Find out the smallest no. which on division by 24, 32 and 36 gives 19, 27 and 31 as remainder respectively.

(a) 283
(b) 273
(c) 281
(d) 285

Anurag Mishra Professor Asked on 30th June 2016 in Maths.
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  • 1 Answer(s)

    Answer: (a) 283 

    Solution:
    The Factors of 24 = 2 x 2 x 2 x 3
    The Factors of 32 = 2 x 2 x 2 x 2 x 2
    The Factors of 36 = 2 x 2 x 3 x 3

    Then, the LCM of 24, 32 and 36 = 2 x 2 x 2 x 2 x 2 x 3 x 3
    = 288

    The difference of 24 – 19 = 32 – 27 = 36 – 31 = 5

    So, the smallest number = 288 – 5
    = 283

    Hence, the correct answer is option (a) 283.

    Anurag Mishra Professor Answered on 5th July 2016.
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