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13 May 2019, 08:59
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:13) correct
40%
(02:06)
wrong
based on 331
sessions
History
Date
Time
Result
Not Attempted Yet
x is a single-digit positive integer. If the units digit of x^2 and the units digit of (x+2)^3 is same. How many values of x are possible?
A. 2
B. 3
C. 4
D. 7
E. 9
A. 2
B. 3
C. 4
D. 7
E. 9
13 May 2019, 10:57
Kudos
Bookmarks
I didn't find any easy or faster method to solve this.
2,4,7,9 satisfy the equation unitdigit(x2) = unitdigit(x+2)3
x^2 (x+2)^3
2 4 4^3 = 64
4 16 6^3 = 216
7 49 9^3 = 729 = 81 * 9 unit digit = 9
9 81 11^3 = unitdigit = 1
Please let me know if you some other method
2,4,7,9 satisfy the equation unitdigit(x2) = unitdigit(x+2)3
x^2 (x+2)^3
2 4 4^3 = 64
4 16 6^3 = 216
7 49 9^3 = 729 = 81 * 9 unit digit = 9
9 81 11^3 = unitdigit = 1
Please let me know if you some other method
14 May 2019, 19:24
Kudos
Bookmarks
mangamma
Let’s test each possible units digit of x^2 and (x + 2)^3 (note: in the calculations below, we only care about the units digit of x^2 and (x + 2)^3):
1^2 = 1 and 3^3 = 7
2^2 = 4 and 4^3 = 4 → Yes!
3^2 = 9 and 5^3 = 5
4^2 = 6 and 6^3 = 6 → Yes!
5^2 = 5 and 7^3 =3
6^2 =6 and 8^3 = 2
7^2 = 9 and 9^3 =9 → Yes!
8^2 = 4 and 10^3 = 0
9^2 = 1 and 11^3 = 1 → Yes
Answer: C
