> 100. Sources-> The following questions are from various sources and quite a few are self made. 1)If z is divisible by all the integers from 1 to 5 inclusive,what is the smallest value of z? 60. 2) Data Sufficiency-> Is the positive integer z divisible by 12? A)z is divisible by 3. B)z is not divisible by 2. B. 3)Data Sufficiency-> Is integer z divisible by 21? A)z is divisible by 7. B)z is divisible by 3. C. 4)Data Sufficiency-> Is integer z divisible by 48? A)z is divisible by 12. B)z is divisible by 4. E. 5)Data Sufficiency-> Is integer z divisible by 12? A)z is divisible by 8. B)z is divisible by 6. C. 6)Data Sufficiency-> Is integer z divisible by 24? A)z is divisible by 6. B)z is divisible by 4. E. 7)How many prime numbers between 1 and 100 are factors of 168? Three. 8)How many Prime factors does 2180 have? Three. 2180=2^2*5*109 Notice 109 is a prime number. Hence, 2180 has three prime factors 9)What is the greatest prime factor of 49^{19}-7^{35} A)2 B)3 C)7 D)13 E)19 E. 49^19=7^38 Thus => 49^19-7^35=7^38-7^35 => 7^35(7^3-1) = 7^35*342=> 7^35*2*3^2*19 Hence greater Prime factor=19. 10)If n is the product of integers from 1 to 12 exclusive,how many prime factors does n have? Five. Product = 2^8*3^4*5^2*7*11 . 11)If 2,3,5 are the prime factors of N,what is the value of N? A)2 B)3 C)5 D)2*3*5 E)cannot be determined. E. N= 2^a*3^b*5^c* P^d Where p is some other prime number. Basically, There are ∞ numbers with the same set of primes . 12)If 2,3,5 are the " ONLY " prime factors of K,what is the value of K? A)2 B)3 C)5 D)2*3*5 E)cannot be determined. E. K =2^a*3^b*5^c Again -> There are ∞ numbers with the same set of primes. 13)Data Sufficiency -> What is the value of positive integer K? A)2,5,7 are prime factors of K. B)K K = 2^a*5^b*7^c*P^z ; for any prime number P. There exist ∞ numbers with the same set of prime numbers. Hence not sufficient. From statement 2-> K Only Value of K possible is 70. 14)If p is a positive integer and 200 multiplies by p is square of an integer,what is the value of p? A)2 B)3 C)5 D)10 E)Cannot be determined. E. Notice that the general expression for p will be --> 2 * Prime^even E.g => p=2 or p=2*5^2 Etc. 15)Data Sufficiency->What is the value of positive integer p? A)200 multiplied by p is square of an integer. B)p is an even number less than 15. E. From statement 1-> p=2 or 2*3^2 or 2*2^2 or 2* Any-prime^even--> not sufficient From statement 2-> p=2 or 4 or any even Not sufficient Combining them -> p=2 or p=2*2^2=8 Hence p can be 2 or 8 Thus not sufficient Hence E. 16)Data Sufficiency->What is the value of positive integer p? A)300 multiplied by p is square of an integer. B)p is an even number less than 15. C. From statement 1-> p=3 or 3*3^2 or 3*7^2 or 3*prime^even--> not sufficient From statement 2-> p=2 or 4 or any even Not sufficient Combining them -> p=3*2^2 =12 is the only possible value. Hence Sufficient. Hence C 17)10^25 – 560 is divisible by all of the following EXCEPT:- A)11 B)8 C)5 D)4 E)3 E. Clearly it is divisible by 8,5,4 The real Test here is choosing between 3 and 11. 10^t for any positive integer t will always leave a remainder 1 with 3. e.g-> 10=> 3*3+1 100=3*33+1 1000=3*333+1 10000=3*3333+1 Hence 10^25=3m+1 Also 560=3k+2 (sum of digits =11,hence remainder with 3-> 2) Thus 3m+1-3k-2 => 3x-1 for some integer x Clearly it will never be divisible by 3. Alternatively,we can also use the "Binomial Expansion". 18)Data Sufficiency->What is the value of positive integer p? A)300 multiplied by p is square of an integer. B)p is a factor of 75 E. Combing the two statements -> p can be 3 or 3*5^2 19)Data Sufficiency->What is the value of positive integer p? A)200 multiplied by p is square of an integer. B)p is a factor of 60. C. Notice that by combing the two statements -> The only value of p possible is 2. 20)Data sufficiency-> If a and b are prime numbers,what is the value of a*b? A)a-b=1 B)a=3 A. Notice that the only consecutive numbers that are prime are -> 2,3 Hence From statement 1-> Product a*b=3*2=6 21)If x and y are prime numbers, each greater than 2, which of the following must be true? I. x+y is an even integer II. xy is an odd integer III. (x/y) is not an integer A. II only B. I and II only C. I and III only D. II and III only E. I, II, and III B. As x and y are prime numbers both greater than 2 => they both must be odd Hence x+y=odd+odd=even x*y=odd*odd=odd Notice that if x=y => x/y will be 1. Hence III is not always true. If x and y would happen to be different primes then III will be always true. 22)Which of the following statements must be true-> A)zero is neither prime nor composite. B)one is neither prime nor composite. C)-3 and 3 are both primes. A and B. Notice -3 is negative and hence cannot be prime 23)Which of the following statements must be true-> 1)A prime number must be positive. 