February 18, 2019 February 18, 2019 10:00 PM PST 11:00 PM PST We don’t care what your relationship status this year  we love you just the way you are. AND we want you to crush the GMAT! February 18, 2019 February 18, 2019 10:00 PM PST 11:00 PM PST Buy "AllInOne Standard ($149)", get free Daily quiz (2 mon). Coupon code : SPECIAL
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 299

In a sequence of 13 consecutive integers, all of which are
[#permalink]
Show Tags
Updated on: 02 Jul 2013, 13:05
Question Stats:
30% (02:46) correct 70% (03:00) wrong based on 138 sessions
HideShow timer Statistics
In a sequence of 13 consecutive integers, all of which are less than 100, there are exactly three multiples of 6. How many integers in the sequence are prime? (1) Both of the multiples of 5 also in the sequence are multiples of either 2 or 3. (2) Only one of the two multiples of 7 also in the sequence is not a multiple of 2 or 3.
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by dimitri92 on 18 May 2010, 07:31.
Last edited by Bunuel on 02 Jul 2013, 13:05, edited 1 time in total.
Edited the OA.




Math Expert
Joined: 02 Sep 2009
Posts: 52938

Re: 13 consecutive integers
[#permalink]
Show Tags
18 May 2010, 08:32
dimitri92 wrote: Is there a faster way to solve this !!!
In a sequence of 13 consecutive integers, all of Which are less than 100, there are exactly three multiples of 6. How many integers in the sequence are prime?
(A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3. (B) Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3. Straight E. Remember only positive integers can be primes, hence if all 13 numbers in the sequence are negative there will be 0 primes but if sequence is positive, it will contain primes, for instance sequence: {6, 7, ..., 18} contains 4 primes (7, 11, 13, 17). Not a good question.
_________________
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Joined: 25 Jun 2009
Posts: 276

Re: 13 consecutive integers
[#permalink]
Show Tags
18 May 2010, 08:35
dimitri92 wrote: Is there a faster way to solve this !!!
In a sequence of 13 consecutive integers, all of Which are less than 100, there are exactly three multiples of 6. How many integers in the sequence are prime?
(A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3. (B) Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3. Bunuel has made a good point but for those who want to tets it out here is one lengthy explanation. First point to note, 13 consecutive integers and there are 3 multiples of 6 that means first number in the sequence and the last number in the sequnce will be multiples of 6 ( for e.g 6, 7, 8, .....18 or 30, 31, 32, ... 42)so total permissible sequnces : 012, 618, 1224, 1830, 2436, 3042, 3648, 4254, 4860, 5466, 6072, 6678, 7284, 7890, 8496Also in any sequence the sequence will be of form, 6x, 6x+1 , 6x +2 ...... 6x + 12. Out of these only 6 numbers will be even and 7 numbers will be odd. and Now, Lets take a look at the statement. St 1. Both of the multiples of 5 are either a multiple of 2 or 3 which means we ignore the sequnce starting from 012, and any sequence in which we have multiples of 5 with prime numbers like 5*7 = 35 so sequence starting from 2436 and 3042 has to be ignored also we need to ignore the sequences in which we have 3 multiples of 5 like 1830, 2436, 3042, 4860, 5466, 6072, 7890, 8496So now we are left with these sequences 618, 1224, 3648, 4254, 6678, 7284,Now the first sequnce 618 has 4 prime numbers and 4254 has 3 prime numbers hence insufficient. St 2 Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3
Which means multiples of 7 like 0*7, 1*7, 5*7, 7*7 are the possible values
these values lie in the sequences of 012, 618, 2436, 3042, 4254, 4860
Also 012 is not valid in this case because it doesnt havethe second multiple of 7
So we are left with 618, 2436, 3042, 4254, 4860again , Insufficient same reasoning as St1 now St 1+ 2, We are only left with 618, 4254, Again insufficient.



Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 299

Re: 13 consecutive integers
[#permalink]
Show Tags
19 May 2010, 18:50
nitishmahajan wrote: dimitri92 wrote: Is there a faster way to solve this !!!
In a sequence of 13 consecutive integers, all of Which are less than 100, there are exactly three multiples of 6. How many integers in the sequence are prime?
(A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3. (B) Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3. Bunuel has made a good point but for those who want to tets it out here is one lengthy explanation. First point to note, 13 consecutive integers and there are 3 multiples of 6 that means first number in the sequence and the last number in the sequnce will be multiples of 6 ( for e.g 6, 7, 8, .....18 or 30, 31, 32, ... 42)so total permissible sequnces : 012, 618, 1224, 1830, 2436, 3042, 3648, 4254, 4860, 5466, 6072, 6678, 7284, 7890, 8496Also in any sequence the sequence will be of form, 6x, 6x+1 , 6x +2 ...... 6x + 12. Out of these only 6 numbers will be even and 7 numbers will be odd. and Now, Lets take a look at the statement. St 1. Both of the multiples of 5 are either a multiple of 2 or 3 which means we ignore the sequnce starting from 012, and any sequence in which we have multiples of 5 with prime numbers like 5*7 = 35 so sequence starting from 2436 and 3042 has to be ignored also we need to ignore the sequences in which we have 3 multiples of 5 like 1830, 2436, 3042, 4860, 5466, 6072, 7890, 8496So now we are left with these sequences 618, 1224, 3648, 4254, 6678, 7284,Now the first sequnce 618 has 4 prime numbers and 4254 has 3 prime numbers hence insufficient. St 2 Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3
Which means multiples of 7 like 0*7, 1*7, 5*7, 7*7 are the possible values
these values lie in the sequences of 012, 618, 2436, 3042, 4254, 4860
Also 012 is not valid in this case because it doesnt havethe second multiple of 7
So we are left with 618, 2436, 3042, 4254, 4860again , Insufficient same reasoning as St1 now St 1+ 2, We are only left with 618, 4254, Again insufficient. see this is what im trying to avoid ...i want an answer which is simpler and faster ....i also arrived at the answer but such a lengthy method is just going to drag me down on gmat



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2577
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: 13 consecutive integers
[#permalink]
Show Tags
13 Jun 2010, 12:49
Lets assume they are +ve. First point to note, 13 consecutive integers and there are 3 multiples of 6 that means first number in the sequence and the last number in the sequnce will be multiples of 6 ( for e.g 6, 7, 8, .....18 or 30, 31, 32, ... 42) Now any number less than 100 will have its sqrt < 10. Only prime numbers less than 10 are 2,3,5,7, If the number is not divisible by any of them then it is prime. This sequence starts and ends with multiple of 6. Since it has 7 even and 6 odd numbers => 137 = 6 are left to be checked for prime condition. (A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3.[/color] In this case 5 is either multiple of 2 or 3 => numbers such as 25,35 wont be included=> the number of 5 will not be simultaneously multiple of 2 and 3. Now between 6k and 6k+12 exclusive there are 3 numbers divisible by 3. as per given condition one of them will be divisible by 5 and other by 2( Between 6k and 6k+12 there will be one number divisible by 6). So we can subtract the count of the third number and that of multiple of 5 i.e. divisible by 3. So total left = 62 = 4. We are left with checking of 7. for numbers >7 it will always be multiple of any other number so prime numbers = 41 =3 If one of the number =7 then the number of prime numbers will remain 4...so not sufficient. (B) Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3.Again starting with the checking of 6 odd numbers. as per the given condition one of the multiple of 7 has to be 7, 49 , 63,77 , 35 etc which is not prime. clearly if one of them is 7 and other is any one of 35,49,63 the number of primes will differ by 1. So not sufficient. If we combine both we have a common case of having 7 or not. Thus E.
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Intern
Joined: 01 Mar 2011
Posts: 5

