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30 Sep 2019, 10:21
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
58% (02:10) correct
42%
(02:32)
wrong
based on 57
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A two digit number is formed such that the last two digits of the cube of the number is 44. What is the sum of all the possible values of the number?
(A) 72
(B) 78
(C) 80
(D) 86
(E) 98
(A) 72
(B) 78
(C) 80
(D) 86
(E) 98
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Updated on: 30 Sep 2019, 20:55
Originally posted by CareerGeek on 30 Sep 2019, 11:47.
Last edited by CareerGeek on 30 Sep 2019, 20:55, edited 1 time in total.
Last edited by CareerGeek on 30 Sep 2019, 20:55, edited 1 time in total.
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Hovkial
Note: If a cube is ending in 4, the number also has its unit digit as 4!!
—> Let the two digit number be \(N = x4\)
—> value of \(N = 10x + 4\)
—> \(N^3 = (10x + 4)^3 = 1000x^3 + 3*(10x)^2*4 + 3*10x*4^2 + 4^3\)
—> \(N^3 = 1000x^3 + 1200x^2 + 480x + 64\)
—> \(N^3 = 100(10*x^3 + 12*x^2 + 4x) + 80x + 64\)
—> Note that, When 80x + 64 gives a 44 as last 2 digits, we get favourable values of x
—> Possible values of x = 1 (80*1 + 64 = 144), 6(80*6 + 64 = 544)
—> Possible values of N = 14, 64
Sum = 14 + 64 = 78
IMO Option B
30 Sep 2019, 16:54
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Hovkial
two digit number must have units digit of 4 to give cube units digit of 4
the cycle for tens digit values of cubes of two digit numbers with units digit of 4 is
4 (14), 2, 0, 8, 6, 4 (64)
14+64=78
B
