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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:09) correct
58%
(02:10)
wrong
based on 69
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive divisors of 6,300 have a units digit of 5?
(A ) three
(B) four
(C) six
(D) nine
(E) twelve
(A ) three
(B) four
(C) six
(D) nine
(E) twelve
19 Jan 2026, 21:30
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Before solving this question, think of the scenarios, as to when would you obtain a units digit of 5? 5*1 = 5, 5*2 = 10, 5*3 = 15, 5*4 = 20, 5*5 = 25, 5*6 = 30 and so on...
If you observe closely, whenever there is some power of 2 being multiplied by 5, we obtain 0 (because of the basic fact that 2*5 = 10), but for odd multiples of 5, we would always obtain 5 as the unit digit.
Hence, we would have to look out for the odd multiples of 5 in this question.
Prime Factorization of 6300 => (2^2)*(3^2)*(5^2)*(7)
Number of odd multiples of 5 = 1*3*2*2 = 12
More Explanation: 1 because only 2^0 needs to be taken into account, 3 because we can take 3^0 or 3^1 or 3^2 here, 2 because we can only take 5^1 or 5^2 and 2 because we consider 7^0 or 7^1 in this case.
Hence, the correct answer is 12 (Option E)
If you observe closely, whenever there is some power of 2 being multiplied by 5, we obtain 0 (because of the basic fact that 2*5 = 10), but for odd multiples of 5, we would always obtain 5 as the unit digit.
Hence, we would have to look out for the odd multiples of 5 in this question.
Prime Factorization of 6300 => (2^2)*(3^2)*(5^2)*(7)
Number of odd multiples of 5 = 1*3*2*2 = 12
More Explanation: 1 because only 2^0 needs to be taken into account, 3 because we can take 3^0 or 3^1 or 3^2 here, 2 because we can only take 5^1 or 5^2 and 2 because we consider 7^0 or 7^1 in this case.
Hence, the correct answer is 12 (Option E)
20 Jan 2026, 10:29
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6300 = (2^2)*(3^2)*(5^2)*(7)
For unit digits to be 5,
at least one 5 should be included and No 2s should be included.
Hence total factors with unit digit 5 = (2+1)*(1+1)*(1+1) = 12
For unit digits to be 5,
at least one 5 should be included and No 2s should be included.
Hence total factors with unit digit 5 = (2+1)*(1+1)*(1+1) = 12
