PUZZLE 8: 25th-31st January (to be e-mailed by the 1st February)
POSTED BY IES CAMPANAR
DIGIT NUMER
A certain four-digit number obeys the following conditions:a) The sum of the squares of the two digits on both sides is = 13.b) The sum of the squares of the two digits in the middle is =85.c) If we substract 1089 from that certain four-digit number, we get another figure this time with the same digits but exactly in the opposite order.Which number is it?
SOLUTION
Because of the conditions:The digits on both sides are: 2, 3The digits in the middle are: 6, 7 or 2,9There are eight possibilities, the numbers: 2673, 2763, 3672, 3762, 2293, 2923, 3292, 3922The solution is the number: 3762 (because 3762-1089 = 2673)
Monday, 25 January 2010
Monday, 11 January 2010
FORTNIGHT'S PUZZLE 7
PUZZLE 7: 11th-17th January (to be e-mailed by the 18th January)
POSTED BY IES CAMPANAR
A FIERCE BATTLE
In a fierce battle, say that at least 70% of the soldiers have lost an eye; at least 75% an ear; at least 80% an arm; and at least 85% a leg.What percentage, at least, must have lost all four? (Lewis Carroll)
SOLUTION
At least 45% lost an eye and an ear (70+75 = 145)At least 65% lost an arm and a leg ( 80 +85 = 165)Therefore at least 10% lost an eye, an ear, an arm and a leg (45 + 65 = 110).
POSTED BY IES CAMPANAR
A FIERCE BATTLE
In a fierce battle, say that at least 70% of the soldiers have lost an eye; at least 75% an ear; at least 80% an arm; and at least 85% a leg.What percentage, at least, must have lost all four? (Lewis Carroll)
SOLUTION
At least 45% lost an eye and an ear (70+75 = 145)At least 65% lost an arm and a leg ( 80 +85 = 165)Therefore at least 10% lost an eye, an ear, an arm and a leg (45 + 65 = 110).
Wednesday, 6 January 2010
Tuesday, 5 January 2010
Monday, 4 January 2010
Sunday, 3 January 2010
Saturday, 2 January 2010
Friday, 1 January 2010
FORTNIGHT'S PUZZLE 7
PUZZLE 7: 7th-13th May (to be e-mailed by the 14th May)
POSTED BY IES CAMPANAR
INHUMAN SIMPLIFICATION
Simplify up to the maximum this big fraction in which both numbers (numerator and denominator), have 2010 digits:
121212...121212
212121...212121
SOLUTION
The correct solution from LUISA GARCÍA is:Simplify up to the maximum:
121212...121212
212121...212121
If you reduce this fraction by 3, you get 4040…0404/7070…0707 (2009 digits). Then you can reduce it by 1010…0101 (2009 digits)
and you get 4/7.
So the solution is 4/7.
POSTED BY IES CAMPANAR
INHUMAN SIMPLIFICATION
Simplify up to the maximum this big fraction in which both numbers (numerator and denominator), have 2010 digits:
121212...121212
212121...212121
SOLUTION
The correct solution from LUISA GARCÍA is:Simplify up to the maximum:
121212...121212
212121...212121
If you reduce this fraction by 3, you get 4040…0404/7070…0707 (2009 digits). Then you can reduce it by 1010…0101 (2009 digits)
and you get 4/7.
So the solution is 4/7.
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