Re: 13 consecutive integers
[#permalink]
Show Tags
13 Mar 2011, 08:41
Bunuel wrote: dimitri92 wrote: Is there a faster way to solve this !!!
In a sequence of 13 consecutive integers, all of Which are less than 100, there are exactly three multiples of 6. How many integers in the sequence are prime?
(A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3. (B) Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3. Straight E. Remember only positive integers can be primes, hence if all 13 numbers in the sequence are negative there will be 0 primes but if sequence is positive, it will contain primes, for instance sequence: {6, 7, ..., 18} contains 4 primes (7, 11, 13, 17). Not a good question. The official answer is C with the following explination: This one is a group problem in disguise. First, there’s a lot of information buried in the question. If there are exactly three multiples of 6 in the sequence, then the sequence must start with a multiple of 6. So, there will always be 7 multiples of 2 and 5 multiples of 3 in the sequence. Those are two of the groups in the group formula. The total is, of course, 13. We’ll also need to subtract the numbers that are both multiples of 2 and 3 (aka the multiples of 6) from the total. That means that the statements need to tell us about the multiples of 5 and 7. Everything else in the sequence (neither in the group formula) will be prime. Statement (1) tells us that we don’t need to count either of the multiples of 5 since they have already been counted but it tells us nothing about the multiples of 7 in the sequence. Hence, BCE. Statement (2) tells us to count one of the multiples of 7 but tells us nothing about the multiples of 5. Cross off B. Finally, when we put the statements together we know that 13 = 7(the multiples of 2) + 5 (the multiples of 3) + 2 (the multiples of 5) + 2 (the multiples of 7) − 3 (the multiples of 6) − 2 (the multiples of 5 and 2 or 3) − 1 (the multiples of 7 and 2 or 3) + Primes. Hence, there are 3 primes in the sequence. The correct answer is C. I got E from sequencies 618 4 primes, 6678 3 primes. I guess it's a fluke then in Princeton CAT, what does everyone think?



Math Expert
Joined: 02 Sep 2009
Posts: 52938

Re: 13 consecutive integers
[#permalink]
Show Tags
13 Mar 2011, 09:00
abogomolov wrote: Bunuel wrote: dimitri92 wrote: Is there a faster way to solve this !!!
In a sequence of 13 consecutive integers, all of Which are less than 100, there are exactly three multiples of 6. How many integers in the sequence are prime?
(A) Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3. (B) Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3. Straight E. Remember only positive integers can be primes, hence if all 13 numbers in the sequence are negative there will be 0 primes but if sequence is positive, it will contain primes, for instance sequence: {6, 7, ..., 18} contains 4 primes (7, 11, 13, 17). Not a good question. The official answer is C with the following explination: This one is a group problem in disguise. First, there’s a lot of information buried in the question. If there are exactly three multiples of 6 in the sequence, then the sequence must start with a multiple of 6. So, there will always be 7 multiples of 2 and 5 multiples of 3 in the sequence. Those are two of the groups in the group formula. The total is, of course, 13. We’ll also need to subtract the numbers that are both multiples of 2 and 3 (aka the multiples of 6) from the total. That means that the statements need to tell us about the multiples of 5 and 7. Everything else in the sequence (neither in the group formula) will be prime. Statement (1) tells us that we don’t need to count either of the multiples of 5 since they have already been counted but it tells us nothing about the multiples of 7 in the sequence. Hence, BCE. Statement (2) tells us to count one of the multiples of 7 but tells us nothing about the multiples of 5. Cross off B. Finally, when we put the statements together we know that 13 = 7(the multiples of 2) + 5 (the multiples of 3) + 2 (the multiples of 5) + 2 (the multiples of 7) − 3 (the multiples of 6) − 2 (the multiples of 5 and 2 or 3) − 1 (the multiples of 7 and 2 or 3) + Primes. Hence, there are 3 primes in the sequence. The correct answer is C. I got E from sequencies 618 4 primes, 6678 3 primes. I guess it's a fluke then in Princeton CAT, what does everyone think? Answer to this question is E, not C (no matter what the OA says). It's not a good question though so I wouldn't worry about it at all.
_________________
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1338