2)For any prime number p,there is no x such that 1<x<p and x is a divisor of p. 3)The product of first ten primes is even. 4)All prime numbers greater than 71 are odd. 5)2 and 3 are the only consecutive integers that are also prime numbers. 6)p is a prime number and x and y are positive integers.If p=x*y then one out of x or y must be one. 7)All the prime numbers greater than 3 can be written as either 4n+1 or 4n-1. 8)All the prime numbers greater than 3 can be written as either 6+1 or 6n-1. All statements are true. 24)Data Sufficiency->If p is a prime number,what is the value of p? A)11 If A and B are positive integers, is A/B an integer? (1) Every factor of B is also a factor of A. (2) Every prime factor of B is also a prime factor of A. A. Statement 1->Using the property => "Every number is a factor of itself " B must be a factor of B Hence B must be a factor of A. Thus A/B must be an integer -> Sufficient Statement 2-> We can use test cases to Prove this insufficient Case 1-> \frac{2*3*7}{2*3*7} =1=Integer Case 2-> \frac{2*3*7*11}{2^5} => Not an integer. Hence insufficient Hence A. Source-> GMAT-Prep 26)Data Sufficiency->Is \frac{n}{14} an integer? (1) n is divisible by 28. (2) n is divisible by 70. D. Statement 1->n=28k = 14*2k=14k' Statement 2->n=70k= 14*5k=14k' Hence D. Alternatively,we can use the property => "A number is divisible by factors as well as factors of its factors" 27)If x is a prime number, what is the value of x? (1) There are a total of 50 prime numbers between 71 and x, inclusive. (2) There is no integer n such that x is divisible by n and 1 Notice that 71 is 20th prime. 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71. Hence x will be 50th prime from 71. Hence sufficient. Statement 2-> This is just another way of saying that x is prime. Hence not sufficient Hence A. 28)Data Sufficiency->If p = (n)(5^x)(3^k), is p divisible by 10? (1) n, x, and k are even. (2) x > n > k > 0 C. Firstly we are not given in p,x,k,n are an integers or not. If p is an integer -> Then for p to be divisible by 10 => It must be dividable by 5 and 2=> It must have both 5 and 2 as its primes. Hence => n must be even and n,k,x must be all pastime integers. Statement 1-> All n,k,x are even. Case 1-> -2*5^{-10}*3^{-10} =>p is not an integer Case 2-> 2*5^2*3^4 => P is an integer and is dividable by 10. Hence not sufficient. Statement 2-> All are positive Again lest use test cases Case 1-> 1*5*3 => p is not divisible by 10 Case 2-> 2*5*3 => p is divisible by 10 Combing the two statements => n must be a positive integer and x,k are positive even numbers. Hence p must be dividable both 2 and 5. Hence p must be dividable by 10. Hence C. 29)Data Sufficiency->If x is a prime number,what is the value of x? 1)There are 3 prime numbers between 11 and x 2 )x+1 is a prime number B. 30)If A is positive integer,then A , A^3 and A^7 will have the exact same prime factors. A)True B)False True. Property -> X and X^n always have the exact same prime factors. 31)Data Sufficiency->If y and x are positive integers, is y divisible by 3? (1) y=x(x+1) (2) y^2 is divisible by 9. B. 32)Data Sufficiency->If y and x are positive integers, is y divisible by x? (1) y = x^2 + x (2) x has the same prime factors as y. A. 33)Data Sufficiency->If z is a positive integer,is (z^31+7)^2 divisible by 4? A)√z has five prime factors. B)All prime factors of z^3 are greater than 7. B. Property in action -> For any number to be odd -> Two must not be its prime. Also-> X and X^n always have the exact same prime factors. 34)If x is a positive odd integer and y is a negative even integer, which of the following must be true? A. x^3 + y is a positive odd integer B. x^2 + y^2 is a negative odd integer C. x^0 + y^11 is a negative odd integer D. x + y is a positive odd integer E. x + y is a negative odd integer C. 35)Data Sufficiency->How many positive prime numbers are less than the integer n? (1) 14 If x and y are positive integers, is x divisible by y? 1) x is divisible by 5. 2) y is divisible by 5. E. 37)Data Sufficiency->How many prime factors does x^{37} have? (1) The number of prime factors of 16x is one more than the number of prime factors of x^4 (2) 2x^{16} has 2 prime factors C.This question is based on a very important property -> X and X^n always have the exact same prime factors. So x^{37} will have the exact same prime factors as x. Thus,here the question is asking us to get the number of prime factors of x . Statement 1-> 16x has one more prime factor than x^4 Now x and x^4 will have the exact same prime factors. Hence 16x has one more prime than x 16x=2^4*x => Hence 2 cannot be the prime factor of x . But we still don't have the number of prime factors of x . Hence not sufficient. Statement 2-> 2*x^{16} has two prime factors Here again x^{16} will have the exact same primes as x . So as 2*x^{16} has two prime factors => x can have either one prime factor if 2 isn't one of its prime . Or two prime factors if 2 is one of its prime. Hence not sufficient. Combining the two statements=> As 2 cannot be one of the primes of x => x must have only one prime factor Hence sufficient. Hence C. 38)If a positive integer x has 3 prime factors,how many prime factors does x^3 have? A)27 B)6 C)3 D)1 E)cannot be determined C. X and X^n always have the exact same prime factors. 