Difference between revisions of "1973 Canadian MO Problems/Problem 1"
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Revision as of 20:22, 16 December 2011
Problem
Solve the simultaneous inequalities,
and
; i.e. find a single inequality equivalent to the two simultaneous inequalities.
What is the greatest integer that satisfies both inequalities
and
.
Give a rational number between
and
.
Express
as a product of two integers neither of which is an integral multiple of
.
Without the use of logarithm tables evaluate
.
Solution

Since from the second inequality , our solution is thus:
.
From
, we get that
. From
, we get:

With these two inequalities, we see that the greatest integer satisfying the requirements is .
. Thus, a rational number in between
and
is
Thus,
See also
1973 Canadian MO (Problems) | ||
Preceded by 1973 Canadian MO Problems |
1 • 2 • 3 • 4 • 5 | Followed by Problem 2 |