If both 11^2 and 3^3 are factors of the number a*4^3*6^3*13^11 then what is the smallest possible value of a
Answers
Answered by
0
Answer:
The continued fraction representations of the limits of the interval are
0.0080120180265
=
[
0
;
124
,
1
,
4
,
2
,
1
,
463872
,
1
,
1
,
12
,
1
,
1
,
41
]
0.0080120180275
=
[
0
;
124
,
1
,
4
,
3
,
545777
,
2
,
13
,
1
,
1
,
1
,
1
,
2
]
The simplest continued fraction (and therefore also the simplest ordinary fraction!) in that interval is
[
0
;
124
,
1
,
4
,
3
]
=
16
1997
=
0.00801201802704056084
…
and the sum of its numerator and denominator is
2013
.
(I used Wolfram Alpha to expand the continued fractions fully. For a pencil-and-paper solution one only needs to carry out the expansion until they start differing, which requires only a handful of long divisions with remainder.)
Answered by
0
Answer:
121
Step-by-step explanation:
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