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

 It is currently 22 Oct 2019, 04:18

### 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

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

### Show Tags

26 Mar 2019, 23:31
00:00

Difficulty:

35% (medium)

Question Stats:

73% (01:53) correct 27% (03:01) wrong based on 11 sessions

### HideShow timer Statistics

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

_________________
Manager
Joined: 05 Oct 2017
Posts: 101
Location: India
Schools: ISB '21, IIMA , IIMB
GPA: 4
WE: Analyst (Energy and Utilities)
Re: Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie [#permalink] ### Show Tags 27 Mar 2019, 02:18 1 Bunuel wrote: Let a, b, and c be digits with a ≠ 0$. 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

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 367 times ]

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8125
Location: United States (CA)
Re: Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie [#permalink] ### Show Tags 29 Mar 2019, 07:44 Bunuel wrote: Let a, b, and c be digits with a ≠ 0. The three-digit integer abc lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer acb lies two thirds of the way between the same two squares. What is a+b+c? (A) 10 (B) 13 (C) 16 (D) 18 (E) 21 The value of the integer abc is 100a + 10b + c and that of acb is 100a + 10c + b. Since abc is ⅓ of the way between two consecutive perfect squares (say n^2 and (n + 1)^2) and acb is ⅔ of the way between the same two perfect squares, the distance (or difference) between abc and acb is ⅓ of the way between the two perfect squares. So we have: (100a + 10c + b) - (100a + 10b + c) = ⅓[(n + 1)^2 - n^2] 10c + b - 10b - c = ⅓[n^2 + 2n + 1 - n^2] 9c - 9b = ⅓(2n + 1) 27c - 27b = 2n + 1 27(c - b) = 2n + 1 We see that 2n + 1 must be a multiple of 27; therefore, n can be 13, 40, 67, etc. However, n can’t be 40 (or more), since n^2 = 40^2 = 1600 and there are no 3-digit numbers that are between 40^2 and 41^2. So n must be 13. In that case, we have: 27(c - b) = 27 c - b = 1 Since n = 13, n^2 = 169 and (n + 1)^2 = 14^2 = 196. So we can see that a must be 1. So acb must be 187 and abc must be 178. In any case, a + b + c = 1 + 7 + 8 = 16. Answer: C _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: Let a, b, and c be digits with a ≠ 0$. The three-digit integer abc lie   [#permalink] 29 Mar 2019, 07:44
Display posts from previous: Sort by