Re: 13 consecutive integers
[#permalink]
Show Tags
13 Mar 2011, 09:09
abogomolov wrote: The official answer is C with the following explination:
This one is a group problem in disguise. First, there’s a lot of information buried in the question. If there are exactly three multiples of 6 in the sequence, then the sequence must start with a multiple of 6. So, there will always be 7 multiples of 2 and 5 multiples of 3 in the sequence. Those are two of the groups in the group formula. The total is, of course, 13. We’ll also need to subtract the numbers that are both multiples of 2 and 3 (aka the multiples of 6) from the total. That means that the statements need to tell us about the multiples of 5 and 7. Everything else in the sequence (neither in the group formula) will be prime. Statement (1) tells us that we don’t need to count either of the multiples of 5 since they have already been counted but it tells us nothing about the multiples of 7 in the sequence. Hence, BCE. Statement (2) tells us to count one of the multiples of 7 but tells us nothing about the multiples of 5. Cross off B. Finally, when we put the statements together we know that 13 = 7(the multiples of 2) + 5 (the multiples of 3) + 2 (the multiples of 5) + 2 (the multiples of 7) − 3 (the multiples of 6) − 2 (the multiples of 5 and 2 or 3) − 1 (the multiples of 7 and 2 or 3) + Primes. Hence, there are 3 primes in the sequence. The correct answer is C.
I got E from sequencies 618 4 primes, 6678 3 primes. I guess it's a fluke then in Princeton CAT, what does everyone think?
Yes, that solution is dead wrong. They seem to be assuming that a number which is a multiple of 7 cannot be prime. Well, there is one multiple of 7 which is prime: 7 itself. So those above who have chosen E are correct, and the OA is not. Regardless, it's a horrible question, since any reasonable approach is far too timeconsuming.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 504

Re: 13 consecutive integers
[#permalink]
Show Tags
13 Mar 2011, 09:59



Manager
Joined: 05 Jul 2010
Posts: 139

Re: 13 consecutive integers
[#permalink]
Show Tags
13 Mar 2011, 19:28
great explanation!



Manager
Joined: 18 Oct 2010
Posts: 70

Re: 13 consecutive integers
[#permalink]
Show Tags
14 Mar 2011, 08:47
to be honest this question is so weird to me, or i didnt understand the whole question. the question is that there are 13 consecutive numbers : a , a+1,a+2...a+12. all these number is <100 1. Both of the multiples of 5 Also in the sequence are multiples of either 2 or 3. it means there are 2 numbers x and y which can divide by 5, 2, and 3 so those number is only 30 and 60 and the distance between 30 and 60 are so far, this is impossible. 2 numbers x and y can divide 5 and 2 so the number could be 10 and 20 and so on. start a =10 so we can find 5 prime numbers 2 numbers x and y can divide 3 and 5 so it could be 15 and 30 and so on start a= 15 so we find only 4 prime number ( this is crazy) insufficient.
statements 2. Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3. numbers can divide 7 is 7,14, and so on,,, the same like statement 1 hmhm it took me 5 mins to think of this solution and yuck! this question is tough