39)Data Sufficiency->If k is a positive integer, how many unique prime factors does 14k have ? (1) k^4 is divisible by 100 (2) 50*k has 2 prime factors C. 14k=2*7*k To get the number of prime factors of 14k => we need the number of prime factors of k Statement 1-> Using the property -> X and X^n always have the exact same primes => k must have both 2 and 5 as its primes. General expression for k= 2^a*5^b*P^z Where a and b are non negative integers and P is any other prime number and z≥0 Hence not sufficient. Statement 2-> Form this statement => k can be 2^a or 5^b or 2^a*5^b Or k=1 Hence not sufficient. Combing the two statements => k must have only 2 and 5 as its prime factors Hence 14k will have 3 prime factors Hence Sufficient Hence C. 40)Data Sufficiency->If x is an integer, is x^3 divisible by 9? (1) x^2 is divisible by 9. (2) x^4 is divisible by 9. D. Property in action -> X and X^n always have the exact same prime factors. 41)Data sufficiency->How many different prime numbers are factors of positive integer n? (1) 4 different prime numbers are factors of 2n (2) 4 different prime numbers are factors of n^2 B. The question is just asking us the prime factors for n. Statement 1-> From this statement we can conclude that n can have either 4 or 3 prime factors. Hence Not sufficient Statement 2-> Using the Property => X and X^n always have the exact same prime factors => As n^2 has 4 prime factors => n myst have 4 prime factors too. Hence B. 42)If x and y are positive integers, are they consecutive? 1) x+y=3 2) x-y=1 D. 43)Data Sufficiency->Is the integer r divisible by 3? (1) r is the product of 4 consecutive integers. (2) r Product of n consecutive integers is always divisible by n factorial. So from statement 1-> r=24k for some integer k. 44)Data Sufficiency->Is x^2*y^4 an integer divisible by 9 ? (1) x is an integer divisible by 3. (2) xy is an integer divisible by 9. E. Using test cases can be extremely helpful here. Clearly statement 1 and 2 are not sufficient on their own. Combining them lets make test cases => Case 1=> x=3 y=3 x^2*y^4 will be divisible by 9. Case 2=> x=18 y=1/2 x^2*y^4 will not be divisible by 9. Hence E. 45)Data Sufficiency->For positive integers x and y, x^2 = 350y. Is y divisible by 28? (1) x is divisible by 4. (2) x^2 is divisible by 28. A. We are told that x and y are both positive integers. x^2=350y As x is an integer => x^2 must be a perfect square. x^2=350y=2*5^2*7*y Hence the general expression for y would be -> 2*7*p^z where p is an integer and z is even. We are asked if y is divisible by 28 or not. Clearly y is already divisible by 7 and 2 Hence for y to be divisible by 28=> p must be even too. Statement 1-> x is divisible by 4 Hence x^2 must have 2^4. As x=2*5^2*7*y => For x^2 to have a 2^4 => y must have a 2^3. Hence y must have a 2^3 and it already has a 7. Thus y must be divisible by 28 Hence sufficient. Statement2-> Here lets use test cases y=2*7 =>y will not be divisible by 28 y=2*7*2^2 => y will be divisible by 28. Hence not sufficient. Hence A. 46)Data Sufficiency-> How many Prime factors does positive integer n have ? A) n^3 has 7 prime factors. B) √n has 7 prime factors. D. X and X^n always have exact same prime factors. In other words => "Exponentiation does not change the prime factors of any positive integer X." 47)If x and y represent the number of factors of 90 and 147 respectively.What is the value of x-y? 6. 90=2*3^2*5 Number of factors of 90 =>2*3*2=> 12 So x=12 147=3*7^2 Number of factors of 147=>2*3=>6 So y=6 Hence x-y=12-6=6. 48)If n is the product of all the integers from 1 to 10 exclusive,how many factors does n have? 160. 49)How many even factors does 21600 have? A)32 B)42 C)60 D)25 E)52 C. 21600=2^5*3^3*5^2 For even factors -> The power of 2 should at-least be 1. Hence possible cases -> 5*4*3=>60. Additionally, for odd factors -> The power of 2 must be zero. Possible cases -> 1*4*3=> 12. 50)How many positive odd divisors does 540 have? A)6 B)8 C)12 D)15 E)24 B. Here 540=2^2*3^3*5 Odd factors => 1*4*2=>8 Additionally Even factors =>2*4*2=>16 Total Factors =>3*4*2=>24. 