PUZZLE 10: 22nd-28th February (to be e-mailed by the 1st March)
POSTED BY IES CAMPANAR
THE MAGIC NUMBER
'Free me, please', Ali begged the genius who had trapped him in a cage.‘I will free you only if you find a number which obeys certain conditions’, the genius answered.And these were the conditions:1.- If the number were multiple of 2, then it would be a number between 50 and 59, both included.2.- If it were not a multiple of 3, then it would be a number between 60 and 69, both included.3.- If it were not a multiple of 4, then it would be a number between 70 and 79, both included.Which was the magic number?
SOLUTION
If the number is a multiple of 2, it could be:50, 52, 54, 56, 58If it is not a multiple of 3, the number could be:61, 62, 64, 65, 67, 68If it is not a multiple of 4, the possibilities are:70, 71, 73, 74, 75, 77, 78, 79By eliminating the only possibility is number 75.
Monday, 22 February 2010
Monday, 8 February 2010
FORTNIGHT'S PUZZLE 9
PUZZLE 9: 8th-14th February (to be e-mailed by the 15th February)
POSTED BY IES CAMPANAR
DIGIT NUMBERS
Try to find all the two-digit numbers which, when divided by the sum of their digits, have a quotient equal to 7 with no remainder.
SOLUTION
The number in polynomial form is 10a+b. Because of the condition:(10a+b)/(a+b) = 710a+b = 7a + 7b3a = 6ba = 2bWith b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84
POSTED BY IES CAMPANAR
DIGIT NUMBERS
Try to find all the two-digit numbers which, when divided by the sum of their digits, have a quotient equal to 7 with no remainder.
SOLUTION
The number in polynomial form is 10a+b. Because of the condition:(10a+b)/(a+b) = 710a+b = 7a + 7b3a = 6ba = 2bWith b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84
Monday, 1 February 2010
Monday, 25 January 2010
FORTNIGHT'S PUZZLE 8
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)
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, 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
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