Senior Manager
Joined: 02 Apr 2014
Posts: 476
Location: India
GPA: 3.5

In a sequence of 13 consecutive integers, all of which are
[#permalink]
Show Tags
22 Sep 2017, 00:44
Yes as Bunuel pointed out, straight E, if no mention of negative or positive integers.
However, let us assume that question mentions that all are positive integers and try to solve.
It is given there are 3 exact multiples of 6 => so 1st , 7th and 13th number is multiple of 6.
Let the starting number be 'a', so sequence is a, a+1, a+2, ....... a+12
Statement 1: Both of the multiples of 5 also in the sequence are multiples of either 2 or 3.
it implies there are only two multiples of 5. so let us examine the positions of multiples of 5. if a is multiple of 5 => (a+5), (a+10) also multiple of 5 => then 3 multiples of 3, only 2 multiples are mentioned, so a not multiple of 5. if a + 1 is multiple of 5 => (a+6), (a+11) also multiple of 5 => then 3 multiples of 3, only 2 multiples are mentioned, so a+1 not multiple of 5. if a + 2 is multiple of 5 => (a+7), (a+12) also multiple of 5 => then 3 multiples of 3, only 2 multiples are mentioned, so a+2 not multiple of 5.
so (a+3) or (a+4) is multiple of 5
Let a + 3 be multiple of 5, then given a is multiple of 6, and a+3 is multiple of 5, one such seq is : (12,13,14,15,...18....20......24), also 20 multiple of 5 and 2, 15 multiple of 5 and 3, as satisfying the statement.
Now next such sequence will occur at 12 + LCM(5,6) => (42......45...48.....50......54) 4 prime numbers 12 + 2 * LCM(5,6) => (72.....75...78....80.....84), 4 prime numbers and stops here as, next seq (12 + 3 * LCM(5,6)) > 100
Now Let a+4 be multiple of 5, so one such sequence is (6......10,.12..15......18)  4 prime numbers Next sequences 6 + LCM(5,6) => (36 ..... 40..42...45....48)  4 prime numbers 6 + 2 * LCM(5,6) => (66 ..... 70..72...75....78)  3 prime numbers and stops here.
So from above, we can see that 3 or 4 prime numbers in the seq satisfying this statement  Not Suff
Statement 2: Only one of the two multiples of 7 Also in the sequence is not a multiple of 2 or 3.
One such sequence is (6,7....12....14....18)  only one multiple of 7 and 2  4 prime numbers Next sequence: 6 + LCM(6,7) > (48, 49...54....56...60)  only one multiple of 7 and 2  2 prime numbers and this pattern stops here.
Another sequence: (24...28..30...35...36)  only one multiple of 7 and 2  2 prime numbers Next sequence 24 + LCM(6,7) > (66,...70,..72,....77..78)  only one multiple of 7 and 2  3 prime numbers
From above, statement 2 cleary insuff as there 2,3 or 4 prime numbers.
Now combining both the statements, common sequences (6,....12....18)  4 prime numbers (66,...72...78)  3 prime numbers.
Still Insuff, So answer E



Intern
Joined: 01 Jan 2019
Posts: 23

Re: In a sequence of 13 consecutive integers, all of which are
[#permalink]
Show Tags
11 Feb 2019, 09:07
If there are exactly 3 multiples of 6 in a sequence of 13 consecutive integers, then the sequence must begin and end with a multiple of 6. So, the first number in the least possible sequence is the number 6. Because all of the integers in the sequence are less than 100, the first number in the greatest possible sequence is 84. Now, determine "What is needed". To count how many integers in the sequence are prime, the statements need to limit the possible sequences to either one sequence, or multiple sequences that contain the same number of prime numbers. Evaluate the statements one at a time. Evaluate Statement (1). Because both of the multiples of 5 in the sequence are also multiples of either 2 or 3, there must be only two multiples of 5 in the sequence. Plug In values that satisfy this statement. If the sequence is 6,7,8,9,10,11,12,13,14,15,16,17,18, then there are 4 prime numbers: 7, 11, 13, and 17. However, if the sequence is 66,67,68,69,70,71,72,73,74,75,76,77,78, then there are 3 prime numbers: 67, 71, and 73. When different numbers that satisfy a statement yield different answers to the question, the statement is insufficient. So, write down BCE. Now, evaluate Statement (2). Because only one of the two multiples of 7 in the sequence is not also a multiple of 2 or 3, one of the two multiples of 7 in the sequence is also a multiple of 2 or 3. Plug In values that satisfy this statement. If the sequence is 6,7,8,9,10,11,12,13,14,15,16,17,18, then there are 4 prime numbers: 7, 11, 13, and 17. However, if the sequence is 66,67,68,69,70,71,72,73,74,75,76,77,78, then there are 3 prime numbers: 67, 71, and 73. When different numbers that satisfy a statement yield different answers to the question, the statement is insufficient. Eliminate choice B. Now evaluate both statements together. Reuse the numbers that were Plugged In for statements (1) and (2) that satisfy both statements. If the sequence is 6,7,8,9,10,11,12,13,14,15,16,17,18, then there are 4 prime numbers. However, if the sequence is 66,67,68,69,70,71,72,73,74,75,76,77,78, then there are 3 prime numbers. When different numbers that satisfy both statements yield different answers to the question, the statements together are insufficient. Eliminate choice C. The correct answer is choice E.
_________________
_________________ Regards K Shrikanth
____________ Please appreciate the efforts by pressing +1 KUDOS (:




Re: In a sequence of 13 consecutive integers, all of which are
[#permalink]
11 Feb 2019, 09:07