51)How many of the factors of 72 are divisible by 2? A)4 B)5 C)6 D)8 E)9 E. 72=2^3*3^2 For Even divisors ->Power of 2 must be at-least 1. Possible cases=>3*3=> 9. Hence E. 52)The number 100^2 has how many factors? A) 28 B) 25 C) 24 D) 20 E) 16 B. 53)How many factors of 330 are odd numbers greater than 1? A)8 B)7 C)6 D)5 E)4 B. 330=2*5*3*11 For Odd factors => Power of 2 must be zero Possible cases -> 1*2*2*2=>8 So 330 has 8 odd divisors including one. Excluding one => 330 must have 7 odd divisors. Hence B. 54)If N=2^7*3^5*5^6*7^8 . How many factors of N are divisible by 50 but NOT by 100? A)240 B)345 C)270 D)120 E)None of these C. For factors to be divisible by 50 and not divisible by 100 --> Allowed powers of 2->one Allowed powers of 5->5 Allowed powers of 3->6 Allowed powers of 7->9 Hence possible cases ->1*5*6*9=>270. Hence C. 55)Data Sufficiency->How many positive factors does the positive integer x have? (1) x is the product of 3 distinct prime numbers. (2) x and 3^7 have the same number of positive factors. D. 56)Data Sufficiency->How many factors does x have, if x is a positive integer? (1) x = p^n , where p is a prime number (2) n^n = n + n , where n is a positive integer C. From statement 1-> Number of factors of x will be (n+1) Since we don't have any clue of n => not sufficient. From statement 2-> n must be 2 as no other value of n will satisfy the equation. Since we don't have any clue of x => not sufficient. Combining the two statements => Number of factors of x-> n+1=2+1=3 Hence C. 57)Data Sufficiency->If 2,3,5 are the only prime factors of a positive integer p,what is the value of p? A)p>100 B)p has exactly 12 factors including 1 and 12. C. Since 2,3,5 are the only prime factors of p=> The general expression for p will be 2^a*3^b*5^c where a,b,c are positive integers. Statement 1->p>100. There exist infinite such cases. In-fact there are only a few cases where p will be less than 100. Hence not sufficient. Statement 2-> Factors =12 Hence (a+1)*(b+1)*(c+1)=12=2*2*3 NOTE-> We can't arrange 12 as 1*1*12 or 2*6*1 or 4*3*1 as a,b,c are positive integers. Hence a,b,c => (1,1,2) or (1,2,1) or (2,1,1) Posible values of p=> 150,90,60 Hence not sufficient. Combining the two statements => p must be 150. Hence C. This is probably the hardest Question in this set. 58)Data Sufficiency->Is the positive integer P prime? (1) 61 A perfect Square always has an odd number of factors If a number has an odd number of factors => It must be a perfect square. 60)How many factors does 64 have? A)8 B)6 C)7 D)6 E)4 C. Method 1=> 64=2^6 Hence it will have 7 factors Hence C. Method 2=> 64 is a perfect square =>Thus, It will always have an of Odd number of factors. The only answer choice that is odd is C. 61)Data Sufficiency->Is the positive integer n divisible by 6 ? (1) \frac{n^2}{180} is an integer. (2) \frac{144}{n^2} is an integer A. For n to be divisible by 6 => it must be divisible by both 2 and 3. Statement 1-> As n^2 is divisible by 180 n^2 must have both 2 and 3 as its prime factors. But X and X^p always have the exact same prime factors. Hence n must also have both 2 and 3 as its primes too. Hence n must be divisible by 6=> Sufficient Statement 2-> Using test cases -> n=1=> \frac{144}{n^2} will be an integer=>NO n is not divisible by 6. n=6=> \frac{144}{n^2} will be an integer => YES n is divisible by 6. Hence not Sufficient. Hence A. 62)If a positive integer x has 5 prime factors, how many prime factors does 3x have? A) 5 B) 6 C) 8 D) 15 E) cannot be determined E. Let x=(p1)^a *(p2)^b *(p3)^c *(p4)^d *(p5)^e Where p1,p2,p3,p4,p5 are prime factors of x and a,b,c,d,e are positive integers. 3x=3*(p1)^a *(p2)^b *(p3)^c *(p4)^d *(p5)^e Hence if any one of p1,p2,p3,p4,p5 is 3 => 3x will have 5 prime factors. Else => 3x will have 6 prime factors. Hence 3x may have 5 or 6 prime factors. So there is no unique answer. Hence E. 63)How many positive distinct prime factors does 5^20 + 5^17 have? A) One B) Two C) Three D) Four E) Five D. 5^{17}+5^{20}=> 5^{17}(5^3+1) = 5^{17}*126=> 2*3^2*5^{17}*7 So it has 4 prime factors. Hence D 64)How many prime numbers exist between 200 and 220? (A) None (B) One (C) Two (D) Three (E) Four B. Property-> In order check whether a given number is prime or not => We just need to divide it by all the prime numbers less than or equal to the square root of that number. Highest value in this set-> 220 √220 =14.something Hence we need to check the divisibility with primes less than 14.something => {2,3,5,7,11,13} All primes >2 are odd So to shorten our calculation we just need to check the odd numbers. 