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Re: Let a 2-placer be a terminating decimal, between 0 and 1, for which [#permalink]
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Essentially, a 2-placer is a fraction of the form a/100 where a is NOT a multiple of 10 (i.e. a cannot have factors 2 and 5 both) and a < 100.
Thus, if 2 such fractions m = a/100 and n = b/100 are multiplied, we get ab/10000

Since this ALSO has to be a 2-placer, ab must be a multiple of 100 so that one '100' cancels out from the denominator
Thus, a must be a multiple of 4 (but not 5) and b must be a multiple of 25 (but not 2) (since 4*25 = 100)

Possible values of a: 4, 8, 12, 16, 24, 28, 32, 36 ... 96
There are 24 multiples of 4 less than 100 of which 4 are multiple of 5 (20, 40, 60 and 80)
=> Required number of terms = 24 - 4 = 20

Possible values of b: 25 and 75
=> Required number of terms = 2

Thus, total combinations = 20 * 2 = 40
Each combination will have a different sum. Thus, there are 40 different summation values

Answer A

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Re: Let a 2-placer be a terminating decimal, between 0 and 1, for which [#permalink]
Thank you Carcass
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Re: Let a 2-placer be a terminating decimal, between 0 and 1, for which [#permalink]
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