Danica drove her new car on a trip for a whole number of hours, averag

Question

Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, abc miles was displayed on the odometer, where abc is a 3-digit number with \(a \geq{1}\) and \(a+b+c \leq{7}\). At the end of the trip, the odometer showed cba miles. What is \(a^2+b^2+c^2\)?.

(A) 26
(B) 27
(C) 36
(D) 37
(E) 41

(A) 26
(B) 27
(C) 36
(D) 37
(E) 41

LevanKhukhunashvili · Answer

Before we dive into the deep darkness of quant:
cba-abc must be divisible by 55=11*5

cba-abc will always be divisible by 11 no matter what the digits.

To be divisible by 5, the units digit of cba-abc must be 0 or 5
0 is impossible as a and c cant be equal, because at this case abc=cba and it's impossible as Danica certainly drove her car
5 is possible if a=1 c=6; 106 satisfies the inequality a+b+c≤7

You can try other combinations but reject all of them as some of them dont satisfy the inequality a+b+c≤7 or some of them dont agree with odometer, as the display on odometer only increases (abc cant be less than cba)

IMO
Ans:  D

KarishmaB · Answer

Distance = Speed * Time = 55 * T (where T is a positive integer)

Diff in odometer reading will be this distance covered.

(100c + 10b + a) - (100a + 10b + c) = 55 * T

99 (c - a) / 55 = T

9/5 * (c - a) = T ( a positive integer)

For T to be an integer, (c - a) must be a multiple of 5. 
So c = 6 and a = 1 satisfies. 
Since sum of a, b and c must be 7 or less, b = 0

a^2+b^2+c^2 = 1^2 + 0 + 6^2 = 37

Answer (D)

ScottTargetTestPrep · Answer

The difference between the three-digit numbers cba and abc is = 100c + 10b + a - (100a + 10b + c) = 100(c - a) + a - c = 100(c - a) - (c - a) = 99(c - a). So the difference is a multiple of 99. However, it is also a multiple of 55 since the problem indicates that Danica drove her car for a whole number of hours at 55 mph. Therefore, the difference between abc and cba must be a common multiple of 99 and 55.

The LCM of 99 and 55 is 495 and their next multiple is 990. However, it can’t be 990 since that would make 99(c - a) = 990 or c - a = 10. However, c and a are only 1-digit numbers. So 99(c - a) must be 495, which makes c - a = 5. So if a = 1, then c = 6 and that would make b = 0 and this is the only option for the values of a, b and c since a ≥ 1 and a + b + c ≤ 7. Therefore, a^2 + b^2 + c^2 = 1^2 + 0^2 + 6^2 = 1 + 0 + 36 = 37.

Answer: D

effatara · Answer

Let cba(odometer reading at the end of the trip)-abc(odometer reading at the start of the trip)=xyz(the number of miles traveled).
Since cba is greater than abc, c>a. Henceforth, we will denote the larger number (cba) as LN and the smaller number (abc) as SN. While subtracting SN from LN, we will first have to subtract the 10ths digit of SN (which is 'c') from the 10ths digit of of LN (which is 'a') to get 'z' (the 10ths digit of xyz). Since c>a, 'z' is equal to  .
Now, xyz=55k (where k is any positive integer). So xyz could be 55*1=55, 55*2=110, 55*3=165 and so on. That is, 'z' must be either 5 or 0. But it can't be 0 because, in that case, c=a which is not possible they are two distinct digits. So z=5. Thus, (a+10-c)=5 or (c-a)=5. The highest value that 'c' could have is 6 because if it is 7 then 'a' has to be 2 and even if 'b' is 0, (a+b+c) becomes 9 which is not possible because it is given that (a+b+c) cannot be more than 7. Also, 'c' cannot be less than 6 because that would make (c-a) less than 5. Therefore, the only possibilities which satisfies all the conditions are: c=6, a=1 and b=0. So, cba- abc=601-106=495=55*9. Thus, a^2+b^2+c^2=1^2+0^2+6"2=1+36=37. Ans: C

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