201-> Divisible by 3 203->Divisible by 7 205->Divisible by 5 207->Divisible by 3 209->Divisible by 11 211-> PRIME=> Not divisible by any prime numbers in our list. 213-> Divisible by 3 215->Divisible by 5 217->Divisible by 7 219->Divisible by 3 Hence only 211 is a prime number. Hence B. 65)Data Sufficiency->If y is a positive odd integer less than 33,does y have a factor p such that 1 Does y have a factor p such that 1 Is y prime? The Question is just asking us if y prime or not. Statement 1-> y=> {1,4,7,10,13,16,19,22,25,28,31} Hence not sufficient. Statement 2-> y can be -> 7-> Prime 17-> Prime 27-> Composite Hence not sufficient Combing the two statements -> y=7 is the only possible value that satisfies both the statements. Hence y=7 So y is prime Hence C. 66)If x is a positive integer less than 100, for how many values of x is x/6 a prime number? A)2 B)6 C)8 D)13 E)17 B. 67)Data Sufficiency->If n is an integer between 10 and 99, is n n=92=> sum=11 which is prime number. So n is not less than 80. n=12=>sum=3 which is a prime number. So n is less than 80. Hence not sufficient. Statement 2-> Both the digits of n are prime. Allowed tens digit of n->{3,5,7} So n must be of the form 3X or 5X or 7X where X is the units digit of n. So,n can never be greater than 80 as both 8 and 9 are non prime numbers. Hence sufficient. Hence B. Source->Gmat-Prep 68)If n is a positive integer, which of the following statements are correct? A)All the prime numbers greater than 3 can be represented in the form 4n +1 or 4n + 3. B)All numbers of the form 4n + 1 and 4n + 3 are prime numbers. C)All the prime numbers greater than 3 can be represented in the form 6n – 1 or 6n + 1. D)All numbers of the form 6n + 1 and 6n - 1 are prime numbers. A and C are correct. An important property of prime numbers ->Every prime number greater than 3 can be written as=> 6n+1 OR 6n-1 6n+5 OR 6n-5 4n+1 OR 4n-1 4n+3 OR 4n-3 This can be verified taking the first few primes. But alternatively every number of these forms isn't prime. E.g-> 25=6n+1 or 4n+1 and it isn't prime. 69)Data Sufficiency->If p is an integer, is p + 4 a prime number less than 50? (1) p + 1 is a prime number. (2) p is a prime number. C. Note=>2 and 3 are the only consecutive integers that are also prime numbers. 70)How many odd factors does the number 2100 have? A)36 B)24 C)12 D)8 E)cannot be determined. C. 2100=2^2*3*5^2*7 For odd factors => Power of 2 must be zero. Possible cases -> 1*2*3*2=>12 . Hence C. 71)Data Sufficiency->If p is an integer greater than 1, is p a prime number? (1) p is a factor of 13. (2) p is a factor of 78. A. We are told that p is an integer greater than 1 and we are asked if p is a prime number or not. Statement 1-> p is a factor of 13. But 13 has got only 2 factors as it is a prime number. Factors => 1 and 13 so p must be either 1 or 13. But p>1 Hence p must be 13. So p is definitely a prime number. Sufficient. Statement 2-> p is factor of 78. Lets use test cases=> p=2->2 is a factor of 78 and 2 is a prime number. p=78->78 is a factor of 78 and clearly 78 isn't a prime number . Hence not sufficient. Hence A. Source->GMAT-prep 72)If the number 13 completely divides x , and x = a^2 * b , where a and b are distinct prime numbers, which of these numbers must be divisible by 169? A) a^2 B) b^2 C) a*b D) a^2*b^2 E) a^3*b D. As 13 is a prime factor of x => either a or b will be 13. Hence a^b*b^2 will always be divisible by 169 73)Data sufficiency->If p is an integer and p>1,is p a prime number? A)p has p factors. B)p is a factor of 26. A. There are only two numbers for which the number of factors = number itself. They are ->{1,2} As p>1 => p must be 2. So p=2 Clearly p is a prime number. Hence Sufficient. Statement 2->Using test cases p=2-> 2 is a factor of 26 and 2 is a prime number. p=26-> 26 is a factor of 26 and 26 is not a prime number. Clearly not sufficient. Hence A. 74)Data Sufficiency->Is the number of distinct prime factors of the positive integer N more than 4? (1) N is a multiple of 42. (2) N is a multiple of 98 E. 75)Data Sufficiency->If k is a positive integer, how many different prime factors does k have? (1) k/30 is an integer (2) k If k is a positive integer, how many different prime factors does k have? (1) k/60 is an integer (2) k If A and B are positive integers, does A have more prime factors than B? (1) B is the square root of A (2) A ÷ B is an integer A. Property in action-> X and X^n always have the exact same prime factors. 78)What is the greatest prime factor of 12^3 – 96 ? A)2 B)3 C)7 D)17 E)19 D. 12^3-96=2^6*3^3-2^5*3=2^5*3(2*3^2-1)=2^5*3(18-1)=2^5*3*17 Clearly here the greatest prime Factor is 17. 79)What is the greatest prime factor of 4^17 - 2^28? A)2 B)3 C)5 D)7 E)11 D. 80)What is the total number of distinct prime factors of 28980? A)2 B)3 C)4 D)5 E)6 D. 28980=2^2*3^2*5*7*23 Hence it has 5 prime factors. Hence D. 81)What is the greatest prime factor of 3^6 - 1 ? A. 2 B. 3 C. 7 D. 13 E. 17 D. using the identity a^2-b^2 = (a+b)(a-b)=> (3^3)^2 -1^2 = 28*26 = 2^3*7*13 Clearly the greatest prime factor will be 13. 