It is given that 15! = 1 A0767436BCDE, where A, B, C, D
and E each stand for a single digit. Find the set which
represents the full collection of all digits, which appear
exactly twice in the expansion of 15!
Answers
Given : 15! = 1 A0767436BCDE
where A, B, C, D and E each stand for a single digit.
To Find : The set which represents the full collection of all digits, which appear exactly twice in the expansion of 15!
3, 6, 7
0, 6, 7
1, 3, 6, 7
0, 1, 7
4, 6, 7
Solution:
15!
Number of factors 2
[15/2} + [15/2²] + [15/2³] + [15/2⁴] + ...
= 7 + 3 + 1 + 0 + 0 + ...
= 11
Number of factors 5
[15/5} + [15/5²] + ...
= 3 + 0 + ...
= 3
Number of factors 5 = 3 , Number of factors 2> 3
Hence (2 * 5)³ is a factor
=> 10³ = 1000 is a factor
Hence CDE must be 0
1A0767436B000
15! has 9 as a factor so sum of digits must be divisible by 9
1+A+0+7+6+7+4+3+6+B+0+0+0
= A + B + 34
A + B = 2 or A+B = 11 are two possibilities
15! has 11 as factor
1 - A + 0 - 7 + 6 - 7 + 4 - 3 + 6 - B + 0 - 0 + 0
-A - B
-(A + B)
A + B must be 0 or 11 to divisible by 11
A+ B = 0 , A + B = 11
from Both cases A + B = 11
1A0767436B000
A + B = 11
A and B can be ( 2 to 9) not 0 or 1
Now check the options:
3, 6, 7 is possible if A and B are 3 and 8 ( as 3 + 8 = 11)
0, 6, 7 not possible ∵ 0 is 4 times
1, 3, 6, 7 not possible ∵ 1 is not 2 times
0, 1, 7 not possible ∵ 1 is not 2 times
4, 6, 7 not possible as if 4 is one of A and B to make 4 twice then there will be 7 also so that will make 7, 3 times
other wise 4 is only one
Hence 3, 6, 7 is the correct answer from given options.
just FYI....
15! = 13,07,67,43,68,000
Learn More:
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