Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
26 Mar 2019, 23:31
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
50% (01:58) correct
50%
(02:39)
wrong
based on 20
sessions
History
Date
Time
Result
Not Attempted Yet
Let a, b, and c be digits with a ≠ 0. The three-digit integer abc lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer acb lies two thirds of the way between the same two squares. What is a+b+c?
(A) 10
(B) 13
(C) 16
(D) 18
(E) 21
(A) 10
(B) 13
(C) 16
(D) 18
(E) 21
shuvodip04
Joined: 05 Oct 2017
Last visit: 03 Mar 2022
Posts: 86
Given Kudos: 103
Location: India
Concentration: Finance, International Business
Schools: IIMB IIMA ISB'22 (S) IIMA PGPX'22 (I)
GMAT 1: 660 Q48 V33
GMAT 2: 700 Q49 V35 (Online)
GPA: 4
WE:Analyst (Energy)
27 Mar 2019, 02:18
Kudos
Bookmarks
Bunuel
As per the question ,
(100a+10c+b) - (100a + 10b + c) = 9(c-b)
From the below diagram we can see that 9(c-b) is 1/3 of the difference between the two consecutive squares.
i.e \(((x+1)^2 - x^2)/3\) = 9(c-b)
=> \(((x+1)^2 - x^2)\)= 27(c-b)...(1)
Now is x is odd then x+1 will be even ,so will be their squares. Similarly if x is even then x+1 will be odd.
So sum of their squares will odd+even or even+odd i.e a odd number
For 27(c-b) to be odd . c-b = 1,3,5,6
For c-b =1 Equation 1 becomes \(((x+1)^2 - x^2)\)=27 => 2x + 1 = 27 => x = 13
So the two squares will be 13^2 and 14^2 i.e 169 and 196
For c-b =3 Equation 1 becomes \(((x+1)^2 - x^2)\)=81 => 2x + 1 = 81 => x = 40
So the two squares will be 40^2 and 41^2 i.e 1600 and 1681 . But here the numbers are greater than 3 digits.
So c-b=1 is only possible i.e x=13
so integer abc =13^2 + (14^2 - 13^2)/3 = 169+ (196 - 169)/3 = 169 +27/3 = 178
So a+b+c = 1 + 7 + 8 = 16. Answer (C).
Attachments
math.jpg [ 6.46 KiB | Viewed 4540 times ]
29 Mar 2019, 07:44
Kudos
Bookmarks
Bunuel
The value of the integer abc is 100a + 10b + c and that of acb is 100a + 10c + b. Since abc is ⅓ of the way between two consecutive perfect squares (say n^2 and (n + 1)^2) and acb is ⅔ of the way between the same two perfect squares, the distance (or difference) between abc and acb is ⅓ of the way between the two perfect squares. So we have:
(100a + 10c + b) - (100a + 10b + c) = ⅓[(n + 1)^2 - n^2]
10c + b - 10b - c = ⅓[n^2 + 2n + 1 - n^2]
9c - 9b = ⅓(2n + 1)
27c - 27b = 2n + 1
27(c - b) = 2n + 1
We see that 2n + 1 must be a multiple of 27; therefore, n can be 13, 40, 67, etc. However, n can’t be 40 (or more), since n^2 = 40^2 = 1600 and there are no 3-digit numbers that are between 40^2 and 41^2. So n must be 13. In that case, we have:
27(c - b) = 27
c - b = 1
Since n = 13, n^2 = 169 and (n + 1)^2 = 14^2 = 196. So we can see that a must be 1. So acb must be 187 and abc must be 178. In any case, a + b + c = 1 + 7 + 8 = 16.
Answer: C