82)Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer? A)1 B)2 C)3 D)4 E)6 E. Method 1-> 180 is a relatively smaller number=> Factors can be seen as follows -> 1*180 2*90 3*60 4*45 5*36 6*30 9*20 10*18 12*15 15*12(The repetition has begun. Hence checking the factors we get => There are 6 factors of the form 4k+2 for k≥0 Method 2-> 180=2^2^3^2*5 Factors of the form -> 4k+2 => 2(2k+1)=> 2*odd number. 2*1 2*3 2*3^2 2*5 2*3*5 2*3^2*5 Hence six values. E. 83)Data Sufficiency->If A and B are positive integers, does B have more prime factors than A? (1) B is the square root of A (2) A ÷ B is an integer D. X and X^n have the exact same primes. So Statement 1 is sufficient. As for statement 2-> If A/B is an integer => Every prime factor of B must be a prime factor of A. Hence B can never have more prime factors than A. Hence Sufficient. Hence D. 84)If N = 10! then what is the minimum number which should be multiplied with N to make it a perfect square? A)7 B)14 C)21 D)28 E)36 A. 10!=1*2*3*4*5*6*7*8*9*10=> 2^8*3^4*5^2*7 In a perfect square the Power of every prime must be even. Thus =>Smallest number to be multiplied will be 7. Hence A. 85)If N = 2^2*3^3*5^5 , find the total number of odd and even factors of N. A)24,96 B)15,30 C)24,48 D)15,96 E)15,48 C. For even factors => Power of 2 must be at-least 1. Hence possible cases => 2*4*6 =>48 odd factors => Power of 2 must be zero. Possible cases =>1*4*6=>24 Hence 48,24 is the correct option. 86)Which of the following numbers is a prime number? A)2343 B)3457 C)4689 D)7731 E)9861 B. Every number except the second one is dividable by 3 so they can never be prime. Hence B must be a prime number. Alternatively prime numbers can be written as either 6n+1 OR 6n-1 Only B can be written as 6n+1 One of the other options can be written as 6n+1 or 6n-1. Hence B is a prime number. 87)The product of all the prime numbers less than 20 is closest to which of the following powers of 10 ? (A) 10^9 (B) 10^8 (C) 10^7 (D) 10^6 (E) 10^5 Product=>2*3*5*7*11*13*17*19=> 10*20*150*300=>9000000=> Close to 1 crore or 10^7 then it is close to 10 lakhs or 10^6 Hence C. Source->Official Guide. 88)If n is the product of the integers from 1 to 8, inclusive, how many different prime factors greater than 1 does n have? (A) four (B) five (C) six (D) seven (E) eight A. Source->Official Guide. 89)Find the number of factors of 240 that are in the form (4*p + 2), where p is a non-negative integer? A)1 B)2 C)3 D)4 E)5 D. 90)Data Sufficiency->What is the value of the integer p ? (1) Each of the integers 2, 3, and 5 is a factor of p. (2) Each of the integers 2, 5, and 7 is a factor of p. E. Basically n can be any integer that has 2,3,5,7 as its prime factors. Eg-> 2*3*5*7 OR 2*3*5*7*13 Etc. Hence E. Source->Official Guide. 91)Data Sufficiency->Does the integer k have at least three different positive prime factors? (1) k/15 is an integer. (2) k/10 is an integer. C. Statement 1->3 and 5 must be the prime factors of K. Statement 2-> 2 and 5 must be the prime factors of K. combing them -> 2,3 and 5 must be the prime factors of K. Hence C. Source->Official Guide. 92)If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18 C. Source->Official Guide. 93)If n = 20! + 17, then n is divisible by which of the following? I. 15 II. 17 III. 19 (A) None (B) I only (C) II only (D) I and II (E) II and II C. Source->Official Guide. 94)If 3 Official Guide. 95)How many integer values are there for x such that 1 (-1.3,4) So possible integer values for x =>-1,0,1,2,3 Thus,five values for x are possible. Hence D. 96)For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301? A. 10,100 B. 20,200 C. 22,650 D. 40,200 E. 45,150 B. Source->Oficial Guide 97)If the sum of the first n positive integers is S, what is the sum of the first n positive even integers, in terms of S ? (A) S/2 (B) S (C) 2S (D) 2S + 2 (E) 4S C. The easiest way here to pick options. Let n=1 Sum=1 Sum of evens = 2 Hence for S=1=> Sum of evens = 2 Only Option C fits that case. NOTE-> Sometimes while using test cases more than one value may satisfy.In that case we must use some additional test cases to arrive at the correct answer. Source->2011 Official Guide. 98)How many prime numbers between 1 and 100 are factors of 7,150 ? (A) One (B) Two (C) Three (D) Four (E) Five D. The question is asking us in a fancy way to get the prime factors of 7150 between 1 and 100. Breaking down 7150=> 2*5^2*11*13 Hence It has 4 prime factors. Hence D. Source-> Official Guide. 99)If n is a prime number between 0 and 100, how many positive divisors does n^3 have? A)1 B)2 C)3 D)4 E)5 D. 100)If y is the smallest positive integer such that 3,150 multiplied by y is the square of an integer, then y must be A)2 B)5 C)6 D)7 E)14 E. Note-> For any perfect square -> Power/exponent of each prime factor must be even. Source-> Official Guide Note-->PM me if you find any answers/solutions inaccurate/inadequate. Any Feedback would be Appreciated.

