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05 Mar 2026, 23:41
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D
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Difficulty:
95%
(hard)
Question Stats:
34% (02:45) correct
66%
(03:20)
wrong
based on 38
sessions
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A positive divisor of 4,200 is selected at random. What is the probability that the divisor leaves a remainder of 7 when divided by 14?
A. 1/24
B. 1/12
C. 1/10
D. 1/8
E. 1/6
A. 1/24
B. 1/12
C. 1/10
D. 1/8
E. 1/6
06 Mar 2026, 00:25
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kevincan
A positive divisor of 4,200 is selected at random, which means the divisor should be a factor of 4200.
The factors of 4200 are = 6*7*100 = 2^3 *3 * 5^2 * 7
Total number of factors = (3+1)*(1+1)*(2+1)*(1+1) = 4*2*3*2 = 48 factors.
we are asked to find out : The divisor leaves a remainder of 7 when divided by 14.
so, the divisor should be of the form 14K + 7.
At k=0, we get 7,
k=1 , we get 21.
k= 2, we get 35. For subsequent values of K, we won’t find any divisor as factor of 4200.
Is this approach right? No. Does it take much time ? Yes.
Taking 7 as common, we get 7*(2k+1).
So, the divisor should be a multiple of 7 as well as ODD multiple ( 2k+1).
Factors of 4200 = 2^3 *3 * 5^2 * 7
Remove all even numbers = 3*5^2 * 7 = 2*3*2 = 12 odd factors.
Out of the 12 factors, factors which are multiple of 7 are 7*[3 * 5^2].
(2*3)= 6 factors, which are multiples of 7.
Probability = 6/48 = 1/8
Option D
07 Mar 2026, 11:41
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kevincan
Approach ->
Step 1 get the prime factorisation of 4200
Step 2 -> get the total factor as probability so we need the total
Step 3 -> Approach we can see we need the divisor leaves a remainder of 7 when divided by 14
Lets lets think this was for a rem of 7 we need the mutiliple to be odd in this case if we are able to count the total no of odd factors apart from 7 as & is needed to make 14(7*2) then we will have our desired result set
4,200 = 2^3 X 3^1 X 5^2 X 7^1
Total factors = (3+1)(1+1)(2+1)(1+1) = 4 X 2 X 3 X 2 = 48
Now the favorable ones :
Power of 7: Must be 7^1
Power of 3: Can be 3^0 or 3^1
Power of 5: Can be 5^0, 5^1, or 5^2
so total = 3 X2 = 6
So Prob = 6/48 or 1/8
