GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 26 Sep 2018, 01:52

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

The sum of 4 consecutive two-digit odd integers, when divided by 10,

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
B
Joined: 31 May 2017
Posts: 29
Location: India
CAT Tests
The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 30 Oct 2017, 08:08
13
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

50% (02:56) correct 50% (02:30) wrong based on 151 sessions

HideShow timer Statistics

The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73
Most Helpful Community Reply
VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1091
The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 21 Jul 2018, 20:07
8
3
swatygi wrote:
The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73


let sequence=x, x+2, x+4, x+6
sum of sequence=4x+12
if 4x+12 is a multiple of 10,
then x must have a units digit of 7
4*17+12=80; 80/10=8 no
4*27+12=120; 120/10=12 no
4*37+12=160; 160/10=16 yes
sequence=37, 39, 41, 43
41
C
General Discussion
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6814
The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 30 Oct 2017, 08:29
3
1
swatygi wrote:
The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67

Can someone please provide me the solution to this question.



the answer depends on properties of number...

first there are five odd digits... 1,3,5,7,9..
SUm of which 4 consecutive gives you multiple of 10.... it is 7+9+1+3=20..
so the numbers will be TWo with one lower TENS digit - X7 and X9 and two with one higher TENS digit - (X+1)1 and (X+1)3
so BASICALLY we are adding SUM of tens digit {x+x+(x+1)+(x+1)}*10
SuM of units digit is 7+9+1+3=20... so when you add this to TENS sum, we get \((x+x+(x+1)+(x+1))*10 + 20 = 10x+10x+10(x+1)+10(x+1) +20=10x+10+10x+10+10(x+1)+10(x+1) = 10(x+1)+10(x+1)+10(x+1)+10(x+1)=40(x+1)\)
so \(\frac{40*(x+1)}{10} = 4*(x+1)\) should be PERFECT SQUARE..
so x+1 can be 4 as 4*4 will be perfect square..
so numbers are 37,39,41,43

ans C
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Intern
Intern
avatar
B
Joined: 15 Aug 2012
Posts: 42
Schools: AGSM '19
GMAT ToolKit User
The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 19 Jul 2018, 20:34
3
I tried the following method which is similar to chetan2u 's solution but maybe takes a little longer.

Let's say the 4 consecutive odd integers are 2k+1, 2k+3, 2k+5 & 2k+7.

It says [(2k+1)+(2k+3)+(2k+5)+(2k+7)]/10 = a^2x*b^2x which is in the form of a perfect square.

> (8k+16)/10=a^2x*b^2y
> 8(K+2)/10=a^2x*b^2y
> 4(K+2)/5=a^2x*b^2y

The first number at which 4(k+2)/5 is a perfect square is when (K+2)/5=4.
This means K=18

So our numbers are 2(18)+1, 2(18)+3, 2(18)+5 & 2(18)+7 = 37,39,41 & 43.


Don't forget the kudos if this helped you.
Director
Director
User avatar
P
Joined: 09 Mar 2016
Posts: 884
The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post Updated on: 24 Jul 2018, 09:00
swatygi wrote:
The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73



pushpitkc, Bunuel

my reasoning was following

say the first three consecutive two-digit odd integers are 11, 13, 15 so if you sum up these numbers we get 39 now i need to try answer choices to find the 4th odd number ...OA is C so if add up 39+41=80 ---> 80/10 = 8 and that is not a perfect square.

i think something wrong is with this question :)

hello generis, my last hope :) why everyone starts counting one digit numbers ? do you have your unique constructivelly structured approach to this question ? :angel: :-) i switched to number properties questions and i think i dont have a good "number sense" though yes i know divisibility rules, i know what is factor and what is multiple ... Factors are what we can multiply to get the number. Multiples are what we get after multiplying the number by an integer etc :) Shine on me crazy diamond :lol: i like begining of the long version of this song :grin: :-)
_________________

In English I speak with a dictionary, and with people I am shy.


Originally posted by dave13 on 22 Jul 2018, 05:14.
Last edited by dave13 on 24 Jul 2018, 09:00, edited 1 time in total.
Intern
Intern
avatar
B
Status: Applying
Affiliations: test
Joined: 16 Apr 2012
Posts: 33
Location: India
Yawer: Yawer
Concentration: Marketing, Technology
GMAT 1: 200 Q33 V33
WE: Consulting (Internet and New Media)
Re: The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 22 Jul 2018, 09:32
rajudantuluri wrote:
I tried the following method which is similar to chetan2u 's solution but maybe takes a little longer.

Let's say the 4 consecutive odd integers are 2k+1, 2k+3, 2k+5 & 2k+7.