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STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.

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stonecold

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loserunderachiever

Stone ,

Can you give me a solution of these two ?

Sure!
Let me try ->
Question 1)
Here, just for the ease of calculation -> let us assume that \(25 = a\) and \(10√6= b\) such that \(T = √a+b + √a-b\)

\(T = √a+b + √a-b\)
\(T^2 = a+b + a - b + 2√a+b * √a-b = 2a + 2√a^2-b^2\)

Now, put back the values of a and b
\(T^2 = 50 + 2√625-600\)
\(T^2 = 50 + 2*√25\)
\(T^2 = 50 + 10 = 60\)
\(T = √60\)
\(T = 2√15\)

SMASH that C!

Question 2) Now, This one is tricky. But notice that this is an infinite loop. So there must be a major trick to solve this.
We can actually replace the value in the root symbol after the first 6 with x.
\(x = √(6+x)\)
\(x^2 = 6+x\)
\(x^2-x-6 = 0\)
\(x^2-3x+2x-6=0\)
\(x(x-3)+2(x-3)=0\)
\((x-3)(x+2)=0\)
\(x=3\) or \(x=-2\)

But x cannot be negative since its a root value and a root is always passive. So x≠-2

Hence x = 3
SMASH that B!

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03 Mar 2018, 02:13

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loserunderachiever

Stone ,

Can you give me a solution of these two ?

First question is discussed here: https://gmatclub.com/forum/new-tough-an ... l#p1029216

Second question is discussed here: https://gmatclub.com/forum/new-tough-an ... l#p1029228

PLEASE FOLLOW THE RULES WHEN POSTING QUESTIONS: https://gmatclub.com/forum/rules-for-po ... 33935.html Questions should NOT be posted in this thread they should be posted in respective forums! Thank you.

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Hi stonecold

just a quick ask... this question is from Mock 1 and I am just wondering if the answer is correct. Please let me know as I am getting 270 total factors.
2^8*3^4*5^2*7^1

so basically = (8+1)(4+1)(2+1)(1+1)

48)If n is the product of all the integers from 1 to 10 exclusive,how many factors does n have?
[Obscure] Spoiler:
160.

stonecold

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Bobzi

Hi,
Your solution is incorrect.

See ->

\(2*3*4*5*6*7*8*9= 2^7*3^4*5*7\)

=> # of factors = \(8*5*2*2 = 160\)

Your mistake -> You did not read the question properly. Notice the word EXCLUSIVE. You must EXCLUDE 1 and 10.

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09 Mar 2018, 06:07

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This is a gold mine of information!

stonecold

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DebUSA

This is a gold mine of information!

I am glad you found it useful. I will be updating the tests soon.

All the best

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14)If p is a positive integer and 200 multiplies by p is square of an integer,what is the value of p?
A)2
B)3
C)5
D)10
E)Cannot be determined.

HI stonecold,

\(x^2 = p*200\) In this question why can't p =2 ?