It says [(2k+1)+(2k+3)+(2k+5)+(2k+7)]/10 = a^2x*b^2x which is in the form of a perfect square.

> (8k+16)/10=a^2x*b^2y
> 8(K+2)/10=a^2x*b^2y
> 4(K+2)/5=a^2x*b^2y

The first number at which 4(k+2)/5 is a perfect square is when (K+2)/5=4.
This means K=18

So our numbers are 2(18)+1, 2(18)+3, 2(18)+5 & 2(18)+7 = 37,39,41 & 43.


Don't forget the kudos if this helped you.


This is a very well thought out approach. Can solve it in a minute .It took me 4 minutes with mine, I got lost in between.
Intern
Intern
avatar
B
Joined: 05 Jun 2018
Posts: 4
CAT Tests
Re: The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 23 Jul 2018, 23:34
gracie wrote:
swatygi wrote:
The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73


let sequence=x, x+2, x+4, x+6
sum of sequence=4x+12
if 4x+12 is a multiple of 10,
thenx must have a units digit of 7
4*17+12=80; 80/10=8 no
4*27+12=120; 120/10=12 no
4*37+12=160; 160/10=16 yes
sequence=37, 39, 41, 43
41
C


Could you please explain the highlighted part?
VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1091
Re: The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 24 Jul 2018, 00:36
1
Swatipupi wrote:
gracie wrote:
swatygi wrote:
The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73


let sequence=x, x+2, x+4, x+6
sum of sequence=4x+12
if 4x+12 is a multiple of 10,
thenx must have a units digit of 7
4*17+12=80; 80/10=8 no
4*27+12=120; 120/10=12 no
4*37+12=160; 160/10=16 yes
sequence=37, 39, 41, 43
41
C


Could you please explain the highlighted part?


hi Swatipupi,
in order for 4x+12 to be a multiple of 10,
then 4x must have a units digit of 8
because x is odd, only a units digit of 7 will produce this
I hope this helps,
gracie
Intern
Intern
avatar
B
Joined: 05 Jun 2018
Posts: 4
CAT Tests
Re: The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 24 Jul 2018, 00:52
Thank you Gracie!! Silly me!

Posted from my mobile device
Intern
Intern
avatar
B
Joined: 17 Jul 2018
Posts: 1
Re: The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 24 Jul 2018, 01:27
Multiply each option choice by 4 to reach close to the sum of the consequtive numbers, than using approximation choose what number upon diving by 10 could possibly be a perfect square. Evaluate the answer choice & you're good to go.

Posted from my mobile device
Intern
Intern
User avatar
B
Joined: 07 Jun 2018
Posts: 20
Location: India
Concentration: Marketing, International Business
GPA: 3.9
WE: Marketing (Insurance)
GMAT ToolKit User
The sum of 4 consecutive two-digit odd integers, when divided by 10,  [#permalink]

Show Tags

New post 29 Aug 2018, 02:18
swatygi wrote:
The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73


I solved it by taking a range of min and max for the sum of 4 two-digit consecutive odd nos, i.e (11+13+15+17 = 10 * 4 + 16 = 56) and (99+97+95+93 = 400 - 16 = 384). Dividing the upper and lower range by 10, the perfect square value should be between 5.6 and 38.4. Means, 9,16,25 and 36.
From the sum of four no, the mean of 4 no. can be found which would be 22.5, 40, 62.5 and 90.
As the four no.s are consecutive odds, the mean would be even, i.e 40 and 90. So, the no. would be any from these 37,39,41,43,87,89, 91 and 93.
From the options, only (C) 41 satisfy.

Interestingly, while solving I read the no. as consecutive rather than consecutive odd. Hence, marked 21 as for regular consecutive no.s the avg would be either 22.5 or 62.5 and only 21 satisfies it.
_________________

Give a Kudo, If I deserve it.



Target 750+
  • 17th June 2018 - Diagnostic Test - Kaplan Premier 2015 - 570
  • 10th Aug 2018 - Veritas 1 - 650 (Q51,V29)
  • 17th Aug 2018 - Veritas 2 - 670 (Q49,V33)
  • 24th Aug 2018 - Veritas 3 - 650 (Q49,V30)
  • 1st Sep 2018 - Veritas 4 - 680 (Q51,V33)
  • 12th Aug 2018 - Veritas 5 - 690 (Q51,V34)

What do you think? Will I achieve my target score?

GMAT Club Bot
The sum of 4 consecutive two-digit odd integers, when divided by 10, &nbs [#permalink] 29 Aug 2018, 02:18
Display posts from previous: Sort by

The sum of 4 consecutive two-digit odd integers, when divided by 10,

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.