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Firstly, thanks for having these questions - they are a great help!

Wanted to ask, though, about the question below:

7)What is the units digit of 23^99∗14^352+9002^1003∗918^437986

I keep getting 7 * 1 + 8 * 4 = 9. I'm using cyclicity here. Thanks in advanced.

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wchin24

Hey wchin24 ,

The mistake you did is highlighted above.

14^352

The pattern of 4 is 4,6

This means cyclicity is 2.

When you divide the power with 2, you will get 0 remainder. That means last digit is last place of the pattern, which is 6 here.

Hence, the last digit of 14^352 is 6.

Does that make sense?

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27 May 2018, 01:16

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abhimahna

wchin24

OMG YES! Thank you! I was tripping up since I was seeing it as 4^0, but in reality this just means that it'll end up at the last place of the pattern here.

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NandishSS

Hi, You are right in saying that p CAN be 2. But you missing the big picture here.
Ask yourself, is that the only possible value of p?
What if p is \(2*5^2\) or \(2*11^2\) or \(2*101^2\) ?

You see, there are various value of the variable p that are possible in thus question. Hence the OA is E.

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121)Data Sufficiency->How many divisors does the positive integer N have?

(1)27N^3 has 16 factors.
(2)90<N^3<200

as explained, how can we say that N is prime from Statement 1. N could be 3^4 as well. If N= 3^4 then 27N^3 = 3^3*((3^4)^3)= 3^3*3^12= 3^15 Hence Factors will be 15+1=16. And Nos of Divisor of N = 3^4 is Five. Not 2 as in the case if N is prime.

Please explain if anything wrong in my understanding

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Hi stonecold

For question 18 It looks like answer should be C and not E as only possible option after combining both the statement is P = 3, Could you please advise if I am missing something.

chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmat

stonecold

18)Data Sufficiency->What is the value of positive integer p?
A)300 multiplied by p is square of an integer.
B)p is a factor of 75

Spoiler: ::
E.
Combing the two statements -> p can be 3 or 3*5^2

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stonecold

18)Data Sufficiency->What is the value of positive integer p?
A)300 multiplied by p is square of an integer.
B)p is a factor of 75

Spoiler: ::
E.
Combing the two statements -> p can be 3 or 3*5^2

No, both 3 and 75 will fit in..
A)300 multiplied by p is square of an integer...300*3=900=30^2 and 300*75=22500=150^2
B)p is a factor of 75... Factors of 75 are 1,3,5,15,25,75, so both 3 and 75 fit in

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Hi chetan2u, Bunuel, VeritasKarishma, Gladiator59, generis

For question 23 statement 7-8

I am only able to find that "All the prime numbers greater than 3 can be written as either 6n+1 or 6n-1." Not able to find anything for 6+1 and 4n+1 or 4n-1.

Could you please review and advise if I am missing something.

stonecold

23)Which of the following statements must be true->
1)A prime number must be positive.
2)For any prime number p,there is no x such that 1<x<p and x is a divisor of p.
3)The product of first ten primes is even.
4)All prime numbers greater than 71 are odd.
5)2 and 3 are the only consecutive integers that are also prime numbers.
6)p is a prime number and x and y are positive integers.If p=x*y then one out of x or y must be one.
7)All the prime numbers greater than 3 can be written as either 4n+1 or 4n-1.
8)All the prime numbers greater than 3 can be written as either 6+1 or 6n-1.

Spoiler: ::
All statements are true.

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stonecold

Hi..

All prime numbers have to be of type of 6n+1 or 6n-1

Now 4n+1 and 4n-1 is nothing but set of all odd numbers..
When n=1, 4n+1 and 4n-1 becomes 3 and 5 and when n=2, 4n+1 and 4n-1 becomes 7 and 9..so 3,5,7,9,11,...
And all primes above 2 are odd numbers..so primes are also of type 4n+1 or 4n-1, basically they will be odd

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Thanks chetan2u for valuable response. It helped me

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As perfectly explained by chetan2u Sir,

All prime numbers greater than 3 can be written as 6n+/-1 or 4n+/-1.

But vice versa is not true
for example, for n =4, 6*4+1= 25 , but it is not prime
Similarly, for n = 2, 4n+1 = 2*4+1= 9, which is not prime

In fact, there is no direct formula, which can find whether a number is prime or not.

All prime numbers greater than 3 can be written as 6n+/-1 or 4n+/-1, but all numbers in form of 6n+/-1 or 4n+/-1 are not prime.

Gmatprep550

stonecold

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@14)If p is a positive integer and 200 multiplies by p is square of an integer,what is the value of p?
A)2
B)3
C)5
D)10
E)Cannot be determined.

Hi, can you explain, how have you written p=2*2^2 expression in the first set of questions, numbered 15-17..
What does this expression mean? i am confused in this and the DS questions that contain 300 and 200 too

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where is the link of mock test 1 and 2?

1 2 3 4 5

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